NCERT Solutions for Class 11 Maths Chapter 10

NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines सरल रेखाएँ, DOWNLOAD in PDF format to use it offline. Now UP Board is also following NCERT Books and Current CBSE Syllabus for intermediate students. Visit to Discussion Forum to ask your doubts and reply to your friends.

 Class: 11 Subject: Maths Chapter 10: Straight Lines

NCERT Solutions for Class 11 Maths Chapter 10

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Class 11 Maths Chapter 10 Straight Lines Solutions

Extra 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]

Important Questions with answers on Straight Lines
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]

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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.

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

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.

2 thoughts on “NCERT Solutions for Class 11 Maths Chapter 10”

1. D.K.Shivakumar Swamy says:

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

2. Dharmaraj Ramasubramanian says:

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