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

To 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  

= 12 |x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)| unit²

Therefore, area of ΔABC

= 12 |-4(7 + 5) + 0(-5 - 5) + 5(5 - 7)| unit²

= 12 |-4(12) + 5(-2)| unit²

= 12 |-48 - 10| unit²

= 12 |58| unit²

= 582 = 29 unit²

Area of ΔACD

= 12 |-4(-5 + 2) + 5(-2 - 5) + (-4)(5 + 5)| unit²

= 12 |-4(-3) + 5(-7) - 4(10)| unit²

= 12 |12 - 35 - 40| unit²

= 12 |-63| unit²

= 632 unit²

Thus, area (ABCD)

= 29 + 632 unit² = 58 + 632 unit² = 1212 unit²

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.

On applying Pythagoras theorem to ΔAOC, we obtain
(AC)² = (OA)²  (OC)²
⇒ (2a)² = (OA)²   a²
⇒ 4a² – a² = (OA)²
⇒ (OA)² = 3a²
⇒ OA =  √3a
∴ Coordinates of point A = (±√3a, 0)
Thus, the vertices of the given equilateral triangle are (0, a), (0, –a), and (√3a, 0) or (0, a), (0, –a), and (-√3a, 0) .

Q3: Find the distance between  P(x₁, y₁) and  Q(x₂,  y₂) when:
(i) PQ is parallel to the y-axis 
(ii) PQ is parallel to the x-axis.
Ans: The given points are P(x₁, y₁) and  Q(x₂,  y₂) .
(i) When PQ is parallel to the y-axis, x₁ = x₂.

In this case, distance between P and Q:

= √((x₂ - x₁)² + (y₂ - y₁)²)

= √((y₂ - y₁)²)

= |y₂ - y₁|

(ii) When PQ is parallel to the x-axis, y₁ = y₂.

In this case, distance between P and Q:

= √((x₂ - x₁)² + (y₂ - y₁)²)

= √((x₂ - x₁)²)

= |x₂ - x₁|

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).
Accordingly,

√((7 - a)² + (6 - 0)²) = √((3 - a)² + (4 - 0)²)

⇒ √(49 + a² - 14a + 36) = √(9 + a² - 6a + 16)

⇒ √(a² - 14a + 85) = √(a² - 6a + 25)

On squaring both sides, we obtain

a² - 14a + 85 = a² - 6a + 25

⇒ -14a + 6a = 25 - 85

⇒ -8a = -60

⇒ a = 608 = 152

Thus, the required point on the x-axis is ( 152 , 0).

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: ( 0 + 82 , −4 + 02 ) = (4, −2)
It is known that the slope (m) of a non-vertical line passing through the points (x₁, y₁) and (x₂, y₂) is given by: m = y₂ − y₁x₂ − x₁ , x₂ ≠ x₁.
Therefore, the slope of the line passing through (0, 0) and (4, −2) is: −2 − 04 − 0 = −24 = −12.
Hence, the required slope of the line is:− 12.

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: m = y₂ − y₁x₂ − x₁ , x₂ ≠ x₁.
∴ Slope of AB (m₁) = 5 − 43 − 4 = 1−1 = −1
Slope of BC (m₂) = −1 − 5−1 − 3 = −6−4 = 32
Slope of CA (m₃) = 4 − 14 − 1 = 55 = 1
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°.

Thus, the slope of the given line is tan 120° = tan (180° – 60°) = –tan 60°  .

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.

Slope of AB = 0 + 14 + 2 = 16

Slope of CD = 2 − 3−3 − 3 = −1−6 = 16

⇒ Slope of AB = Slope of CD

⇒ AB and CD are parallel to each other.

Now, slope of BC = 3 − 03 − 4 = 3−1 = −3

Slope of AD = 2 + 1−3 + 2 = 3−1 = −3

⇒ 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 m = -2 - (-1)4 - 3 = -2 + 1 = −1
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 the lines.
Ans: Let be the slopes of the two given lines such that m¹ = 2m .
We know that if θ is the angle between the lines l₁ and l₂ with slopes m₁ and m₂, then: tan θ = |m₂ − m₁||1 + m₁m₂|.
It is given that the tangent of the angle between the two lines is: 13.

∴ 13 = |m − 2m||1 + (2m)·m|

⇒ 13 = |−m||1 + 2m²|
⇒ 13 = − m1 + 2m².

Case I
13 = −m1 + 2m²
⇒ 1 + 2m² = −3m
⇒ 2m² + 3m + 1 = 0
⇒ 2m(m + 1) + (m + 1) = 0
⇒ (m + 1)(2m + 1) = 0
⇒ m = −1 or m = −12
If m = −1, then the slopes of the lines are −1 and −2.
If m = −12, then the slopes of the lines are −12 and −1.

Case II
13 = m1 + 2m²
⇒ 2m² + 1 = 3m
⇒ 2m² − 3m + 1 = 0
⇒ 2m(m − 1) − (m − 1) = 0
⇒ (m − 1)(2m − 1) = 0
⇒ m = 1 or m = 12

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

If m = 12 , then the slopes of the lines are 12  and 1.

Hence, the slopes of the lines are –1 and –2 or  -12 and –1 or 1 and 2 or 12  and 1 .

Q11: A line passes through (x₁, y₁) and (h, k) . If slope of the line is m, show that k - y₁ = m(h -x₁).

Ans: The slope of the line passing through (x₁, y₁) and (h, k) is: k − y₁h − x₁.
It is given that the slope of the line is m.
k − y₁h − x₁ = m
⇒ k − y₁ = m(h − x₁)
Hence, k − y₁ = m(h − x₁).

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 slope 12 .
Ans: We know that the equation of the line passing through point (x₀, y₀)  , whose slope is m, is (y - y₀) = m(x - x₀) .
Thus, the equation of the line passing through point (–4, 3), whose slope is 12 , is
(y − 3) = 12 (x + 4)
2(y − 3) = x + 4
2y − 6 = x + 4
i.e., x − 2y + 10 = 0

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 (x₀, y₀)  , whose slope is m, is (y - y₀) = m(x - x₀) .
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  (2, 2√3) 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°

We know that the equation of the line passing through point (x₀, y₀)  , whose slope is m, is (y - y₀) = m(x - x₀) .
Thus, if a line passes though (2, 2√3) and inclines with the x-axis at an angle of 75°, then the equation of the line is given as

(y − 2√3) = √3 + 1√3 − 1 (x − 2)
(y − 2√3)(√3 − 1) = (√3 + 1)(x − 2)
y(√3 − 1) − 2√3(√3 − 1) = x(√3 + 1) − 2(√3 + 1)
(√3 − 1)x − (√3 − 1)y = 2√3 − 2 − 6 + 2√3
(√3 + 1)x − (√3 − 1)y = 4√3 − 4
i.e., (√3 + 1)x − (√3 − 1)y = 4(√3 − 1)

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° = 1√3.
Thus, the required equation of the given line is:
y = 1√3 x + 2
y = x + 2√3√3
√3y = x + 2√3
i.e., x − √3y + 2√3 = 0

Q7: Find the equation of the line which passes through the points (–1, 1) and (2, –4).
Ans: It is known that the equation of the line passing through points (x₁, y₁) and (x₂, y₂) is: y − y₁ = y₂ − y₁x₂ − x₁ (x − x₁).
Therefore, the equation of the line passing through the points (−1, 1) and (2, −4) is:
(y − 1) = −4 − 12 + 1 (x + 1)
(y − 1) = −53 (x + 1)
3(y − 1) = −5(x + 1)
3y − 3 = −5x − 5
i.e., 5x + 3y + 2 = 0

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 the midpoint formula, the coordinates of point L are given by: L = ( 2 − 22 , 1 + 32 ) = (0, 2)
It is known that the equation of the line passing through points (x₁, y₁) and (x₂, y₂) is: y − y₁ = y₂ − y₁x₂ − x₁ (x − x₁).Therefore, the equation of RL can be determined by substituting (x₁, y₁) = (4, 5) and (x₂, y₂) = (0, 2).

Hence,
y − 5 = 2 − 50 − 4 (x − 4)
y − 5 = −3−4 (x − 4)
4(y − 5) = 3(x − 4)
4y − 20 = 3x − 12
⇒ 3x − 4y + 8 = 0

Thus, the required equation of the median through vertex R is 3x − 4y + 8 = 0.

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: m = 6 − 5−3 − 2 = 1−5 = − 15
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, the slope of the line perpendicular to the line through the points (2, 5) and (−3, 6) is: − 1m = − 1− 15 = 5
Now, the equation of the line passing through point (−3, 5), whose slope is 5, is:
(y − 5) = 5(x + 3)
y − 5 = 5x + 15
y = 5x + 20
i.e., 5x − y + 20 = 0

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

The slope of the line joining the points (1, 0) and (2, 3) is: m = 3 − 02 − 1 = 3
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, the slope of the line that is perpendicular to the line joining the points (1, 0) and (2, 3) is: − 1m = − 13
Now, the equation of the line passing through: ( n + 2n + 1 , 3n + 1 ) 
and whose slope is: − 13 is given by: y − 3n + 1 = − 13 (x − n + 2n + 1)
⇒ 3( (n + 1)y − 3n + 1 ) = −[(n + 1)(x) − (n + 2)]
⇒ 3(n + 1)y − 9 = −(n + 1)x + n + 2
⇒ (n + 1)x + 3(n + 1)y = n + 11
Thus, the required equation is: (n + 1)x + 3(n + 1)y = n + 11

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
xa + yb = 1 .......(i)

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
xa + yb = 1
⇒ x + y = a .......(ii)

Since the given line passes through point (2, 3), equation (ii) reduces to
2 + 3 = a ⇒ a = 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

xa + yb = 1 .......(i)
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
xa + y9 - a = 1 .......(iii)

It is given that the line passes through point (2, 2). Therefore, equation (iii) reduces to
2a + 29 − a = 1

⇒2( 1a + 19 − a) = 1

⇒( 9 − a + aa(9 − a)) = 1

⇒ 189a − a² = 1

⇒ 18 = 9a − a²

⇒ a² − 9a + 18 = 0

⇒ (a − 6)(a − 3) = 0

⇒ a = 6 or a = 3

If a = 6 and b = 9 − 6 = 3, then the equation of the line is:

x6 + y3 = 1 ⇒ x + 2y − 6 = 0

If a = 3 and b = 9 − 3 = 6, then the equation of the line is:

x3 + y6 = 1 ⇒ 2x + y − 6 = 0

Q13: Find equation of the line through the point (0, 2) making an angle 2π3  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 2π3 with the positive x-axis is m = tan( 2π3 ) = −√3.
Now, the equation of the line passing through point (0, 2) and having a slope −√3 is: y − 2 = −√3(x − 0)
i.e., √3x + y − 2 = 0
The slope of the line parallel to line √3x + y − 2 = 0 is −√3.
It is given that the line parallel to line √3x + y − 2 = 0 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 −√3 is:
y − (−2) = −√3(x − 0)
y + 2 = −√3x
√3x + y + 2 = 0

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: m₁ = 9 − 0−2 − 0 = −92.
Accordingly, the slope of the line perpendicular to the line joining the origin and point (−2, 9) is: m₂ = − 1m₁ = − 1−92 = 29.
Now, the equation of the line passing through point (−2, 9) and having a slope m₂ is:
y − 9 = 29 (x + 2)
9(y − 9) = 2(x + 2)
9y − 81 = 2x + 4
i.e., 2x − 9y + 85 = 0

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) = 125.134 − 124.942110 − 20 (C − 20)

L − 124.942 = 0.19290 (C − 20)

i.e., L = 0.19290 (C − 20) + 124.942, which is the required linear relation.

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).
y − 980 = 1220 − 98016 − 14 (x − 14)
y − 980 = 2402 (x − 14)
y − 980 = 120(x − 14)
i.e., y = 120(x − 14) + 980
When x = Rs 17/litre,
y = 120(17 − 14) + 980
⇒ y = 120 × 3 + 980
⇒ y = 360 + 980 = 1340

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 xa + yb = 2.
Ans: Let AB be the line segment between the axes and let P (a, b) be its mid-point.

Let the coordinates of A and B be (0, y) and (x, 0) respectively.
Since P (a, b) is the mid-point of AB,
( 0 + x2 , y + 02 ) = (a, b)
⇒ ( x2 , y2 ) = (a, b)
⇒ x2 = a and y2 = b
∴ x = 2a and y = 2b

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:
y − 2b = 0 − 2b2a − 0 (x − 0)
y − 2b = −2b2a x
a(y − 2b) = −bx
ay − 2ab = −bx
i.e., bx + ay = 2ab
On dividing both sides by ab, we obtain:

bxab + ayab = 2abab

⇒ xa + yb = 2
Thus,the equation of the line is xa + yb = 2 .

Q18: Point R (h, k) divides a line segment between the axes in the ratio 1:2. Find the 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,

(h, k) = ( 1 × 0 + 2 × x1 + 2 , 1 × y + 2 × 01 + 2 )
⇒ (h, k) = ( 2x3 , y3 )
⇒ h = 2x3 and k = y3
⇒ x = 3h2 and y = 3k
Therefore, the respective coordinates of A and B are ( 3h2 , 0) and (0, 3k).
Now, the equation of line AB passing through points ( 3h2 , 0) and (0, 3k) is:
(y − 0) = 3k − 00 − 3h2 (x − 3h2)
y = 2kh (x − 3h2)
hy = −2kx + 3hk
i.e., 2kx + hy = 3hk
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
(y − 0) = −2 − 0−2 − 3 (x − 3)
y = −25 (x − 3)

5y = 2x − 6
i.e., 2x − 5y = 6
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
y = −17 x + 0 ... (1)
This equation is of the form y = mx + c, where m = −17 and c = 0.
Therefore, equation (1) is in the slope-intercept form, where the slope and the y-intercept are: −17 and 0, respectively.
(ii) The given equation is 6x + 3y − 5 = 0.
It can be written as:
y = 13 (−6x + 5)
y = −2x + 53 ... (2)
This equation is of the form y = mx + c, where m = −2 and c = 53.
Therefore, equation (2) is in the slope-intercept form, where the slope and the y-intercept are –2 and 53 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
3x + 2y = 12
3x12 + 2y12 = 1
i.e., x4 + y6 = 1 ... (1)
This equation is of the form xa + yb = 1, 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:
4x6 − 3y6 = 1
2x3 + y−2 = 1
i.e., x32 + y−2 = 1 ... (2)
This equation is of the form xa + yb = 1, where a = 32 and b = −2.
Therefore, equation (2) is in the intercept form, where the intercepts on the x and y axes are 32  and –2 respectively.

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

It can be written as
3y  - 2 = 0.
i.e., y-23 = 1    ... (3)

This equation is of the form xa + yb = 1 , where a = 0 and b =  -23 .

Therefore, equation (3) is in the intercept form, where the intercept on the y-axis is -23 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: d = |Ax₁ + By₁ + C|√(A² + B²).
The given point is (x₁, y₁) = (−1, 1).
Therefore, the distance of point (−1, 1) from the given line:
= |12(−1) + (−5)(1) + 82|√((12)² + (−5)²) units
= |−12 − 5 + 82|√(169) units
= |65|13 units = 5 units

Question 4: Find the points on the x-axis, whose distances from the line  x3 + y4 = 1  are 4 units.

ANSWER : - The given equation of line is
x3 + y4 = 1
or, 4x + 3y - 12 = 0   ....(1)

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 d = |Ax₁ + By₁ + C|√(A² + B²).
Therefore,
4 = |4a + 3 × 0 − 12|√(4² + 3²)
⇒ 4 = |4a − 12|5
⇒ |4a − 12| = 20
⇒ ±(4a − 12) = 20
⇒ (4a − 12) = 20 or −(4a − 12) = 20
⇒ 4a = 20 + 12 or 4a = −20 + 12
⇒ a = 8 or a = −2
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: d = |C₁ − C₂|√(A² + B²).
(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:

d = |−34 − 31|√(15² + 8²) units = |−65|17 units = 6517 units.

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

l(x + y) + p = 0 and l(x + y) − r = 0

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

Therefore, the distance between the parallel lines is:

d = |C₁ − C₂|√(A² + B²) = |p + r|√(l² + l²) = |p + r|√2l = |p + r|l√2 = 1√2 |p + r|/l units. 

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
3x − 4y + 2 = 0

or y = 3x4 + 24

or y = 34 x + 12 , which is of the form y = mx + c

∴ Slope of the given line = 34

It is known that parallel lines have the same slope.

∴ Slope of the other line = m = 34

Now, the equation of the line that has a slope of 34 and passes through the point (−2, 3) is:

(y − 3) = 34 (x − (−2))

4y − 12 = 3x + 6

i.e., 3x − 4y + 18 = 0

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 the line is x − 7y + 5 = 0.

Or, y = 17 x + 57, which is of the form y = mx + c.

∴ Slope of the given line = 17

The slope of the line perpendicular to the line having a slope of 17 is:

m = − 117 = −7

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

y = m (x – d)

⇒ y = –7 (x – 3)

⇒ y = –7x  21

⇒ 7x + y = 21

Question 8: Find angles between the lines √3x + y = 1 and x + √3y = 1

ANSWER : - The given lines are √3x + y = 1 and x + √3y = 1.
y = −√3x + 1 ... (1)

and y = 1√3 x + 1√3 ... (2)

The slope of line (1) is m₁ = −√3, while the slope of line (2) is m₂ = 1√3.

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

tan θ = |m₁ − m₂|1 + m₁m₂

= |−√3 − 1√3|1 + (−√3) × 1√3

tan θ = |−3 + 1|1 + 1 = 22 × √3

tan θ = 1√3

θ = 30°

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
m₁ = 1 − 34 − h = −24 − h

The slope of the line 7x − 9y − 19 = 0 or y = 79 x − 199 is m₂ = 79

It is given that the two lines are perpendicular.

∴ m₁ × m₂ = −1

⇒ −24 − h × 79 = −1

⇒ −1436 − 9h = −1

⇒ −14 = −36 + 9h

⇒ 14 = 36 − 9h

⇒ 9h = 36 − 14

⇒ h = 229

Thus, the value of h is 229

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 Ax + By + C = 0 or y = −AB x + −CB is m = −A B

It is known that parallel lines have the same slope.
∴ Slope of the other line = m = −AB
The equation of the line passing through point (x₁, y₁) and having a slope m = −AB is:
y − y₁ = m(x − x₁)
y − y₁ = −AB (x − x₁)
B(y − y₁) = −A(x − x₁)
A(x − x₁) + B(y − y₁) = 0
Hence, the line through point (x₁, y₁) and parallel to line Ax + By + 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°.
∴ tan 60° = |m₁ − m₂|1 + m₁m₂
⇒ √3 = |2 − m₂|1 + 2m₂
⇒ √3 = ± 2 − m₂1 + 2m₂
⇒ √3 = 2 − m₂1 + 2m₂ or √3 = −(2 − m₂)1 + 2m₂
⇒ √3(1 + 2m₂) = 2 − m₂ or √3(1 + 2m₂) = −(2 − m₂)
⇒ √3 + 2√3m₂ = 2 − m₂ or √3 + 2√3m₂ = −2 + m₂
⇒ √3 + (2√3 + 1)m₂ = 2 or √3 + (2√3 + 1)m₂ = −2
⇒ m₂ = 2 − √3(2√3 + 1) or m₂ = −(2 + √3)(2√3 + 1)
Case 1: m₂ = (2 − √32√3 + 1)

The equation of the line passing through point (2, 3) and having a slope of   (2 − √3)(2√3 + 1) is
(y − 3) = 2 − √32√3 + 1 (x − 2)
(2√3 + 1)y − 3(2√3 + 1) = (2 − √3)x − 2(2 − √3)
(2√3 + 1)y − 6√3 − 3 = 4 − 2√3 − 4√3 + 3
(2√3 + 1)y = 4 + 2√3 + 6√3 + 3
(2√3 + 1)y = −1 + 8√3
In this case, the equation of the other line is:
(√3 − 2)x + (2√3 + 1)y = −1 + 8√3

Case II: m₂ = −(2 + √3)2√3 − 1
The equation of the line passing through point (2, 3) and having a slope of:
(y − 3) = −(2 + √3)2√3 − 1 (x − 2)
(2√3 − 1)y − 3(2√3 − 1) = (−(2 + √3))x + 2(2 + √3)
(2√3 − 1)y + 6√3 − 3 = −4 − 2√3 + 4√3 − 3
(2√3 − 1)y = −1 + 8√3
In this case, the equation of the other line is: (2 + √3)x + (2√3 − 1)y = −1 + 8√3

Thus, the required equation of the other line is: (√3 − 2)x + (2√3 + 1)y = −1 + 8√3 or (2 + √3)x + (2√3 − 1)y = −1 + 8√3

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, the midpoint of AB is: ( 3 − 12 , 4 + 22 ) = (1, 3)
Slope of AB = 2 − 4−1 − 3 = −2−4 = 12
∴ Slope of the line perpendicular to AB = − 112 = −2

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.

Slope of the line joining (−1, 3) and (a, b), m₁ = b − 3a + 1
Slope of the line 3x − 4y − 16 = 0 or y = 34 x − 4, m₂ = 34
Since these two lines are perpendicular, m₁m₂ = −1
∴ (b − 3)(a + 1) × 34 = −1
⇒ 3b − 9 = −4a − 4
⇒ 3b − 9 = −4a − 4
⇒ 4a + 3b = 5 ... (1)
Point (a, b) lies on line 3x − 4y = 16.
∴ 3a − 4b = 16 ... (2)
On solving equations (1) and (2), we obtain:
a = 6825 and b = 4925
Thus, the required coordinates of the foot of the perpendicular are: ( 6825 , 4925 )

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) = 2−1 = −2

The slope of the given line is m.

∴ m × −2 = −1 [The two lines are perpendicular]

⇒ m = 12

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

∴ 2 = m(−1) + c

⇒ 2 = 12 (−1) + c

⇒ c = 2 + 12 = 52

Thus, the respective values of m and c are 12 and 52.

Question 15: If p and q are the lengths of perpendiculars from the origin to the lines x cos θ – y 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 θ – y 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 d = |Ax₁ + By₁ + C|√(A² + B²).

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).

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).

 q = |A(0) + B(0) + C|√(A² + B²) = |C|√(A² + B²) = |−k|√(sec²θ + cosec²θ) ... (4)

From (3) and (4), we have:
p² + 4q² = (k cos 2θ)² + 4k²(sec²θ + cosec²θ)
= k² cos² 2θ + 4k²1cos²θ + 1sin²θ
= k² cos² 2θ + 4k²sin²θ cos²θ
= k² cos² 2θ + 4k² sin θ cos θ
= k²(cos² 2θ + sin² 2θ)
= k²

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

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

(y – 3) = 1(x – 2)

⇒ x – y + 1 = 0

⇒ y – x = 1

Therefore, equation of the altitude from vertex A = y – x = 1.

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

The equation of BC is

(y + 1) = 2 + 11 − 4 (x − 4)

⇒ (y + 1) = −1(x − 4)

⇒ y + 1 = −x + 4

⇒ x + y − 3 = 0 ... (1)

The perpendicular distance (d) of a line Ax + By + C = 0 from a point (x₁, y₁) is given by:

d = |Ax₁ + By₁ + C|√(A² + B²)

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 = |1 × 2 + 1 × 3 − 3|√(1² + 1²) units

= |2|√2 units

= 2√2 units

= √2 units

Thus, the equation and the length of the altitude from vertex A are y – x = 1 and  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 

1p² = 1a² + 1b² .
ANSWER: It is known that the equation of a line whose intercepts on the axes are a and b is:

xa + yb = 1

or bx + ay = ab

or bx + ay − ab = 0 ... (1)

The perpendicular distance (d) of a line Ax + By + C = 0 from a point (x₁, y₁) is given by:

d = |Ax₁ + By₁ + C|√(A² + B²)

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:

p = |A(0) + B(0) + C|√(b² + a²)

⇒ p = |−ab|√(a² + b²)

On squaring both sides, we obtain:

p² = (−ab)²(a² + b²)

⇒ p²(a² + b²) = a²b²

⇒ 1p² = 1a² + 1b²

Hence, we showed that  1p² = 1a² + 1b² .

Exercise Miscellaneous

Question 1: Find the values of k for which the line (k – 3) x – (4 – k²) y + k² – 7k + 6 = 0 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

 ,  

which is of the form y = mx  + c.

∴Slope of the given line = (k − 3)(4 − k²)

Slope of the x-axis = 0

∴ (k − 3)(4 − k²) = 0

⇒ k − 3 = 0

⇒ k = 3

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: (k − 3)(4 − k²).

Now, (k − 3)(4 − k²) is undefined at 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 the line.

(k − 3)(0) − (4 − k²)(0) + k² − 7k + 6 = 0

⇒ k² − 7k + 6 = 0

⇒ (k − 6)(k − 1) = 0

⇒ k = 6 or k = 1

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  xcosθ + ysinθ = p is the normal form of the line √3x + y + 2 = 0. (out of NCERT
ANSWER : - The equation of the given line is √3x + y + 2 = 0 . 

This equation can be reduced as
√3x + y + 2 = 0
⇒ -√3x - y = 2

On dividing both sides by  , we obtain
− √32 x − 12 y = 22

⇒ − √32 x + −12 y = 1 ... (1)

On comparing equation (1) to x cos θ + y sin θ = p, we obtain:

cos θ = − √32, sin θ = − 12, and p = 1

Since the values of sin θ and cos θ are negative,

θ = π + π6 = 7π6

Thus, the respective values of θ and p are: θ = 7π6 and p = 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
xa + yb = 1 or bx +ay - ab = 0

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  x3 + y4 = 1 is 4 units.

ANSWER : -  Let (0, b) be the point on the y-axis whose distance from line  x3 + y4 = 1  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 d = |Ax₁ + By₁ + C|√(A² + B²) .

Therefore, if (0, b) is the point on the y-axis whose distance from line  x3 + y4 = 1 is 4 units, then:
4 = |4(0) + 3(b) − 12|√(4² + 3²)

⇒ 4 = |3b − 12|5

⇒ 20 = |3b − 12|

⇒ 20 = ±(3b − 12)

⇒ 20 = 3b − 12 or 20 = −(3b − 12)

⇒ 3b = 20 + 12 or 3b = −20 + 12

⇒ b = 323 or b = −83

Thus, the required points are: (0, 323 ) and (0, −83 )

Question 5: Find the perpendicular distance from the origin to the line joining the points (cos θ, sin θ) and (cos ϕ, sin ϕ).
ANSWER: The equation of the line joining the points (cos θ, sin θ) and (cos ϕ, sin ϕ) is given by:

y − sin θ = sin ϕ − sin θcos ϕ − cos θ (x − cos θ)

y(cos ϕ − cos θ) − sin θ(cos ϕ − cos θ) = x(sin ϕ − sin θ) − cos θ(sin ϕ − sin θ)

x(sin θ − sin ϕ) + y(cos ϕ − cos θ) + cos θ sin ϕ − sin θ cos ϕ = 0

x(sin θ − sin ϕ) + y(cos ϕ − cos θ) + sin(ϕ − θ) = 0

A_x + B_y + C = 0, where:

• A = sin θ − sin ϕ
• B = cos ϕ − cos θ
• C = sin(ϕ − θ)

It is known that the perpendicular distance (d) of a line Ax + By + C = 0 from a point (x₁, y₁) is given by d = |Ax₁ + By₁ + C|√(A² + B²) .
Therefore, the perpendicular distance (d) of the given line from point (x₁, y₁) = (0, 0) is

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:
x = −522 and y = 1522.
Therefore, ( −522, 1522 ) is the point of intersection of lines (2) and (3).
Since line x = a passes through point ( −522, 1522 ), a = −522.
Thus, the required equation of the line is: x = −522.

Question 7: Find the equation of a line drawn perpendicular to the line x4 + y6 = 1  through the point, where it meets the y-axis.
ANSWER: The equation of the given line is: x4 + y6 = 1

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

y = −32 x + 6, which is of the form y = mx + c
∴ Slope of the given line = −32
∴ Slope of the line perpendicular to the given line = − 1−32 = 23
Let the given line intersect the y-axis at (0, y).

On substituting x = 0 in the equation of the given line, we obtain: y6 = 1 ⇒ y = 6

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

The equation of the line that has a slope of 23 and passes through point (0, 6) is:
(y − 6) = 23 (x − 0)
3y − 18 = 2x
2x − 3y + 18 = 0
Thus, the required equation of the line is: 2x − 3y + 18 = 0

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

ANSWER : -  The equations of the given lines are

y – x = 0 … (1)

x + y = 0 … (2)

x – k = 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: 12 |x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|

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

= 12 |0(−k − k) + k(k − 0) + k(0 + k)| square units

= 12 |k² + k²| square units

= 12 |2k²| square units

= k² square units

Question 9: Find the value of p so that the three lines 3x + y – 2 = 0, px + 2y – 3 = 0 and 2x – y – 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)

2x – y – 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 are: y = m₁x + c₁, y = m₂x + c₂, and y = m₃x + c₃ are concurrent, then show that: m₁(c₂ − c₃) + m₂(c₃ − c₁) + m₃(c₁ − c₂) = 0
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
0 = (m₂ − m₁)x + (c₂ − c₁)
⇒ (m₂ − m₁)x = c₁ − c₂
⇒ x = c₂ − c₁m₁ − m₂
On substituting this value of x in (1), we obtain:
y = m₁ c₂ − c₁m₁ − m₂ + c₁
y = m₁c₂ − m₁c₁m₁ − m₂ + c₁
y = m₁c₂ − m₁c₁ + c₁(m₁ − m₂)m₁ − m₂
y = m₁c₂ − m₂c₁m₁ − m₂
∴ ( c₂ − c₁m₁ − m₂, m₁c₂ − m₂c₁m₁ − m₂ ) 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).
Substituting into (3):
m₃ c₂ − c₁m₁ − m₂ + m₁c₂ − m₂c₁m₁ − m₂ + c₃ = 0
Combining terms and simplifying:
m₁(c₂ − c₃) + m₂(c₃ − c₁) + m₃(c₁ − c₂) = 0

Hence, the condition is proven.

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  y = 12x - 32 , which is of the form y = mx +  c

∴Slope of the given line =  

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  .

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

3x – y = 7

Case II: m₁ = - 13

The equation of the line passing through (3, 2) and having a slope of - 13 is:
y - 2 = - 13 (x - 3)
3y - 6 = -x + 3
x + 3y = 9

Thus, the equations of the lines are 3x – y = 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  xa + yb = 1 Or x + y = a  ....... (1)

On solving equations 4x + 7y – 3 = 0 and 2x – 3y + 1 = 0, we obtain x = 113  and y = 513.
∴ ( 113, 513 )   is the point of intersection of the two given lines.

Since equation (1) passes through point ( 113, 513 ) ,
113 +  513   = a

⇒ a = 613

∴ Equation (1) becomes x + y = 613  i.e., 13x + 13y = 6

Thus, the required equation of the line is 13x + 13y = 6 .

Question 13: Show that the equation of the line passing through the origin and making an angle θ with the line .

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

Case I:

Case II:

Therefore, the required line is given by  .

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

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 the section formula:

(1, 3) = k(−1) + 1(5)1 + k, k(1) + 1(7)1 + k

⇒ (1, 3) = −k + 51 + k k + 71 + k

⇒ −k + 51 + k = 1 and k + 71 + k = 3

From the first equation:

−k + 5 = 1 + k

⇒ 2k = 4

⇒ k = 2

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 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 the slope of the required line PQ be m.
The equation of the line PQ, which passes through the point P(-1, 2) and has slope m, is
y - y₁ = m(x - x₁) 
y - 2 = m(x + 1) or 
mx - y + m + 2 = 0 .......(i)
Equation of line AB: x + y = 4 
∴ y = 4 - x
Putting the value of y in equation (1),
mx - (4 - x) + m + 2 = 0 
or (m + 1)x + m - 2 = 0 
∴ x = (m - 2)/(m + 1)
Now y = 4 - x 
y = 4 - (m - 2)/(m + 1) = (4m + 4 - m + 2)/(m + 1) = (5m + 2)/(m + 1)
Given: PQ = 3 or PQ² = 9
∴ ((m - 2)/(m + 1))² + ((5m + 2 - 2)/(m + 1))² = 9 
or ((-m + 2)/(m + 1))² + ((5m + 2 - 2m - 2)/(m + 1))² = 9 
or ((m - 2)/(m + 1))² + ((3m)/(m + 1))² = 9 
or 9 + 9m²/(m + 1)² = 9
or (m + 1)² = 1 + m² + 2m + m² 
∴ 1 + m² = 1 + (m + 1)²
1 + m² = 1 + m²+2m
or 2m = 0
Hence m = 0

Question 16: The hypotenuse of a right-angled triangle has its ends at the points (1, 3) and (– 4, 1). Find an equation of the legs (perpendicular sides) of the triangle which are parallel to the axes.

ANSWER: - Let triangle ABC be a right-angled triangle whose hypotenuse is AB. The coordinates of A and B are (1, 3) and (-4, 1) respectively.
Let the slope of BC be m.
AC is perpendicular to BC.
So, the slope of AC = -1/m.
Equation of line BC:
y - y₁ = m(x - x₁)
y - 1 = m(x + 4)
or mx - y + 4m + 1 = 0 ----(i)
Equation of line AC:
y - 3 = (-1/m) * (x - 1)
or my - 3m = -x + 1
or x + my - 3m - 1 = 0 ----(ii)
The equation of both these lines can be found from the given value of m.
If side BC is parallel to the x-axis, then m = 0.
Equation of BC:
y - 1 = 0
or y = 1.
AC is parallel to the y-axis and it passes through A(1, 3).
So, the equation of AC is x = 1.
Thus, the equations of BC and AC are y = 1 and x = 1.

 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.
Slope of AB = b − 8a − 3, while the slope of line (1) = −13.

Since line (1) is perpendicular to AB,

b − 8a − 3 × −13 = −1

⇒ b − 8 = 3a − 9

⇒ b = 3a − 1 ... (2)

Mid-point of AB = ( a + 32, b + 82 )

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

Hence, from equation (1), we have:

a + 32 + 3 × b + 82 = 7

⇒ a + 3 + 3b + 24 = 14
⇒ a + 3b = -13 ... (3)

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),  

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).

∴ |m₂ − m₁||1 + m₁m₂| = |m₃ − m₂||1 + m₂m₃|

⇒ 3 − m1 + 3m = 1 − 2mm + 2

⇒ 3 − m1 + 3m = ± 1 − 2mm + 2

If 3 − m1 + 3m = 1 − 2mm + 2,

then:

(3 − m)(m + 2) = (1 − 2m)(1 + 3m)

⇒ 3m² + 5m + 6 = 1 + m − 6m²

⇒ 5m² + 5 = 0

⇒ m² + 1 = 0

⇒ m = √−1, which is not real.
Hence the case is not possible.
If 3 − m1 + 3m = 1 − 2mm + 2, then:

(3 − m)(m + 2) = (1 − 2m)(1 + 3m)

⇒ −m² + m + 6 = −(1 + m − 6m²)

⇒ 7m² − 2m − 7 = 0

Using the quadratic formula:

m = 2 ± √[4 − 4(7)(−7)]2(7)

⇒ m = 2 ± √[4 + 49]14

⇒ m = 2 ± 714

⇒ m = 1 ± 5√27

Thus, the required value of m is: 1 ± 5√27.

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
d₁ = |x + y − 5|√(1² + 1²) = |x + y − 5|√2

d₂ = |3x − 2y + 7|√(3² + (−2)²) = |3x − 2y + 7|√13

It is given that d₁ + d₂ = 10.

∴ |x + y − 5|√2 + |3x − 2y + 7|√13 = 10

⇒ √13(x + y − 5) + √2(3x − 2y + 7) − 10√26 = 0

[Assuming (x + y − 5) and (3x − 2y + 7) are positive]

⇒ x(√13 + 3√2) + y(√13 − 2√2) + (7√2 − 5√13 − 10√26) = 0, which is the equation of a line.

Similarly, we can obtain the equation of a line for any signs of (x + y − 5) and (3x − 2y + 7).

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

d₁ = |9h + 6k − 7|√(9² + 6²) = |9h + 6k − 7|√117 = |9h + 6k − 7|3√13

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

d₂ = |3h + 2k + 6|√(3² + 2²) = |3h + 2k + 6|√13

Since P(h, k) is equidistant from lines (1) and (2), d₁ = d₂.

∴ |9h + 6k − 7|3√13 = |3h + 2k + 6|√13

⇒ |9h + 6k − 7| = 3|3h + 2k + 6|

⇒ 9h + 6k − 7 = ±3(3h + 2k + 6)

Case 1: 9h + 6k − 7 = 3(3h + 2k + 6)

⇒ 9h + 6k − 7 = 9h + 6k + 18 (which is absurd).

Case 2: 9h + 6k − 7 = −3(3h + 2k + 6)

⇒ 9h + 6k − 7 = −9h − 6k − 18

∴ 9h + 6k − 7 = −3(3h + 2k + 6).

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

Let ∠CAX = θ

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

= 180° – θ – 180°+ 2θ

= θ

∴∠BAX = 180° – θ

Now, slope of line AC = 3 − 05 − a

⇒ tan θ = 35 − a ... (1)

Slope of line AB = 2 − 01 − a

⇒ tan(180° − θ) = 21 − a

⇒ tan θ = 2a − 1 ... (2)

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

35 − a = 2a − 1

⇒ 3(a − 1) = 10 − 2a

⇒ 3a − 3 = 10 − 2a

⇒ 5a = 13

⇒ a = 135

Thus, the coordinates of point A are: ( 135, 0)

Question 22: Prove that the product of the lengths of the perpendiculars drawn from the points

ANSWER : -  The equation of the given line is

Length of the perpendicular from point   to line (1) is

Length of the perpendicular from point    to line (2) is

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

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:

x = −117 and y = 2217.

Thus, the person is standing at point:  ( −117, 2217).

The person can reach path (3) in the least time 
if they walk along the perpendicular to line (3) from point ( −117, 2217).

Slope of the line (3) = 67

∴ Slope of the line perpendicular to line (3) = −76

The equation of the line passing through ( −117, 2217 ) and having a slope of −76 is given by:

y − 2217 = −76(x + 117)

⇒ 6(17y − 22) = −7(17x + 1)

⇒ 102y − 132 = −119x − 7

⇒ 119x + 102y = 125

Hence, the path that the person should follow is 119x + 102y = 125.