NCERT Solutions
Class 11 - Mathematics - Chapter 9: STRAIGHT LINES
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) \(y = -\dfrac{1}{7}x + 0,\; -\dfrac{1}{7},\; 0\)

(ii) \(y = -2x + \dfrac{5}{3},\; -2,\; \dfrac{5}{3}\)

(iii) \(y = 0x + 0,\; 0,\; 0\)

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) \(\dfrac{x}{4} + \dfrac{y}{6} = 1,\; 4,\; 6\)

(ii) \(\dfrac{x}{3} + \dfrac{y}{-2} = 1,\; \dfrac{3}{2},\; -2\)

(iii) \(y = -\dfrac{2}{3},\; \) intercept with \(y\)-axis \(= -\dfrac{2}{3}\) and no intercept with \(x\)-axis

Question. 3

Find the distance of the point \((-1, 1)\) from the line \(12(x + 6) = 5(y - 2)\).

Answer:

5 units

Question. 4

Find the points on the \(x\)-axis whose distances from the line \(\dfrac{x}{3} + \dfrac{y}{4} = 1\) are 4 units.

Answer:

\((-2, 0)\) 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:

(i) \(\dfrac{65}{17}\) units

(ii) \(\dfrac{1}{\sqrt{2}} \left| \dfrac{p + r}{l} \right|\) units

Question. 6

Find equation of the line parallel to the line \(3x - 4y + 2 = 0\) and passing through the point \((-2, 3)\).

Answer:

\(3x - 4y + 18 = 0\)

Question. 7

Find equation of the line perpendicular to the line \(x - 7y + 5 = 0\) and having \(x\)-intercept 3.

Answer:

\(y + 7x = 21\)

Question. 8

Find angles between the lines \(\sqrt{3}x + y = 1\) and \(x + \sqrt{3}y = 1\).

Answer:

\(30^\circ\) and \(150^\circ\)

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:

\(\dfrac{22}{9}\)

Question. 10

Prove that the line through the point \((x_1, y_1)\) and parallel to the line \(Ax + By + C = 0\) is \(A(x - x_1) + B(y - y_1) = 0\).

Question. 11

Two lines passing through the point \((2, 3)\) intersect each other at an angle of \(60^\circ\). If slope of one line is 2, find equation of the other line.

Answer:

\((\sqrt{3} + 2)x + (2\sqrt{3} - 1)y = 8\sqrt{3} + 1\)

or

\((\sqrt{3} - 2)x + (1 + 2\sqrt{3})y = -1 + 8\sqrt{3}\)

Question. 12

Find the equation of the right bisector of the line segment joining the points \((3, 4)\) and \((-1, 2)\).

Answer:

\(2x + y = 5\)

Question. 13

Find the coordinates of the foot of perpendicular from the point \((-1, 3)\) to the line \(3x - 4y - 16 = 0\).

Answer:

\(\left(\dfrac{68}{25}, -\dfrac{49}{25}\right)\)

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:

\(m = \dfrac{1}{2},\; c = \dfrac{5}{2}\)

Question. 15

If \(p\) and \(q\) are the lengths of perpendiculars from the origin to the lines \(x \cos \theta - y \sin \theta = k \cos 2\theta\) and \(x \sec \theta + y \csc \theta = k\), respectively, prove that \(p^2 + 4q^2 = k^2\).

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:

\(y - x = 1,\; \sqrt{2}\)

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

\(\displaystyle \dfrac{1}{p^2} = \dfrac{1}{a^2} + \dfrac{1}{b^2}\).

NCERT Solutions Class 11 – Mathematics – Chapter 9: STRAIGHT LINES – EXERCISE 9.3 | Detailed Answers