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NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

Exercise 9.1

Q1: Draw a quadrilateral in the Cartesian plane, whose vertices are (–4, 5), (0, 7), (5, –5) and (–4, –2). Also, find its area.
Ans: Let ABCD be the given quadrilateral with vertices A (–4, 5), B (0, 7), C (5, –5), and D (–4, –2).
Then, by plotting A, B, C, and D on the Cartesian plane and joining AB, BC, CD, and DA, the given quadrilateral can be drawn as

NCERT Solutions Class 11 Maths Chapter 9 - Straight LinesTo find the area of quadrilateral ABCD, we draw one diagonal, say AC.
Accordingly, area (ABCD) = area (ΔABC) + area (ΔACD)
We know that the area of a triangle whose vertices are (x1, y1), (x2, y2), and (x3, y3) is  NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines
Therefore, area of ΔABC

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

Area of ΔACD

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

Thus, area (ABCD) NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

Q2: The base of an equilateral triangle with side 2a lies along they y-axis such that the mid point of the base is at the origin. Find vertices of the triangle.
Ans: Let ABC be the given equilateral triangle with side 2a.
Accordingly, AB = BC = CA = 2a
Assume that base BC lies along the y-axis such that the mid-point of BC is at the origin.
i.e., BO = OC = a, where O is the origin.
Now, it is clear that the coordinates of point C are (0, a), while the coordinates of point B are (0, –a).
It is known that the line joining a vertex of an equilateral triangle with the mid-point of its opposite side is perpendicular.
Hence, vertex A lies on the y-axis.

NCERT Solutions Class 11 Maths Chapter 9 - Straight LinesOn applying Pythagoras theorem to ΔAOC, we obtain
(AC)2 = (OA)2  (OC)2
⇒ (2a)2 = (OA)2   a2
⇒ 4a2a2 = (OA)2
⇒ (OA)2 = 3a2
⇒ OA = NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines
∴ Coordinates of point A = NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines
Thus, the vertices of the given equilateral triangle are (0, a), (0, –a), and  NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines  or (0, a), (0, –a), and NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .

Q3: Find the distance between  NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines and  NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines when:
(i) PQ is parallel to the y-axis 
(ii) PQ is parallel to the x-axis.
Ans: The given points are  NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines and NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .
(i) When PQ is parallel to the y-axis, x1 = x2.
In this case, distance between P and Q  NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines
(ii) When PQ is parallel to the x-axis, y1 = y2.
In this case, distance between P and Q  NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

Q4: Find a point on the x-axis, which is equidistant from the points (7, 6) and (3, 4).
Ans: Let (a, 0) be the point on the x axis that is equidistant from the points (7, 6) and (3, 4).

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

On squaring both sides, we obtain
a2 – 14a + 85 = a2 – 6a + 25
⇒ –14a + 6a = 25 – 85
⇒ –8a = –60

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines
Thus, the required point on the x-axis is NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .

Q5: Find the slope of a line, which passes through the origin, and the mid-point of the line segment joining the points P (0, –4) and B (8, 0).
Ans: The coordinates of the mid-point of the line segment joining the points
P (0, –4) and B (8, 0) are NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines
It is known that the slope (m) of a non-vertical line passing through the points (x1, y1) and (x2, y2) is given by NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .
Therefore, the slope of the line passing through (0, 0) and (4, –2) is

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .
Hence, the required slope of the line is NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .

Q6: Without using the Pythagoras theorem, show that the points (4, 4), (3, 5) and (–1, –1) are the vertices of a right angled triangle.
Ans: The vertices of the given triangle are A (4, 4), B (3, 5), and C (–1, –1).
It is known that the slope (m) of a non-vertical line passing through the points (x1, y1) and (x2, y2) is given by NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .
∴ Slope of AB (m1) NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines
Slope of BC (m2) NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines
Slope of CA (m3) NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines
It is observed that m1m3 = –1
This shows that line segments AB and CA are perpendicular to each other
i.e., the given triangle is right-angled at A (4, 4).
Thus, the points (4, 4), (3, 5), and (–1, –1) are the vertices of a right-angled triangle.

Q7: Find the slope of the line, which makes an angle of 30° with the positive direction of y-axis measured anticlockwise.
Ans: If a line makes an angle of 30° with the positive direction of the y-axis measured anticlockwise, then the angle made by the line with the positive direction of the x-axis measured anticlockwise is 90° +  30° = 120°.

NCERT Solutions Class 11 Maths Chapter 9 - Straight LinesThus, the slope of the given line is tan 120° = tan (180° – 60°) = –tan 60°  NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines.

Q8: Without using distance formula, show that points (–2, –1), (4, 0), (3, 3) and (–3, 2) are vertices of a parallelogram.
Ans: Let points (–2, –1), (4, 0), (3, 3), and (–3, 2) be respectively denoted by A, B, C, and D.

NCERT Solutions Class 11 Maths Chapter 9 - Straight LinesSlope of AB NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines
Slope of CD = NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines
⇒ Slope of AB = Slope of CD
⇒ AB and CD are parallel to each other.
Now, slope of BC = NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines
Slope of AD = NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines
⇒ Slope of BC = Slope of AD
⇒ BC and AD are parallel to each other.
Therefore, both pairs of opposite sides of quadrilateral ABCD are parallel. Hence, ABCD is a parallelogram.
Thus, points (–2, –1), (4, 0), (3, 3), and (–3, 2) are the vertices of a parallelogram.

Q9: Find the angle between the x-axis and the line joining the points (3, –1) and (4, –2).
Ans: The slope of the line joining the points (3, –1) and (4, –2) is  NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines
Now, the inclination (θ ) of the line joining the points (3, –1) and (4, – 2) is given by  tan θ= –1
⇒ θ = (90° + 45°) = 135°
Thus, the angle between the x-axis and the line joining the points (3, –1) and (4, –2) is 135°.

Q10: The slope of a line is double of the slope of another line. If tangent of the angle between them is 1/3 , find the slopes of he lines.
Ans: Let NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines  be the slopes of the two given lines such that NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .
We know that if θ is the angle between the lines l1 and l2 with slopes m1 and m2, then NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .
It is given that the tangent of the angle between the two lines is NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

Case I

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines
If m = –1, then the slopes of the lines are –1 and –2.
If m = NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines , then the slopes of the lines are NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines and –1.

Case II

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

If m = 1, then the slopes of the lines are 1 and 2.

If m = NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines , then the slopes of the lines are NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .

Hence, the slopes of the lines are –1 and –2 or  NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines and –1 or 1 and 2 or NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .

Q11: A line passes through NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines . If slope of the line is m, show that NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines.
Ans: The slope of the line passing through  NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines is NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .
It is given that the slope of the line is m.

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines
Hence, NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

Exercise 9.2

Q1: Write the equations for the x and y-axes.
Ans: The y-coordinate of every point on the x-axis is 0.
Therefore, the equation of the x-axis is y = 0.
The x-coordinate of every point on the y-axis is 0.
Therefore, the equation of the y-axis is y = 0.

Q2: Find the equation of the line which passes through the point (–4, 3) with slopeNCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .
Ans: We know that the equation of the line passing through point NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines , whose slope is m, is NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .
Thus, the equation of the line passing through point (–4, 3), whose slope is NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines , is

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

Q3: Find the equation of the line which passes though (0, 0) with slope m.
Ans: We know that the equation of the line passing through point NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines , whose slope is m, is NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .
Thus, the equation of the line passing through point (0, 0), whose slope is m,is
(y – 0) = m(x – 0)
i.e., y = mx

Q4: Find the equation of the line which passes though  NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines and is inclined with the x-axis at an angle of 75°.
Ans: The slope of the line that inclines with the x-axis at an angle of 75° is m = tan 75°

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines
We know that the equation of the line passing through point NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines , whose slope is m, is NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .
Thus, if a line passes though NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines and inclines with the x-axis at an angle of 75°, then the equation of the line is given asNCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

Q5: Find the equation of the line which intersects the x-axis at a distance of 3 units to the left of origin with slope –2.
Ans: It is known that if a line with slope m makes x-intercept d, then the equation of the line is given as
y = m(x – d)
For the line intersecting the x-axis at a distance of 3 units to the left of the origin, d = –3.
The slope of the line is given as m = –2
Thus, the required equation of the given line is
y = –2 [x – (–3)]
y = –2x – 6
i.e., 2x +  y + 6 = 0

Q6: Find the equation of the line which intersects the y-axis at a distance of 2 units above the origin and makes an angle of 30° with the positive direction of the x-axis.
Ans: It is known that if a line with slope m makes y-intercept c, then the equation of the line is given as
y = mx + c
Here, c = 2 and m = tan 30° NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .
Thus, the required equation of the given line is

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

Q7: Find the equation of the line which passes through the points (–1, 1) and (2, –4).
Ans: t is known that the equation of the line passing through points (x1, y1) and (x2, y2) is NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .
Therefore, the equation of the line passing through the points (–1, 1) and
(2, –4) is

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

Q8: The vertices of ΔPQR are P (2, 1), Q (–2, 3) and R (4, 5). Find equation of the median through the vertex R.
Ans: It is given that the vertices of ΔPQR are P (2, 1), Q (–2, 3), and R (4, 5).
Let RL be the median through vertex R.
Accordingly, L is the mid-point of PQ.
By mid-point formula, the coordinates of point L are given by NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines
It is known that the equation of the line passing through points (x1, y1) and (x2, y2) is NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .
Therefore, the equation of RL can be determined by substituting (x1, y1) = (4, 5) and (x2, y2) = (0, 2).

NCERT Solutions Class 11 Maths Chapter 9 - Straight LinesHence, 

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines
Thus, the required equation of the median through vertex R is NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .

Q9: Find the equation of the line passing through (–3, 5) and perpendicular to the line through the points (2, 5) and (–3, 6).
Ans: The slope of the line joining the points (2, 5) and (–3, 6) is NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines
We know that two non-vertical lines are perpendicular to each other if and only if their slopes are negative reciprocals of each other.
Therefore, slope of the line perpendicular to the line through the points (2, 5) and (–3, 6)   NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines
Now, the equation of the line passing through point (–3, 5), whose slope is 5, is

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

Q10: A line perpendicular to the line segment joining the points (1, 0) and (2, 3) divides it in the ratio 1:n. Find the equation of the line.
Ans: According to the section formula, the coordinates of the point that divides the line segment joining the points (1, 0) and (2, 3) in the ratio 1: n is given by

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines
The slope of the line joining the points (1, 0) and (2, 3) is

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines
We know that two non-vertical lines are perpendicular to each other if and only if their slopes are negative reciprocals of each other.
Therefore, slope of the line that is perpendicular to the line joining the points (1, 0) and (2, 3)   NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines
Now, the equation of the line passing through NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines  and whose slope is NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines is given by

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

Q11: Find the equation of a line that cuts off equal intercepts on the coordinate axes and passes through the point (2, 3).
Ans: The equation of a line in the intercept form is

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines
Here, a and b are the intercepts on x and y axes respectively.
It is given that the line cuts off equal intercepts on both the axes. This means that a = b.
Accordingly, equation (i) reduces to

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines
Since the given line passes through point (2, 3), equation (ii) reduces to
2 + 3 = aa = 5
On substituting the value of a in equation (ii), we obtain
x  + y = 5, which is the required equation of the line

Q12: Find equation of the line passing through the point (2, 2) and cutting off intercepts on the axes whose sum is 9.
Ans: The equation of a line in the intercept form is

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines
Here, a and b are the intercepts on x and y axes respectively.
It is given that a  + b = 9 ⇒ b = 9 – a … (ii)
From equations (i) and (ii), we obtain

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines
It is given that the line passes through point (2, 2). Therefore, equation (iii) reduces to

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines
If a = 6 and b = 9 – 6 = 3, then the equation of the line is

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines
If a = 3 and b = 9 – 3 = 6, then the equation of the line is

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

Q13: Find equation of the line through the point (0, 2) making an angle  NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines with the positive x-axis. Also, find the equation of line parallel to it and crossing the y-axis at a distance of 2 units below the origin.
Ans: The slope of the line making an angle NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines with the positive x-axis is  NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines
Now, the equation of the line passing through point (0, 2) and having a slope  NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines  is NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines
The slope of line parallel to line NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines  is NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .
It is given that the line parallel to line NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines crosses the y-axis 2 units below the origin i.e., it passes through point (0, –2).
Hence, the equation of the line passing through point (0, –2) and having a slope NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines  is

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

Q14: The perpendicular from the origin to a line meets it at the point (– 2, 9), find the equation of the line.
Ans: The slope of the line joining the origin (0, 0) and point (–2, 9) is  NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines
Accordingly, the slope of the line perpendicular to the line joining the origin and point (– 2, 9) is

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines
Now, the equation of the line passing through point (–2, 9) and having a slope m2 is

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

Q15: The length L (in centimetre) of a copper rod is a linear function of its Celsius temperature C. In an experiment, if L = 124.942 when C = 20 and L = 125.134 when C = 110, express L in terms of C.
Ans: It is given that when C = 20, the value of L is 124.942, whereas when C = 110, the value of L is 125.134.
Accordingly, points (20, 124.942) and (110, 125.134) satisfy the linear relation between L and C.
Now, assuming C along the x-axis and L along the y-axis, we have two points i.e., (20, 124.942) and (110, 125.134) in the XY plane.
Therefore, the linear relation between L and C is the equation of the line passing through points (20, 124.942) and (110, 125.134).
(L – 124.942) =  NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

Q16: The owner of a milk store finds that, he can sell 980 litres of milk each week at Rs 14/litre and 1220 litres of milk each week at Rs 16/litre. Assuming a linear relationship between selling price and demand, how many litres could he sell weekly at Rs 17/litre?
Ans: The relationship between selling price and demand is linear.
Assuming selling price per litre along the x-axis and demand along the y-axis, we have two points i.e., (14, 980) and (16, 1220) in the XY plane that satisfy the linear relationship between selling price and demand.
Therefore, the linear relationship between selling price per litre and demand is the equation of the line passing through points (14, 980) and (16, 1220).

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

When x = Rs 17/litre,

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

Thus, the owner of the milk store could sell 1340 litres of milk weekly at Rs 17/litre.

Q17: P (a, b) is the mid-point of a line segment between axes. Show that equation of the line is  NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines
Ans: Let AB be the line segment between the axes and let P (a, b) be its mid-point.

NCERT Solutions Class 11 Maths Chapter 9 - Straight LinesLet the coordinates of A and B be (0, y) and (x, 0) respectively.
Since P (a, b) is the mid-point of AB,

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

Thus, the respective coordinates of A and B are (0, 2b) and (2a, 0).
The equation of the line passing through points (0, 2b) and (2a, 0) is

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines
On dividing both sides by ab, we obtain

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines
Thus,the equation of the line is NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .

Q18: Point R (h, k) divides a line segment between the axes in the ratio 1:2. Find equation of the line.
Ans: Let AB be the line segment between the axes such that point R (h, k) divides AB in the ratio 1: 2.
Let the respective coordinates of A and B be (x, 0) and (0, y).
Since point R (h, k) divides AB in the ratio 1: 2, according to the section formula,

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines
Therefore, the respective coordinates of A and B are NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines  and (0, 3k).
Now, the equation of line AB passing through points  NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines  and (0, 3k) is

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines
Thus, the required equation of the line is 2kx + hy = 3hk.

Q19: By using the concept of equation of a line, prove that the three points (3, 0), (–2, –2) and (8, 2) are collinear.
Ans: In order to show that points (3, 0), (–2, –2), and (8, 2) are collinear, it suffices to show that the line passing through points (3, 0) and (–2, –2) also passes through point (8, 2).
The equation of the line passing through points (3, 0) and (–2, –2) is

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines
It is observed that at x = 8 and y = 2,
L.H.S. = 2 × 8 – 5 × 2 = 16 – 10 = 6 = R.H.S.
Therefore, the line passing through points (3, 0) and (–2, –2) also passes through point (8, 2). Hence, points (3, 0), (–2, –2), and (8, 2) are collinear.

Exercise 9.3

Question 1: Reduce the following equations into slope-intercept form and find their slopes and the y-intercepts.

(i) x + 7y = 0  
 (ii) 6x + 3y – 5 = 0    
 (iii) y = 0

ANSWER : - (i) The given equation is x + 7y = 0.

It can be written as

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

This equation is of the form y = mx + c, where NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .

Therefore, equation (1) is in the slope-intercept form, where the slope and the y-intercept are  NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines and 0 respectively.

(ii) The given equation is 6x + 3y – 5 = 0.

It can be written as

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

Therefore, equation (2) is in the slope-intercept form, where the slope and the y-intercept are –2 and NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines  respectively.

(iii) The given equation is y = 0.

It can be written as

y = 0.x + 0 … (3)

This equation is of the form y = mx + c, where m = 0 and c = 0.

Therefore, equation (3) is in the slope-intercept form, where the slope and the y-intercept are 0 and 0 respectively.

Question 2: Reduce the following equations into intercept form and find their intercepts on the axes.

(i) 3x + 2y – 12 = 0
 (ii) 4x – 3y = 6
 (iii) 3y + 2 = 0.

ANSWER : - (i) The given equation is 3x + 2y – 12 = 0.

It can be written as

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

This equation is of the form NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines , where a = 4 and b = 6.

Therefore, equation (1) is in the intercept form, where the intercepts on the x and y axes are 4 and 6 respectively.

(ii) The given equation is 4x – 3y = 6.

It can be written as

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

This equation is of the form NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines , where a =  NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines  and b = –2.

Therefore, equation (2) is in the intercept form, where the intercepts on the x and y axes are  NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines  and –2 respectively.

(iii) The given equation is 3y + 2 = 0.

It can be written as

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

This equation is of the form NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines , where a = 0 and b =  NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .

Therefore, equation (3) is in the intercept form, where the intercept on the y-axis is NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines  and it has no intercept on the x-axis.

Question 3: Find the distance of the point (–1, 1) from the line 12(x + 6) = 5(y – 2).

ANSWER : - The given equation of the line is 12(x + 6) = 5(y – 2).

⇒ 12x + 72 = 5y – 10

⇒12x – 5y + 82 = 0 … (1)

On comparing equation (1) with general equation of line Ax  + By + C = 0, we obtain A = 12, B = –5, and C = 82.

It is known that the perpendicular distance (d) of a line Ax  + By  + C = 0 from a point (x1, y1) is given by NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .

The given point is (x1, y1) = (–1, 1).

Therefore, the distance of point (–1, 1) from the given line

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

 

Question 4: Find the points on the x-axis, whose distances from the line  NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines  are 4 units.

ANSWER : - The given equation of line is

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

On comparing equation (1) with general equation of line Ax  + By +  C = 0, we obtain A = 4, B = 3, and C = –12.

Let (a, 0) be the point on the x-axis whose distance from the given line is 4 units.

It is known that the perpendicular distance (d) of a line Ax  + By  + C = 0 from a point (x1, y1) is given by NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .

Therefore,

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

Thus, the required points on the x-axis are (–2, 0) and (8, 0).

Question 5: Find the distance between parallel lines

(i) 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0

(ii) l (x +  y) + p = 0 and l (x +  y) – r = 0

ANSWER : - It is known that the distance (d) between parallel lines Ax + By + C1 = 0 and Ax + By + C2 = 0 is given by NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .

(i) The given parallel lines are 15x + 8y – 34 = 0 and 15x + 8y +31 = 0.

Here, A = 15, B = 8, C1 = –34, and C2 = 31.

Therefore, the distance between the parallel lines is

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

(ii) The given parallel lines are l (x +  y) + p = 0 and l (x +  y) – r = 0.

lx +  ly +  p = 0 and lx  + lyr = 0

Here, A = l, B = l, C1 = p, and C2 = –r.

Therefore, the distance between the parallel lines is

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

 

Question 6: Find equation of the line parallel to the line 3x – 4y + 2 = 0 and passing through the point (–2, 3).

ANSWER : - The equation of the given line is

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines , which is of the form y = mx + c

∴ Slope of the given line  NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

It is known that parallel lines have the same slope.

∴ Slope of the other line =  NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

Now, the equation of the line that has a slope of  NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines  and passes through the point (–2, 3) is

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

Question 7: Find equation of the line perpendicular to the line x – 7y + 5 = 0 and having x intercept 3.

ANSWER : - The given equation of line is NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines , which is of the form y = mx + c

∴Slope of the given line NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

The slope of the line perpendicular to the line having a slope of  NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines  is  NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

The equation of the line with slope –7 and x-intercept 3 is given by

y = m (xd)

y = –7 (x – 3)

y = –7x  21

⇒ 7x + y = 21

 

Question 8: Find angles between the lines  NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

ANSWER : - The given lines are NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

The slope of line (1) is NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines , while the slope of line (2) is NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .

The acute angle i.e., θ between the two lines is given by

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

Thus, the angle between the given lines is either 30° or 180° – 30° = 150°.

Question 9: The line through the points (h, 3) and (4, 1) intersects the line 7x – 9y – 19 = 0. at right angle. Find the value of h.

ANSWER : - The slope of the line passing through points (h, 3) and (4, 1) is

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

The slope of line 7x – 9y – 19 = 0 or  NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines  is NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .

It is given that the two lines are perpendicular.

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

Thus, the value of h is NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .

Question 10: Prove that the line through the point (x1, y1) and parallel to the line Ax + By + C = 0 is A (x –x1)  + B (y – y1) = 0.

ANSWER : -  The slope of line A+ B C = 0 or  NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines  is  NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

It is known that parallel lines have the same slope.

∴ Slope of the other line =  NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

The equation of the line passing through point (x1, y1) and having a slope  NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines is

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

Hence, the line through point (x1, y1) and parallel to line A+ B+ C = 0 is

A (x –x1)  + B (y – y1) = 0

Question 11: Two lines passing through the point (2, 3) intersects each other at an angle of 60°. If slope of one line is 2, find equation of the other line.

ANSWER : -  It is given that the slope of the first line, m1 = 2.

Let the slope of the other line be m2.

The angle between the two lines is 60°.

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

The equation of the line passing through point (2, 3) and having a slope of  NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines is

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

In this case, the equation of the other line is NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

The equation of the line passing through point (2, 3) and having a slope of NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines  is

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

In this case, the equation of the other line is NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .

Thus, the required equation of the other line is NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines  or  NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .

 

Question 12: Find the equation of the right bisector of the line segment joining the points (3, 4) and   (1, 2).

ANSWER : -  The right bisector of a line segment bisects the line segment at 90°.

The end-points of the line segment are given as A (3, 4) and B (–1, 2).

Accordingly, mid-point of AB  NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

Slope of AB NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

∴Slope of the line perpendicular to AB = NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

The equation of the line passing through (1, 3) and having a slope of –2 is

(y – 3) = –2 (x – 1)

y – 3 = –2x + 2

2x +  y = 5

Thus, the required equation of the line is 2x + y = 5.

 

Question 13: Find the coordinates of the foot of perpendicular from the point (1, 3) to the line 3x – 4y – 16 = 0.

ANSWER : -  Let (a, b) be the coordinates of the foot of the perpendicular from the point (–1, 3) to the line 3x – 4y – 16 = 0.

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

Slope of the line joining (–1, 3) and (a, b), m1NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

Slope of the line 3x – 4y – 16 = 0 or  NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

Since these two lines are perpendicular, m1m2 = –1

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

Point (a, b) lies on line 3x – 4y = 16.

∴3a – 4b = 16 … (2)

On solving equations (1) and (2), we obtain

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

Thus, the required coordinates of the foot of the perpendicular are NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .

 

Question 14: The perpendicular from the origin to the line y = mx + c meets it at the point (1, 2). Find the values of m and c.

ANSWER : -  The given equation of line is y = mx + c.

It is given that the perpendicular from the origin meets the given line at (–1, 2).

Therefore, the line joining the points (0, 0) and (–1, 2) is perpendicular to the given line.

∴Slope of the line joining (0, 0) and (–1, 2)  NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

The slope of the given line is m.

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

Since point (–1, 2) lies on the given line, it satisfies the equation y = mx +  c.

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

Thus, the respective values of m and c are NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .

 

Question 15: If p and q are the lengths of perpendiculars from the origin to the lines x cos θy sin θ = k cos 2θ and x sec θ + y cosec θ = k, respectively, prove that p2 + 4q2 = k2

ANSWER : -  The equations of given lines are

x cos θy sinθ = k cos 2θ … (1)

x secθ + y cosec θ= k … (2)

The perpendicular distance (d) of a line Ax +  By + C = 0 from a point (x1, y1) is given by NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .

On comparing equation (1) to the general equation of line i.e., Ax  + By +  C = 0, we obtain A = cosθ, B = –sinθ, and C = –k cos 2θ.

It is given that p is the length of the perpendicular from (0, 0) to line (1).

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

On comparing equation (2) to the general equation of line i.e., Ax + By + C = 0, we obtain A = secθ, B = cosecθ, and C = ­–k.

It is given that q is the length of the perpendicular from (0, 0) to line (2).

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

From (3) and (4), we have

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

Hence, we proved that p2 + 4q2 = k2.

Question 16: In the triangle ABC with vertices A (2, 3), B (4, 1) and C (1, 2), find the equation and length of altitude from the vertex A.

ANSWER : -  Let AD be the altitude of triangle ABC from vertex A.

Accordingly, AD⊥BC

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

The equation of the line passing through point (2, 3) and having a slope of 1 is

(y – 3) = 1(x – 2)

xy + 1 = 0

yx = 1

Therefore, equation of the altitude from vertex A = yx = 1.

Length of AD = Length of the perpendicular from A (2, 3) to BC

The equation of BC is

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

The perpendicular distance (d) of a line Ax  + By  + C = 0 from a point (x1, y1) is given by NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .

On comparing equation (1) to the general equation of line Ax + By + C = 0, we obtain A = 1, B = 1, and C = –3.

∴Length of AD  NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

Thus, the equation and the length of the altitude from vertex A are yx = 1 and  NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines units respectively.

 

Question 17: If p is the length of perpendicular from the origin to the line whose intercepts on the axes are a and b, then show that NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .

ANSWER : -  It is known that the equation of a line whose intercepts on the axes are a and b is

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

The perpendicular distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is given by NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .

On comparing equation (1) to the general equation of line Ax + By + C = 0, we obtain A = b, B = a, and C = –ab.

Therefore, if p is the length of the perpendicular from point (x1, y1) = (0, 0) to line (1), we obtain

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

On squaring both sides, we obtain

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

Hence, we showed that NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .

Exercise Miscellaneous

Question 1: Find the values of k for which the line NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines is

(a) Parallel to the x-axis,
 (b) Parallel to the y-axis,
 (c) Passing through the origin.

ANSWER : -  The given equation of line is

(k – 3) x – (4 – k2) y + k2 – 7k + 6 = 0 … (1)

(a) If the given line is parallel to the x-axis, then

Slope of the given line = Slope of the x-axis

The given line can be written as

(4 – k2) y = (k – 3) x + k2 – 7k  6 = 0

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines ,  

which is of the form y = mx  + c.

∴Slope of the given line = NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

Slope of the x-axis = 0

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

Thus, if the given line is parallel to the x-axis, then the value of k is 3.

(b) If the given line is parallel to the y-axis, it is vertical. Hence, its slope will be undefined.

The slope of the given line is NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .

Now,  NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines is undefined at k2 = 4

k2 = 4

k = ±2

Thus, if the given line is parallel to the y-axis, then the value of k is ±2.

(c) If the given line is passing through the origin, then point (0, 0) satisfies the

given equation of line.

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

Thus, if the given line is passing through the origin, then the value of k is either 1 or 6.

Question 2: Find the values of θ and p, if the equation  NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines is the normal form of the line NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .

ANSWER : -  The equation of the given line is NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .

This equation can be reduced as

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

On dividing both sides by NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines , we obtain

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

On comparing equation (1) to NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines , we obtain

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

Since the values of sin θ and cos θ are negative,  NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

Thus, the respective values of θ and p are  NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines  and 1

 

Question 3: Find the equations of the lines, which cut-off intercepts on the axes whose sum and product are 1 and –6, respectively.

ANSWER : -  Let the intercepts cut by the given lines on the axes be a and b.

It is given that

a + b = 1 … (1)

ab = –6 … (2)

On solving equations (1) and (2), we obtain

a = 3 and b = –2 or a = –2 and b = 3

It is known that the equation of the line whose intercepts on the axes are a and b is

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

Case I: a = 3 and b = –2

In this case, the equation of the line is –2x + 3y + 6 = 0, i.e., 2x – 3y = 6.

Case II: a = –2 and b = 3

In this case, the equation of the line is 3x – 2y + 6 = 0, i.e., –3x + 2y = 6.

Thus, the required equation of the lines are 2x – 3y = 6 and –3x + 2y = 6.

Question 4: What are the points on the y-axis whose distance from the line  NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines is 4 units.

ANSWER : -  Let (0, b) be the point on the y-axis whose distance from line  NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines  is 4 units.

The given line can be written as 4x + 3y – 12 = 0 … (1)

On comparing equation (1) to the general equation of line Ax  + By +  C = 0, we obtain A = 4, B = 3, and C = –12.

It is known that the perpendicular distance (d) of a line Ax  + By +  C = 0 from a point (x1, y1) is given by NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .

Therefore, if (0, b) is the point on the y-axis whose distance from line  NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines  is 4 units, then:

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

Thus, the required points are  NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines and NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .

 

Question 5: Find the perpendicular distance from the origin to the line joining the points  NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

 

ANSWER : -  The equation of the line joining the points  NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines is given by

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

It is known that the perpendicular distance (d) of a line Ax + By + C = 0 from a point (x1, y1) is given by NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .

Therefore, the perpendicular distance (d) of the given line from point (x1, y1) = (0, 0) is

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

 

Question 6: Find the equation of the line parallel to y-axis and drawn through the point of intersection of the lines x – 7y + 5 = 0 and 3x + y = 0.

ANSWER : -  The equation of any line parallel to the y-axis is of the form

x = a … (1)

The two given lines are

x – 7y + 5 = 0 … (2)

3x + y = 0 … (3)

On solving equations (2) and (3), we obtain NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .

Therefore,  NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines is the point of intersection of lines (2) and (3).

Since line x = a passes through point NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines ,  NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .

Thus, the required equation of the line is NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .

Question 7: Find the equation of a line drawn perpendicular to the line  NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines through the point, where it meets the y-axis.

ANSWER : -  The equation of the given line is NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .

This equation can also be written as 3x + 2y – 12 = 0

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines , which is of the form y = mx + c

∴Slope of the given line  NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

∴Slope of line perpendicular to the given line NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

Let the given line intersect the y-axis at (0, y).

On substituting x with 0 in the equation of the given line, we obtain NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

∴The given line intersects the y-axis at (0, 6).

The equation of the line that has a slope of NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines  and passes through point (0, 6) is

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

Thus, the required equation of the line is NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .

Question 8: Find the area of the triangle formed by the lines yx = 0, x + y = 0 and xk = 0.

ANSWER : -  The equations of the given lines are

yx = 0 … (1)

x + y = 0 … (2)

xk = 0 … (3)

The point of intersection of lines (1) and (2) is given by

x = 0 and y = 0

The point of intersection of lines (2) and (3) is given by

x = k and y = –k

The point of intersection of lines (3) and (1) is given by

x = k and y = k

Thus, the vertices of the triangle formed by the three given lines are (0, 0), (k, –k), and (k, k).

We know that the area of a triangle whose vertices are (x1, y1), (x2, y2), and (x3, y3) is NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .

Therefore, area of the triangle formed by the three given lines

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

Question 9: Find the value of p so that the three lines 3x + y – 2 = 0, px + 2y – 3 = 0 and 2xy – 3 = 0 may intersect at one point.

ANSWER : -  The equations of the given lines are

3x +  y – 2 = 0 … (1)

px + 2y – 3 = 0 … (2)

2xy – 3 = 0 … (3)

On solving equations (1) and (3), we obtain

x = 1 and y = –1

Since these three lines may intersect at one point, the point of intersection of lines (1) and (3) will also satisfy line (2).

p (1) + 2 (–1) – 3 = 0

p – 2 – 3 = 0

p = 5

Thus, the required value of p is 5.

Question 10: If three lines whose equations areNCERT Solutions Class 11 Maths Chapter 9 - Straight Lines concurrent, then show that  NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

 

ANSWER : -  The equations of the given lines are

y = m1x +  c1 … (1)

y = m2x + c2 … (2)

y = m3x + c3 … (3)

On subtracting equation (1) from (2), we obtain

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

On substituting this value of x in (1), we obtain

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines  is the point of intersection of lines (1) and (2).

It is given that lines (1), (2), and (3) are concurrent. Hence, the point of intersection of lines (1) and (2) will also satisfy equation (3).

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

Hence, NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

 

Question 11: Find the equation of the lines through the point (3, 2) which make an angle of 45° with the line x –2y = 3.

 

ANSWER : -  Let the slope of the required line be m1.

The given line can be written as  NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines , which is of the form y = mx +  c

∴Slope of the given line =  NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

It is given that the angle between the required line and line x – 2y = 3 is 45°.

We know that if θ is the acute angle between lines l1 and l2 with slopes m1 and m2 respectively, then NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

Case I: m1 = 3

The equation of the line passing through (3, 2) and having a slope of 3 is:

y – 2 = 3 (x – 3)

y – 2 = 3x – 9

3xy = 7

Case II: m1 =  NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

The equation of the line passing through (3, 2) and having a slope of NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines  is:

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

Thus, the equations of the lines are 3xy = 7 and x + 3y = 9.

Question 12: Find the equation of the line passing through the point of intersection of the lines  4x + 7y – 3 = 0 and 2x – 3y + 1 = 0 that has equal intercepts on the axes.

ANSWER : -  Let the equation of the line having equal intercepts on the axes be

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

On solving equations 4x + 7y – 3 = 0 and 2x – 3y + 1 = 0, we obtain NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines  is the point of intersection of the two given lines.

Since equation (1) passes through point NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines ,

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

∴ Equation (1) becomes NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

Thus, the required equation of the line is NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .

Question 13: Show that the equation of the line passing through the origin and making an angle θ with the line NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .

ANSWER : -  Let the equation of the line passing through the origin be y = m1x.

If this line makes an angle of θ with line y = mx + c, then angle θ is given by

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

Case I:  NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

Case II:  NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

Therefore, the required line is given by NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .

 

Question 14: In what ratio, the line joining (–1, 1) and (5, 7) is divided by the line x + y = 4?

ANSWER : -  The equation of the line joining the points (–1, 1) and (5, 7) is given by

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

The equation of the given line is

x + y – 4 = 0 … (2)

The point of intersection of lines (1) and (2) is given by

x = 1 and y = 3

Let point (1, 3) divide the line segment joining (–1, 1) and (5, 7) in the ratio 1:k. Accordingly, by section formula,

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

Thus, the line joining the points (–1, 1) and (5, 7) is divided by line

x  + y = 4 in the ratio 1:2.

 

Question 15: Find the distance of the line 4x + 7y + 5 = 0 from the point (1, 2) along the line 2xy= 0.

ANSWER : -  The given lines are

2xy = 0 … (1)

4x + 7y + 5 = 0 … (2)

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

A (1, 2) is a point on line (1).

Let B be the point of intersection of lines (1) and (2).

On solving equations (1) and (2), we obtain NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .

∴Coordinates of point B are NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .

By using distance formula, the distance between points A and B can be obtained as

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

Thus, the required distance is NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .

 

Question 16: Find the direction in which a straight line must be drawn through the point (–1, 2) so that its point of intersection with the line x   y = 4 may be at a distance of 3 units from this point.

ANSWER : -  Let y = mx + c be the line through point (–1, 2).

Accordingly, 2 = m (–1) + c.

⇒ 2 = –m + c

c = m + 2

y = mx + m + 2 … (1)

The given line is

x + y = 4 … (2)

On solving equations (1) and (2), we obtain

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines  is the point of intersection of lines (1) and (2).

Since this point is at a distance of 3 units from point (– 1, 2), according to distance formula,

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

Thus, the slope of the required line must be zero i.e., the line must be parallel to the x-axis.

 

Question 17: Find the image of the point (3, 8) with respect to the line x +3y = 7 assuming the line to be a plane mirror.

ANSWER : -  The equation of the given line is

x + 3y = 7 … (1)

Let point B (a, b) be the image of point A (3, 8).

Accordingly, line (1) is the perpendicular bisector of AB.

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

Since line (1) is perpendicular to AB,

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

 

The mid-point of line segment AB will also satisfy line (1).

Hence, from equation (1), we have

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

On solving equations (2) and (3), we obtain a = –1 and b = –4.

Thus, the image of the given point with respect to the given line is (–1, –4).

 

Question 18: If the lines y = 3x + 1 and 2y = x + 3 are equally inclined to the line y = mx + 4, find the value of m.

ANSWER : -  The equations of the given lines are

y = 3x + 1 … (1)

2y = x + 3 … (2)

y = mx + 4 … (3)

Slope of line (1), m1 = 3

Slope of line (2),  NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

Slope of line (3), m3 = m

It is given that lines (1) and (2) are equally inclined to line (3). This means that

the angle between lines (1) and (3) equals the angle between lines (2) and (3).

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

Thus, the required value of m is NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .

Question19: If sum of the perpendicular distances of a variable point P (x, y) from the lines x + y – 5 = 0 and 3x – 2y + 7 = 0 is always 10. Show that P must move on a line.

ANSWER : -  The equations of the given lines are

x + y – 5 = 0 … (1)

3x – 2y + 7 = 0 … (2)

The perpendicular distances of P (x, y) from lines (1) and (2) are respectively given by

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

It is given that NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines , which is the equation of a line.

Similarly, we can obtain the equation of line for any signs of NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .

Thus, point P must move on a line.

 

Question 20: Find equation of the line which is equidistant from parallel lines 9x + 6y – 7 = 0 and 3x + 2y + 6 = 0.

ANSWER : -  The equations of the given lines are

9x + 6y – 7 = 0 … (1)

3x + 2y + 6 = 0 … (2)

Let P (h, k) be the arbitrary point that is equidistant from lines (1) and (2). The perpendicular distance of P (h, k) from line (1) is given by

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

The perpendicular distance of P (h, k) from line (2) is given by

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

Since P (h, k) is equidistant from lines (1) and (2),  NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

9h + 6k – 7 = – 9h – 6k – 18

⇒ 18h + 12k + 11 = 0

Thus, the required equation of the line is 18x + 12y + 11 = 0.

 

Question 21: A ray of light passing through the point (1, 2) reflects on the x-axis at point A and the reflected ray passes through the point (5, 3). Find the coordinates of A.

ANSWER : -   Let the coordinates of point A be (a, 0).

Draw a line (AL) perpendicular to the x-axis.

We know that angle of incidence is equal to angle of reflection. Hence, let

∠BAL = ∠CAL = Φ

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

Let ∠CAX = θ

∴∠OAB = 180° – (θ + 2Φ) = 180° – [θ + 2(90° – θ)]

= 180° – θ – 180°+ 2θ

= θ

∴∠BAX = 180° – θ

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

From equations (1) and (2), we obtain

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

Thus, the coordinates of point A are NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .

Question 22: Prove that the product of the lengths of the perpendiculars drawn from the points  NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

ANSWER : -  The equation of the given line is

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

Length of the perpendicular from point  NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines to line (1) is 

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

Length of the perpendicular from point  NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines  to line (2) is

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

On multiplying equations (2) and (3), we obtain

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

Hence, proved.

Question 23: A person standing at the junction (crossing) of two straight paths represented by the equations 2x – 3y + 4 = 0 and 3x + 4y – 5 = 0 wants to reach the path whose equation is 6x – 7y + 8 = 0 in the least time. Find equation of the path that he should follow.

ANSWER : -  The equations of the given lines are

2x – 3y + 4 = 0 … (1)

3x + 4y – 5 = 0 … (2)

6x – 7y + 8 = 0 … (3)

The person is standing at the junction of the paths represented by lines (1) and (2).

On solving equations (1) and (2), we obtain NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .

Thus, the person is standing at point NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .

The person can reach path (3) in the least time if he walks along the perpendicular line to (3) from point NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

∴Slope of the line perpendicular to line (3)  NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

The equation of the line passing through  NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines  and having a slope of NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines  is given by

NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

Hence, the path that the person should follow is NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines .

The document NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines is a part of the JEE Course Mathematics (Maths) for JEE Main & Advanced.
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FAQs on NCERT Solutions Class 11 Maths Chapter 9 - Straight Lines

1. What are the properties of a straight line?
Ans. A straight line has properties such as infinite length, constant slope, and the equation of the form y = mx + c, where m is the slope and c is the y-intercept.
2. How do you find the slope of a straight line?
Ans. The slope of a straight line can be calculated using the formula (y2 - y1) / (x2 - x1), where (x1, y1) and (x2, y2) are two points on the line.
3. What is the significance of the y-intercept of a straight line?
Ans. The y-intercept of a straight line is the point where it intersects the y-axis. It represents the value of y when x is zero and is used to determine the starting point of the line.
4. How can you determine if two lines are parallel or perpendicular?
Ans. Two lines are parallel if they have the same slope, and they are perpendicular if the product of their slopes is -1.
5. Can a straight line have a negative slope?
Ans. Yes, a straight line can have a negative slope if it is sloping downwards from left to right on a graph.
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