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Class: | 11 |

Subject: | Maths |

Chapter 10: | Straight Lines |

Table of Contents

- 1 NCERT Solutions for Class 11 Maths Chapter 10
- 1.1 Class 11 Maths Chapter 10 Straight Lines Solutions
- 1.1.1 Extra Questions on Straight Lines
- 1.1.2 Write the equations for the x and y-axes.
- 1.1.3 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.
- 1.1.4 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.

- 1.1 Class 11 Maths Chapter 10 Straight Lines Solutions

## NCERT Solutions for Class 11 Maths Chapter 10

Go Back to Class 11 Maths Main Page

### Class 11 Maths Chapter 10 Straight Lines Solutions

- Download Exercise 10.1
- Download Exercise 10.2
- Download Exercise 10.3
- Download Miscellaneous Exercise 10
- NCERT Books for Class 11
- Revision Books for Class 11
- Hindi Medium Solutions will be uploaded very soon.

#### Extra Questions on Straight Lines

- On shifting the origin to (p, q), the coordinates of point (2, –1) changes to (5, 2). Find p and q. [Answer: p = -3, q = – 3]
- Determine the equation of line through a point (–4, –3) and parallel to x-axis. [Answer: y + 3 = 0]
- If the image of the point (3, 8) in the line px + 3y – 7 = 0 is the point (–1, –4), then find the value of p. [Answer: p = 1]
- Find the distance of the point (3, 2) from the straight line whose slope is 5 and is passing through the point of intersection of lines x + 2y = 5 and x – 3y + 5 = 0. [Answer: 10/√26]
- The line 2x – 3y = 4 is the perpendicular bisector of the line segment AB. If coordinates of A are (–3, 1) find coordinates of B. [Answer: (1, -5)]

###### Questions for Practice

- If a vertex of a triangle is (1, 1) and the midpoints of two sides through this vertex are (–1, 2) and (3, 2). Then find the centroid of the triangle. [Answer: (1, 7/3)]
- The points (1, 3) and (5, 1) are two opposite vertices of a rectangle. The other two vertices lie on line y = 2x + c. Find c and remaining two vertices. [Answer: c = – 4, (2, 0), (4, 4)]
- If two sides of a square are along 5x – 12y + 26 = 0 and 5x – 12y – 65 = 0 then find its area. [Answer: 49 square units]
- In what ratio, the line joining (–1, 1) and (5, 7) is divided by the line x + y = 4? [Answer: 1:2]
- Find the equation of a line with slope –1 and whose perpendicular distance from the origin is equal to 5. [Answer: x + y + 5√2 = 0 and x + y – 5√2 = 0]

##### Important Questions with answers on Straight Lines

- If a vertex of a square is at (1, –1) and one of its side lie along the line 3x – 4y – 17 = 0 then find the area of the square. [Answer: 4 square units]
- Find the area of the triangle formed by the lines y = x, y = 2x, y = 3x + 4. [Answer: 4 square units]
- Find the coordinates of the orthocentre of a triangle whose vertices are (–1, 3) (2, –1) and (0, 0). [Orthocentre is the point of concurrency of three altitudes]. [Answer: (-4, -3)]
- What is the value of y so that line through (3, y) and (2, 7) is parallel to the line through (–1, 4) and (0, 6)? [Answer: y = 9]
- Find the equation of the lines which cut-off intercepts on the axes whose sum and product are 1 and –6 respectively. [Answer: 2x – 3y – 6 = 0 and – 3x + 2y – 6 = 0]

###### Try These

- Find the equation of a straight line which passes through the point of intersection of 3x + 4y – 1 = 0 and 2x – 5y + 7 = 0 and which is perpendicular to 4x – 2y + 7 = 0. [Answer: x + 2y = 1]
- If the image of the point (2, 1) in a line is (4, 3) then find the equation of line. [Answer: x + y – 5 = 0]

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Table of Contents

#### Write the equations for the x and y-axes.

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.

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.

#### 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.

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

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

#### 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.

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.

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.

A very good website but there is no solutions after 9 chapter we cant copy it we have to first download it then we have to save it we cant copy it

Quite useful. I just wanted a clarifcation regarding Question 4 of Miscellaneous Exercise on Chapter 10. The y intercept in this case is (0,4). So points (0,8) and (0,0) would also be at a distance of 4 units from the line. Is this answer wrong? Request your revert.

Thanks