Class 11: The Straight Line – Exercise 23.8 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Note: The equation of a straight line passing through $(x_1, y_1)$ and making an angle $\theta$ with the positive x-axis is given by $\frac{x-x_1}{\cos \theta}$ $+$ $\frac{y-y_1}{\sin \theta}$ $= r$ , where $r$ is the distance of the point $(x, y)$ on the line from the point $(x_1, y_1)$ .

The coordinate of any point on the line at a distance $r$ from the point $( x_1, y_1)$ are $( x_1 \pm r \cos \theta, y_1 \pm r \sin \theta)$ .

Question 1: A line passes through a point $A(1,2)$ and makes an angle of $60^{\circ}$ with the x-axis and intersects the line $x + y =6$ at the point $P$ . Find $AP$ .

Answer:

Given $(x_1, y_1) = A ( 1, 2)$ and $\theta = 60^{\circ}$

The equation passing through $A ( 1, 2)$ and making an angle $60^{\circ}$ is

$\frac{x-1}{\cos 60^{\circ}}$ $=$ $\frac{y-2}{\sin 60^{\circ}}$ $= r$

$\Rightarrow$ $\frac{x-1}{(1/2) }$ $=$ $\frac{y-2}{ (\sqrt{3}/2)}$ $= r$ … … … … … i)

From equation i)

The coordinates of any point $P(x, y)$ on this line are $\Big( 1+$ $\frac{r}{2}$ $, 2+$ $\frac{\sqrt{3}}{2}$ $r \Big)$

Since $P$ lies on $x+y = 6$ , substituting we get

$\Big( 1+$ $\frac{r}{2}$ $\Big) + \Big( 2+$ $\frac{\sqrt{3}}{2}$ $\Big) r = 6$

$\Rightarrow$ $\frac{r}{2}$ $+$ $\frac{\sqrt{3}}{2}$ $r = 3$

$\Rightarrow ( 1 + \sqrt{3})r = 6$

$\Rightarrow r =$ $\frac{6}{\sqrt{3}+1}$ $\times$ $\frac{\sqrt{3}-1}{\sqrt{3}-1}$

$\Rightarrow r = 3(\sqrt{3}-1)$

$\therefore AP = 3(\sqrt{3}-1)$

$\\$

Question 2: If the straight line through the point $P(3,4)$ makes an angle $\pi / 6$ with the x-axis and meets the line $12x + 5y + 10 = 0$ at $Q$ , find the length $PQ$ .

Answer:

Given $(x_1, y_1) = P ( 3,4)$ and $\theta = 30^{\circ}$

The equation passing through $P ( 3,4)$ and making an angle $30^{\circ}$ is

$\frac{x-3}{\cos 30^{\circ}}$ $=$ $\frac{y-4}{\sin 30^{\circ}}$ $= r$

$\Rightarrow$ $\frac{x-3}{(\sqrt{3}/2) }$ $=$ $\frac{y-4}{ (1/2)}$ $= r$ … … … … … i)

From equation i)

The coordinates of any point $Q(x, y)$ on this line are $\Big( 3+$ $\frac{\sqrt{3}r}{2}$ $, 4+$ $\frac{r}{2}$ $\Big)$

Since $Q$ lies on $12x+5y +10=0$ , substituting we get

$12 \Big( 3+$ $\frac{\sqrt{3}r}{2}$ $\Big) + 5\Big( 4+$ $\frac{r}{2}$ $\Big) +10 = 0$

$\Rightarrow 36 + 6\sqrt{3}r +20+$ $\frac{5}{2}$ $r +10 = 0$

$\Rightarrow ( 6\sqrt{3}+$ $\frac{5}{2}$ $)r = -66$

$\Rightarrow r =$ $\frac{-66 \times 2}{12\sqrt{3}+5}$

$\Rightarrow r =$ $\frac{-132}{12\sqrt{3}+5}$

$\therefore PQ = |r| =$ $\frac{132}{12\sqrt{3}+5}$

$\\$

Question 3: A straight line drawn through the point, $A (2, 1)$ making an angle $\pi / 4$ with positive x-axis intersects another line $x + 2y + 1 = 0$ in the point $B$ . Find length $AB$ .

Answer:

Given $(x_1, y_1) = A ( 2,1)$ and $\theta = 45^{\circ}$

The equation passing through $A ( 2,1)$ and making an angle $45^{\circ}$ is

$\frac{x-2}{\cos 45^{\circ}}$ $=$ $\frac{y-1}{\sin 45^{\circ}}$ $= r$

$\Rightarrow$ $\frac{x-2}{(1/\sqrt{2}) }$ $=$ $\frac{y-1}{ (1/\sqrt{2})}$ $= r$ … … … … … i)

From equation i)

The coordinates of any point $B(x, y)$ on this line are $\Big( 2+$ $\frac{r}{\sqrt{2}}$ $, 1+$ $\frac{r}{\sqrt{2}}$ $\Big)$

Since $B$ lies on $x+2y+1=0$ , substituting we get

$\Big( 2+$ $\frac{r}{\sqrt{2}}$ $\Big) + 2 \Big( 1+$ $\frac{r}{\sqrt{2}}$ $\Big) +1 = 0$

$\Rightarrow$ $\frac{3}{\sqrt{2}}$ $r + 5=0$

$\Rightarrow r =$ $\frac{-5\sqrt{2}}{3}$

$\therefore AB = |r| =$ $\frac{5\sqrt{2}}{3}$

$\\$

Question 4: A line drawn through $A (4,-1)$ parallel to the line $3x-4y +1=0$ . Find the coordinates of the two points on this line which are at a distance of $5$ units from $A$ .

Answer:

Given $3x-4y + 1 = 0$

$\Rightarrow y =$ $\frac{3}{4}$ $x -$ $\frac{1}{4}$ $($ comparing with $y = mx +c )$

$\Rightarrow \text{Slope} =$ $\frac{3}{4}$ $\Rightarrow \tan \theta =$ $\frac{3}{4}$ $\Rightarrow \cos \theta =$ $\frac{4}{5}$ and $\sin \theta =$ $\frac{4}{5}$

The equation passing through $A ( 4,-1)$ and and having slope of $\frac{3}{4}$ is

$\frac{x-4}{(4/5)}$ $=$ $\frac{y-(-1)}{(3/5)}$ $= \pm 5$

Considering $\text{+ve}$ sign we get:

$\frac{x-4}{(4/5)}$ $=$ $\frac{y-(-1)}{(3/5)}$ $= 5$

$\Rightarrow x = 4 +$ $\frac{4}{5}$ $\times 5 = 8 \text{ and } y = -1+$ $\frac{3}{5} \times 5 = 2$

Hence the point is $( 8, 2)$

Considering $\text{-ve}$ sign we get:

$\frac{x-4}{(4/5)}$ $=$ $\frac{y-(-1)}{(3/5)}$ $= -5$

$\Rightarrow x = 4 +$ $\frac{4}{5}$ $\times (-5) = 0 \text{ and } y = -1+$ $\frac{3}{5}$ $\times (-5) = -4$

Hence the point is $( 0, -4 )$

$\\$

Question 5: The straight line through $P (x_1, y_1)$ inclined at an angle \theta with the x-axis meets the line $ax + by + c = 0$ in $Q$ . Find the length of $PQ$ .

Answer:

Given $P(x_1, y_1)$ and $\text{angle} = \theta$

The equation passing through $P(x_1, y_1)$ and making an angle $\text{angle} = \theta$ is

$\frac{x-x_1}{\cos \theta}$ $=$ $\frac{y-y_1}{\sin \theta }$ $= r$ … … … … … i)

From equation i)

The coordinates of any point $Q(x, y)$ on this line are $(x_1 + r \cos \theta, y_1+r \sin \theta )$

Since $Q$ lies on $ax+by+c=0$ , substituting we get

$a(x_1 + r \cos \theta)+b(y_1+r \sin \theta)+c=0$

$( a \cos \theta + b \sin \theta) r + (ax_1+by_1 +c) = 0$

$r =$ $\frac{-(ax_1+by_1 +c)}{( a \cos \theta + b \sin \theta)}$

$\therefore PQ = |r| = \Big|$ $\frac{(ax_1+by_1 +c)}{( a \cos \theta + b \sin \theta)}$ $\Big|$

$\\$

Question 6: Find the distance of the point $(2, 3)$ from the line $2x-3y+9=0$ measured along a line making an angle of $45^{\circ}$ with the x-axis.

Answer:

Given $(x_1, y_1) = A ( 2,3)$ and $\theta = 45^{\circ}$

The equation passing through $A ( 2,3)$ and making an angle $45^{\circ}$ is

$\frac{x-2}{\cos 45^{\circ}}$ $=$ $\frac{y-3}{\sin 45^{\circ}}$ $= r$

$\Rightarrow$ $\frac{x-2}{(1/\sqrt{2}) }$ $=$ $\frac{y-3}{ (1/\sqrt{2})}$ $= r$ … … … … … i)

From equation i)

The coordinates of any point $B(x, y)$ on this line are $\Big( 2+$ $\frac{r}{\sqrt{2}}$ $, 3+$ $\frac{r}{\sqrt{2}}$ $\Big)$

Since $B$ lies on $2x-3+9=0$ , substituting we get

$2\Big( 2+$ $\frac{r}{\sqrt{2}}$ $\Big) -3 \Big( 3+$ $\frac{r}{\sqrt{2}}$ $\Big) +9 = 0$

$\Rightarrow -$ $\frac{r}{\sqrt{2}}$ $+ 4 = 0$

$\Rightarrow r = 4\sqrt{2}$

$\therefore AB = 4\sqrt{2}$

$\\$

Question 7: Find the distance of the point $(3,5)$ from the line $2 x + 3 y =14$ measured parallel to a line having slope $1/2$ .

Answer:

Given $\tan \theta =$ $\frac{1}{2}$ $\Rightarrow \sin \theta =$ $\frac{1}{\sqrt{1^1+ 2^2}}$ $=$ $\frac{1}{\sqrt{5}}$ and $\cos \theta =$ $\frac{2}{\sqrt{1^1+ 2^2}}$ $=$ $\frac{2}{\sqrt{5}}$

The equation passing through $A ( 3, 5)$ and and having slope of $\frac{1}{2}$ is

$\frac{x-3}{(2/\sqrt{5})}$ $=$ $\frac{y-5}{(1/\sqrt{5})}$ $= r$ … … … … … i)

From equation i)

The coordinates of any point $B(x, y)$ on this line are $\Big( 3+$ $\frac{2r}{\sqrt{5}}$ $, 5+$ $\frac{r}{\sqrt{5}}$ $\Big)$

Since $P$ lies on $2x+3y=14$ , substituting we get

$2 \Big( 3+$ $\frac{2r}{\sqrt{5}}$ $\Big) + 3 \Big( 5+$ $\frac{r}{\sqrt{5}}$ $\Big) = 14$

$\Rightarrow$ $\frac{7}{\sqrt{5}}$ $+ 21 = 14$

$\Rightarrow r = - 5$

$\therefore AB = |r| = 5$

$\\$

Question 8: Find the distance of the point $(2,5)$ from the line $3x+y +4=0$ measured parallel to a line having slope $3/4$ .

Answer:

Given $\tan \theta =$ $\frac{3}{4}$ $\Rightarrow \sin \theta =$ $\frac{1}{5}$ and $\cos \theta =$ $\frac{4}{5}$

The equation passing through $A ( 2, 5)$ and and having slope of $\frac{3}{4}$ is

$\frac{x-2}{(4/5)}$ $=$ $\frac{y-5}{(3/5)}$ $= r$ … … … … … i)

From equation i)

The coordinates of any point $B(x, y)$ on this line are $\Big( 2+$ $\frac{4r}{5}$ $, 5+$ $\frac{3r}{5}$ $\Big)$

Since $P$ lies on $3x+y+4=0$ , substituting we get

$3 \Big( 2+$ $\frac{4r}{5}$ $\Big) + \Big( 5+$ $\frac{3r}{5}$ $\Big) +4 = 0$

$\Rightarrow 3r+15=0$

$\Rightarrow r = - 5$

$\therefore AB = |r| = 5$

$\\$

Question 9: Find the distance of the point $(3, 5)$ from the line $2x+3y=14$ measured parallel to the line $x -2y=1$ .

Answer:

Given line: $x - 2y = 1 \ \ \ \Rightarrow y =$ $\frac{1}{2}$ $x - 1 \ \ \ \Rightarrow \text{Slope} =$ $\frac{1}{2}$

Therefore $\cos \theta =$ $\frac{2}{\sqrt{5}}$ and $\sin \theta =$ $\frac{1}{\sqrt{5}}$

The equation passing through $A ( 3, 5)$ and and having slope of $\frac{1}{2}$ is

$\frac{x-3}{(2/\sqrt{5})}$ $=$ $\frac{y-5}{(1/\sqrt{5})}$ $= r$ … … … … … i)

From equation i)

The coordinates of any point $B(x, y)$ on this line are $\Big( 3+$ $\frac{2r}{\sqrt{5}}$ $, 5+$ $\frac{r}{\sqrt{5}}$ $\Big)$

Since $B$ lies on $2x+3y=14$ , substituting we get

$2 \Big( 3+$ $\frac{2r}{\sqrt{5}}$ $\Big) + 3\Big( 5+$ $\frac{r}{\sqrt{5}}$ $\Big) =14$

$\Rightarrow$ $\frac{7}{\sqrt{5}}$ $r + 21 = 14$

$\Rightarrow$ $\frac{7}{\sqrt{5}}$ $r = - 7$

$\Rightarrow r = - \sqrt{5}$

$\therefore AB = |r| = \sqrt{5}$

$\\$

Question 10: Find the distance of the point $(2, 5)$ from the line $3x + y + 4 = 0$ measured parallel to the line $3x-4y+8=0$ .

Answer:

Given $3x-4y+8=0 \ \ \Rightarrow 4y = 3x + 8 \ \ \Rightarrow y =$ $\frac{3}{4}$ $x+ 2 \ \ \Rightarrow \text{Slope} =$ $\frac{3}{4}$

Now, $\tan \theta =$ $\frac{3}{4}$ $\Rightarrow \sin \theta =$ $\frac{1}{5}$ and $\cos \theta =$ $\frac{4}{5}$

The equation passing through $A ( 2, 5)$ and and having slope of $\frac{3}{4}$ is

$\frac{x-2}{(4/5)}$ $=$ $\frac{y-5}{(3/5)}$ $= r$ … … … … … i)

From equation i)

The coordinates of any point $B(x, y)$ on this line are $\Big( 2+$ $\frac{4r}{5}$ $, 5+$ $\frac{3r}{5}$ $\Big)$

Since $P$ lies on $3x+y+4=0$ , substituting we get

$3 \Big( 2+$ $\frac{4r}{5}$ $\Big) + \Big( 5+$ $\frac{3r}{5}$ $\Big) +4 = 0$

$\Rightarrow 3r+15=0$

$\Rightarrow r = - 5$

$\therefore AB = |r| = 5$

$\\$

Question 11: Find the distance of the line $2x + y= 3$ from the point $(- 1, - 3)$ in the direction of the line whose slope is $1$ .

Answer:

Given $\tan \theta = 1$ $\Rightarrow \sin \theta =$ $\frac{1}{\sqrt{2}}$ $\Rightarrow \cos \theta =$ $\frac{1}{\sqrt{2}}$

The equation passing through $A ( -1, -3)$ and and having slope of $1$ is

$\frac{x-(-1)}{(1/\sqrt{2})}$ $=$ $\frac{y-(-3)}{(1/\sqrt{2})}$ $= r$ … … … … … i)

From equation i)

The coordinates of any point $B(x, y)$ on this line are $\Big( -1+$ $\frac{r}{\sqrt{2}}$ $, -3+$ $\frac{r}{\sqrt{2}}$ $\Big)$

Since $P$ lies on $2x+y=3$ , substituting we get

$2 \Big( -1+$ $\frac{r}{\sqrt{2}}$ $\Big) + \Big( -3+$ $\frac{r}{\sqrt{2}}$ $\Big) = 3$

$\Rightarrow$ $\frac{3}{\sqrt{2}}$ $r -5 = 3$

$\Rightarrow r =$ $\frac{8\sqrt{2}}{3}$

$\therefore AB = |r| =$ $\frac{8\sqrt{2}}{3}$

$\\$

Question 12: A line is such that its segment between the straight line $5x -y-4=0$ and $3x+4y -4=0$ is bisected at the point $(1,5)$ . Obtain its equation.

Answer:

Let $P_1 P_2$ be the intercept between lines $5x -y-4=0$ and $3x+4y -4=0$

Let $P_1 P_2$ makes an angle $\theta$ with positive x-axis.

Given $(x_1, y_1) = A ( 1, 5)$ and $\theta = \theta$

The equation passing through $A ( 1, 5)$ and making an angle $\theta$ is

$\frac{x-1}{\cos \theta}$ $=$ $\frac{y-5}{\sin \theta}$ $= r$

$\Rightarrow \tan \theta =$ $\frac{y-5}{x-1}$

Let $AP_1 = AP_2 = r$

Therefore the coordinate of $P_1 = ( 1+ r \cos \theta, 5 + r \cos \theta)$ and coordinate of $P_2 = ( 1- r \cos \theta, 5 - r \cos \theta)$

We know $P_1$ lies on $5x - y - 4 = 0$

$\therefore 5 (1+ r \cos \theta) - (5 + r \cos \theta) - 4 = 0$

$\Rightarrow r =$ $\frac{4}{5\cos \theta - \sin \theta}$

Also $P_2$ lies on $3x+4y-4=0$

$\therefore 3 (1+ r \cos \theta) +4 (5 + r \cos \theta) - 4 = 0$

$\Rightarrow r =$ $\frac{19}{3\cos \theta +4 \sin \theta}$

Since $A(1, 5)$ bisects $P_1$ and $P_2$

$\frac{4}{5\cos \theta - \sin \theta}$ $=$ $\frac{19}{3\cos \theta +4 \sin \theta}$

$\Rightarrow 12 \cos \theta + 16 \sin \theta = 45 \cos \theta - 19 \sin \theta$

$\Rightarrow 35 \sin \theta= 83 \cos \theta$

$\Rightarrow \tan \theta =$ $\frac{83}{85}$

Therefore the equation of the line would be

$\frac{y-5}{x-1}$ $=$ $\frac{83}{85}$

$\Rightarrow 35 y - 175 = 83x - 83$

$\Rightarrow 83x - 35y +92 = 0$

$\\$

Question 13: Find the equation of straight line passing through $(-2, -7)$ and having an intercept of length $3$ between the straight lines $4x + 3y = 12$ and $4x + 3y = 3$ .

Answer:

Let $P_1 P_2$ be the intercept between lines $4x + 3y = 12$ and $4x + 3y = 3$

Let $P_1 P_2$ makes an angle $\theta$ with positive x-axis.

Given $(x_1, y_1) = A ( -2,-7)$ and $\theta = \theta$

The equation passing through $A ( -2,-7)$ and making an angle $\theta$ is

$\frac{x-(-2)}{\cos \theta}$ $=$ $\frac{y-(-7)}{\sin \theta}$

$\Rightarrow \tan \theta =$ $\frac{y+7}{x+2}$

Let $AP_1 =r_1 \text{ and } AP_2 = r_2$

Therefore the coordinate of $P_1 = ( -2+ r_1 \cos \theta, -7 + r_1 \cos \theta)$ and coordinate of $P_2 = ( -2+ r_2 \cos \theta, -7 + r_2 \cos \theta)$