# NCERT Solutions for Class 11 Maths Chapter 10

NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines सरल रेखाएँ, DOWNLOAD in PDF file format to use it offline updated for new academic session 2020-2021.

Now UP Board is also following NCERT Books 2020-2021 and Current CBSE Syllabus 2020-2021 for intermediate students. So, download UP Board Solutions for Class 11 Maths Chapter 10 in PDF format. Visit to Discussion Forum to ask your doubts and reply to your friends.## NCERT Solutions for Class 11 Maths Chapter 10

Class: | 11 |

Subject: | Maths |

Chapter 10: | Straight Lines |

### 11th Maths Chapter 10 Solutions

NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines is given below to free download in PDF form. NCERT Solutions and Offline Apps are updated for new session 2020-21 following CBSE Syllabus 2020-2021.

### 11th Maths Chapter 10 Solutions in PDF

#### Important Questions on Straight Lines

1. 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]

2. Determine the equation of line through a point (–4, –3) and parallel to x-axis. [Answer: y + 3 = 0]

3. 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]

4. 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]

5. 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

1. 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)]

2. 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)]

3. 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]

4. In what ratio, the line joining (–1, 1) and (5, 7) is divided by the line x + y = 4? [Answer: 1:2]

5. 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]

###### Questions with Answers for Final Exams

1. 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]

2. Find the area of the triangle formed by the lines y = x, y = 2x, y = 3x + 4. [Answer: 4 square units]

3. 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)]

4. 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]

5. 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

1. 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]

2. 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]

### Important Questions on 11th Maths Chapter 10

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.

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

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.