NCERT Solutions: Exercise Miscellaneous- Straight Lines

# Exercise Miscellaneous- Straight Lines NCERT Solutions | Mathematics (Maths) Class 11 - Commerce PDF Download

Question 1: Find the values of k for which the line  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 – k2yk2 – 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 – k2y = (k – 3) xk2 – 7k  6 = 0

,

which is of the form ymx  + c.

∴Slope of the given line =

Slope of the x-axis = 0

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  .

Now,   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.

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   is the normal form of the line  .

ANSWER : -  The equation of the given line is  .

This equation can be reduced as

On dividing both sides by  , we obtain

On comparing equation (1) to  , we obtain

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

Thus, the respective values of θ and p are    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

ab = 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

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   is 4 units.

ANSWER : -  Let (0, b) be the point on the y-axis whose distance from line    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 (x1y1) is given by  .

Therefore, if (0, b) is the point on the y-axis whose distance from line    is 4 units, then:

Thus, the required points are   and  .

Question 5: Find the perpendicular distance from the origin to the line joining the points

ANSWER : -  The equation of the line joining the points   is given by

It is known that the perpendicular distance (d) of a line AxByC = 0 from a point (x1y1) is given by  .

Therefore, the perpendicular distance (d) of the given line from point (x1y1) = (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

xa … (1)

The two given lines are

x – 7y + 5 = 0 … (2)

3xy = 0 … (3)

On solving equations (2) and (3), we obtain  .

Therefore,   is the point of intersection of lines (2) and (3).

Since line xa passes through point  ,   .

Thus, the required equation of the line is  .

Question 7: Find the equation of a line drawn perpendicular to the line   through the point, where it meets the y-axis.

ANSWER : -  The equation of the given line is  .

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

, which is of the form ymxc

∴Slope of the given line

∴Slope of line perpendicular to the given line

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

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

The equation of the line that has a slope of   and passes through point (0, 6) is

Thus, the required equation of the line is  .

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

y – x = 0 … (1)

xy = 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

xk and y = –k

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

xk and yk

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

We know that the area of a triangle whose vertices are (x1y1), (x2y2), and (x3y3) is  .

Therefore, area of the triangle formed by the three given 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 are concurrent, then show that

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

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

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).

Hence,

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   , 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

3xy = 7

Case II: m1 =

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

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

On solving equations 4x + 7y – 3 = 0 and 2x – 3y + 1 = 0, we obtain  .

is the point of intersection of the two given lines.

Since equation (1) passes through point  ,

∴ Equation (1) becomes

Thus, the required equation of the line is  .

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 section formula,

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)

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  .

∴Coordinates of point B are  .

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

Thus, the required distance is  .

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

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,

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.

Since line (1) is perpendicular to AB,

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

Hence, from equation (1), we have

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).

Thus, the required value of m is  .

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

It is given that  .

, which is the equation of a line.

Similarly, we can obtain the equation of line for any signs of  .

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

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

Since P (h, k) is equidistant from lines (1) and (2),

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

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

Thus, the coordinates of point A are  .

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  .

Thus, the person is standing at point  .

The person can reach path (3) in the least time if he walks along the perpendicular line to (3) from point  .

∴Slope of the line perpendicular to line (3)

The equation of the line passing through    and having a slope of   is given by

Hence, the path that the person should follow is  .

The document Exercise Miscellaneous- Straight Lines NCERT Solutions | Mathematics (Maths) Class 11 - Commerce is a part of the Commerce Course Mathematics (Maths) Class 11.
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## Mathematics (Maths) Class 11

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## FAQs on Exercise Miscellaneous- Straight Lines NCERT Solutions - Mathematics (Maths) Class 11 - Commerce

 1. How do I solve miscellaneous exercises on straight lines for the JEE exam?
Ans. To solve miscellaneous exercises on straight lines for the JEE exam, you can follow these steps: 1. Read the given problem statement carefully and identify the information provided. 2. Use the equation of a straight line, y = mx + c, to form an equation based on the given conditions. 3. Simplify the equation and bring it to the standard form, Ax + By + C = 0, if required. 4. Solve the equation by applying the appropriate concepts of straight lines, such as finding the slope, distance between points, or intersection of lines. 5. Verify your solution and check if it satisfies all the given conditions in the problem statement. 6. If required, draw a figure or graph to visualize the problem and better understand the solution.
 2. What are the important concepts and formulas to remember while solving straight lines for the JEE exam?
Ans. Some important concepts and formulas to remember while solving straight lines for the JEE exam are: 1. Slope of a line: The slope of a line passing through two points (x1, y1) and (x2, y2) is given by m = (y2 - y1) / (x2 - x1). 2. Point-slope form: The equation of a line passing through a point (x1, y1) with slope m is given by y - y1 = m(x - x1). 3. Slope-intercept form: The equation of a line with slope m and y-intercept c is given by y = mx + c. 4. Distance between two points: The distance between two points (x1, y1) and (x2, y2) is given by d = sqrt((x2 - x1)^2 + (y2 - y1)^2). 5. Parallel and perpendicular lines: Two lines are parallel if their slopes are equal, and they are perpendicular if the product of their slopes is -1. 6. Intersection of lines: The intersection point of two lines can be found by solving their equations simultaneously.
 3. Are there any shortcut techniques or tricks to solve straight line problems quickly for the JEE exam?
Ans. While there are no specific shortcut techniques or tricks to solve straight line problems quickly, you can follow some strategies to improve your speed and accuracy: 1. Practice regularly to enhance your problem-solving skills and speed. 2. Understand and memorize the important concepts and formulas related to straight lines. 3. Break down complex problems into smaller, manageable parts to solve them step by step. 4. Visualize the problem by drawing figures or graphs to gain better insights into the solution. 5. Use logical reasoning and common sense to eliminate incorrect options in multiple-choice questions. 6. Solve previous year JEE question papers and mock tests to familiarize yourself with the exam pattern and time management.
 4. Can I score well in the JEE exam by focusing only on the miscellaneous exercises of straight lines?
Ans. While solving miscellaneous exercises on straight lines is essential for a comprehensive understanding of the topic, it is not sufficient to score well in the JEE exam. The JEE syllabus covers various other subjects and topics, and it is important to have a balanced preparation strategy. You should allocate sufficient time to all subjects, including physics, chemistry, and mathematics, and practice a wide range of problem-solving techniques. Additionally, it is crucial to solve previous year question papers, take mock tests, and seek guidance from experienced teachers or mentors to enhance your overall performance in the JEE exam.
 5. Is it necessary to draw graphs while solving miscellaneous exercises on straight lines for the JEE exam?
Ans. Drawing graphs can be a helpful visual aid while solving miscellaneous exercises on straight lines for the JEE exam, especially when dealing with geometric interpretations or understanding the behavior of lines. However, it is not always necessary to draw graphs. If the problem statement provides sufficient information and you can visualize the scenario mentally, you can proceed with solving the equations and finding the solution algebraically. Drawing graphs can be a personal preference or a technique to enhance understanding, but it is not a mandatory step in every problem-solving process.

## Mathematics (Maths) Class 11

75 videos|238 docs|91 tests

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