# NCERT Solutions For Class 11 Maths Chapter 10 Exercise 10.3

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## Chapter 10 Ex.10.3 Question 1

Reduce the following equation into slope-intercept form and find their slopes and the $$y$$-intercepts.

(i) $$x + 7y = 0$$

(ii) $$6x + 3y - 5 = 0$$

(iii) $$y = 0$$

### Solution

(i) The given equation is $$x + 7y = 0$$

It can be written as

$y = - \frac{1}{7}x + 0$

This equation is of the form $$y = mx + c$$, where $$m = - \frac{1}{7}$$ and $$c = 0$$

Therefore, equation $$x + 7y = 0$$ is the slope-intercept form, where the slope and the $$y$$-intercept are $$- \frac{1}{7}$$ and 0 respectively.

(ii) The given equation is $$6x + 3y - 5 = 0$$

It can be written as

\begin{align}y = \frac{1}{3}\left( { - 6x + 5} \right)\\y = - 2x + \frac{5}{3}\end{align}

This equation is of the form $$y = mx + c$$, where $$m = - 2$$ and $$c = \frac{5}{3}$$

Therefore, equation $$6x + 3y - 5 = 0$$ is the slope-intercept form, where the slope and the $$y$$-intercept are $$- 2$$ and $$\frac{5}{3}$$ respectively.

(iii) The given equation is $$y = 0$$.

It can be written as $$y = 0.x + 0$$

This equation is of the form $$y = mx + c$$, where $$m = 0$$ and $$c = 0$$.

Therefore, equation $$y = 0$$ is in the slope-intercept form, where the slope and the $$y$$-intercept are $$0$$ and $$0$$ respectively.

## Chapter 10 Ex.10.3 Question 2

Reduce the following equations into intercept form and find their intercepts on the axis.

(i) $$3x + 2y-12 = 0$$

(ii) $$4x-3y = 6$$

(iii) $$3y + 2 = 0$$

### Solution

(i) The given equation is $$3x + 2y-12 = 0$$

It can be written as

\begin{align}3x + 2y &= 12\\\frac{{3x}}{{12}} + \frac{{2y}}{{12}} &= 1\\\frac{x}{4} + \frac{y}{6} &= 1 \qquad \quad \ldots \left( 1 \right)\end{align}

This equation is of the form $$\frac{x}{a} + \frac{y}{b} = 1$$, where $$a = 4$$ and $$b = 6$$.

Therefore, equation (1) is in the intercept form, where the intercepts on the $$x$$ and $$y$$ axes are $$4$$ and $$6$$ respectively.

(ii) The given equation is $$4x-3y = 6$$.

It can be written as

\begin{align}\frac{{4x}}{6} - \frac{{3y}}{6} &= 1\\\frac{{2x}}{3} - \frac{y}{2} &= 1\\\frac{x}{{\frac{3}{2}}} + \frac{y}{{\left( { - 2} \right)}} &= 1 \qquad \quad \ldots \left( 2 \right)\end{align}

This equation is of the form $$\frac{x}{a} + \frac{y}{b} = 1$$, where $$a = \frac{3}{2}$$ and $$b = - 2$$ .

Therefore, equation (2) is in the intercept form, where the intercepts on $$x$$ and $$y$$-axes are $$\frac{3}{2}$$ and $$- 2$$ respectively.

(iii) The given equation is $$3y + 2 = 0$$

It can be written as

\begin{align}3y &= - 2\\\frac{y}{{\left( { - \frac{2}{3}} \right)}}& = 1 \qquad \quad \ldots \left( 3 \right)\end{align}

This equation is of the form $$\frac{x}{a} + \frac{y}{b} = 1$$, where $$a = 0$$ and $$b = - \frac{2}{3}$$ .

Therefore, equation (3) is in the intercept form, where the intercepts on the $$y$$-axis is $$- \frac{2}{3}$$ and it has no intercept on the $$x$$-axis.

## Chapter 10 Ex.10.3 Question 3

Reduce the following equations into normal form. Find their perpendicular distance from the origin and angle between perpendicular and the positive $$x$$-axis.

(i) $$x - 3\sqrt y + 8 = 0$$

(ii) $$y-2 = 0$$

(iii) $$x-y = 4$$

### Solution

(i) The given equation is $$x - 3\sqrt y + 8 = 0$$

It can be written as

\begin{align}x - 3\sqrt y &= - 8\\- x + 3\sqrt y &= 8\end{align}

On dividing both sides by $$\sqrt {{{( - 1)}^2} + {{\left( {\sqrt 3 } \right)}^2}} = \sqrt 4 = 2$$, we obtain

\begin{align}- \frac{x}{2} + \frac{{\sqrt 3 }}{2}y &= \frac{8}{2}\\\left( { - \frac{1}{2}} \right)x + \left( {\frac{{\sqrt 3 }}{2}} \right)y &= 4\\x\cos 120^\circ + y\sin 120^\circ &= 4 \qquad \quad \ldots \left( 1 \right)\end{align}

Equation (1) is in the normal form.

On comparing equation (1) with the normal form of equation of the line

$$x\cos \omega + y\sin \omega = p$$, we obtain $$\omega = 120^\circ$$ and $$p = 4$$.

Thus, the perpendicular distance of the line from the origin is $$4$$, while the angle between the perpendicular and the positive x-axis is $$120^\circ.$$

(ii) The given equation is $$y-2 = 0$$

It can be represented as $$0.x + 1.y = 2$$

On dividing both sides by $$\sqrt {{0^2} + {1^2}} = 1$$, we obtain

\begin{align}&0.x + 1.y = 2\\& \Rightarrow\; x\cos 90^\circ + y\sin 90^\circ = 2 \qquad \quad \ldots \left( 2 \right) \end{align}

Equation (2) is in the normal form.

On comparing equation (2) with the normal form of equation of line

$$x\cos \omega + y\sin \omega = p$$, we obtain $$\omega = 90^\circ$$ and $$p = 2$$.

Thus, the perpendicular distance of the line from the origin is $$2$$, while the angle between the perpendicular and the positive x-axis is $$90^\circ$$.

(iii) The given equation is $$x-y = 4$$.

It can be reduced as $$1.x + \left( { - 1} \right)y = 4$$

On dividing both sides by $$\sqrt {{1^2} + {{\left( { - 1} \right)}^2}} = \sqrt 2$$, we obtain

\begin{align}\frac{1}{{\sqrt 2 }}x + \left( { - \frac{1}{{\sqrt 2 }}} \right)y &= \frac{4}{{\sqrt 2 }}\\\Rightarrow x\cos \left( {2\pi - \frac{\pi }{4}} \right) + y\sin \left( {2\pi - \frac{\pi }{4}} \right) &= 2\sqrt 2 \\\Rightarrow x\cos 315^\circ + y\sin 315^\circ &= 2\sqrt 2 \qquad \quad \ldots \left( 3 \right)\end{align}

Equation (3) is in the normal form.

On comparing equation (3) with the normal form of the equation of the line

$$x\cos \omega + y\sin \omega = p$$, we obtain $$\omega = 315^\circ$$ and $$p = 2\sqrt 2$$ .

Thus, the perpendicular distance of the line from the origin is $$2\sqrt 2$$, while the angle between the perpendicular and the positive $$x$$-axis is $$315^\circ$$.

## Chapter 10 Ex.10.3 Question 4

Find the distance of the points $$\left( { - 1,1} \right)$$ from the line $$12\left( {x + 6} \right) = 5\left( {y-2} \right)$$.

### Solution

The given equation of the line is $$12\left( {x + 6} \right) = 5\left( {y-2} \right)$$

\begin{align}&\Rightarrow \;12x + 72 = 5y - 10\\&\Rightarrow \;12x - 5y + 82 = 0 \qquad \quad \ldots \left( 1 \right)\end{align}

On comparing equation (1) with general equation of line $$Ax + By + C = 0$$ we obtain $$A = 12,\;B = - 5$$ and $$C = 82$$.

It is known that the perpendicular distance (d) of a line $$Ax + By + C = 0$$ from a point

$$\left( {{x_1},{y_1}} \right)$$ is given by $$d = \frac{{|A{x_1} + B{y_1} + C|}}{{\sqrt {{A^2} + {B^2}} }}$$

The given point is $$\left( {{x_1},{y_1}} \right) = \left( { - 1,1} \right)$$.

Therefore, the distance of point $$\left( { - 1,1} \right)$$ from the given line is

\begin{align}\frac{{|12\left( { - 1} \right) + \left( { - 5} \right)\left( 1 \right) + 82}}{{\sqrt {{{12}^2} + {{\left( { - 5} \right)}^2}} }} &= \frac{{| - 12 - 5 + 82|}}{{\sqrt {169} }}\\&= \frac{{|65|}}{{13}}\\&= 5\end{align}

Hence, the distance of point $$\left( { - 1,1} \right)$$ from the given line is $$5$$ units.

## Chapter 10 Ex.10.3 Question 5

Find the points on the $$x$$-axis whose distance from the line $$\frac{x}{3} + \frac{y}{4} = 1$$ are $$4$$ units.

### Solution

The given equation of line is

\begin{align}\frac{x}{3} + \frac{y}{4} &= 1\\4x + 3y - 12 &= 0 \qquad \quad \ldots \left( 1 \right)\end{align}

On comparing equation (1) with general equation of line $$Ax + By + C = 0$$ we obtain $$A=4$$ , $$B=3$$ , and $$C=-12$$.

Let $$\left( a,0 \right)$$ be the point on the $$x$$-axis whose distance from the given line is $$4$$ units.

It is known that the perpendicular distance (d) of a line $$Ax + By + C = 0$$ from a point

$$\left( {{x_1},{y_1}} \right)$$ is given by $$d = \frac{{|A{x_1} + B{y_1} + C|}}{{\sqrt {{A^2} + {B^2}} }}$$

Therefore,

$$\Rightarrow \left( {4a - 12} \right) = 20$$ or $$- \left( {4a - 12} \right) = 20$$

$$\Rightarrow 4a = 20 + 12$$ or $$4a = - 20 + 12$$

$$\Rightarrow a = 8$$ or $$a = - 2$$

Thus, the required points on $$x$$-axis are $$\left( { - 2,0} \right)$$ and $$\left( {8,0} \right)$$.

## Chapter 10 Ex.10.3 Question 6

Find the distance between parallel lines

(i) $$15x + 8y--34 = 0$$ and $$15x + 8y + 31 = 0$$

(ii) $$l\left( x+y \right)+p=0$$ and $$l\left( {x + y} \right)-r = 0$$

### Solution

It is known that the distance (d) between parall lines $$Ax + By + {C_2} = 0$$ and $$Ax + By + {C_2} = 0$$ is given by $$d = \frac{{\left| {{C_1} - {C_2}} \right|}}{{\sqrt {{A^2} + {B^2}} }}$$

(i) The given parallel lines are $$15x + 8y-34 = 0$$ and $$15x + 8y + 31 = 0$$

Here, $$A = 15,\;B = 8,\;{C_1} = - 34$$ and $${C_2} = - 31$$

Therefore, the distance between the parallel lines is

\begin{align}d &= \frac{{\left| {{C_1} - {C_2}} \right|}}{{\sqrt {{A^2} + {B^2}} }}\\ &= \frac{{\left| {34 - 31} \right|}}{{\sqrt {{{\left( {15} \right)}^2} + {{\left( 8 \right)}^2}} }}\rm{units}\\ &= \frac{{\left| { - 65} \right|}}{{\sqrt {289} }}\rm{units}\\ &= \frac{{65}}{{17}}\rm{units}\end{align}

(ii) The given parallel lines are $$l\left( {x + y} \right) + p = 0$$ and $$l\left( {x + y} \right)-r = 0$$

Here, $$A = B = l,\;{C_1} = p$$ and $${C_2} = - r$$

Therefore, the distance between the parallel lines is

\begin{align}d &= \frac{{|{C_1} - {C_2}|}}{{\sqrt {{A^2} + {B^2}} }}\\&= \frac{{|p + r|}}{{\sqrt {{l^2} + {l^2}} }}units\\&= \frac{{|p + r|}}{{\sqrt {2{l^2}} }}units\\&= \frac{{|p + r|}}{{l\sqrt 2 }}units\\&= \frac{1}{{\sqrt 2 }}\frac{{|p + r|}}{l}units\end{align}

## Chapter 10 Ex.10.3 Question 7

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

### Solution

The equation of the given line is

\begin{align} &3x-4y + 2= 0\\& y= \frac{{3x}}{4} + \frac{2}{4}\end{align}

$$y = \frac{3}{4}x + \frac{2}{4}$$, which is of the form $$y = mx + c$$

Therefore, slope of the given line is $$\frac{3}{4}$$

It is known that parallel lines have the same slope.

Slope of the other line is $$m = \frac{3}{4}$$

Now, the equation of the line that has a slope of $$\frac{3}{4}$$ and passes through the points $$\left( { - 2,3} \right)$$is

\begin{align}&\left( {y - 3} \right) = \frac{3}{4}\left\{ {x - \left( { - 2} \right)} \right\}\\&4y - 12 = 3x + 6\\&3x - 4y + 18 = 0\end{align}

## Chapter 10 Ex.10.3 Question 8

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

### Solution

The given equation of the line is $$x - 7y + 5 = 0$$

Or $$y = \frac{1}{7}x + \frac{5}{7}$$, which is of the form $$y = mx + c$$

Therefore, slope of the given line is $$\frac{1}{7}$$

The slope of the line perpendicular to the line having a slope is $$m = - \frac{1}{{\left( {\frac{1}{7}} \right)}} = - 7$$

The equation of the line with slope $$- 7$$ and $$x$$-intercept $$3$$ is given by

\begin{align}&y = m\left( {x - d} \right)\\&y = - 7\left( {x - 3} \right)\\ &y= - 7x + 21\\ &7x + y - 21 = 0\end{align}

Hence, the required equation of the line is $$7x + y - 21 = 0$$

## Chapter 10 Ex.10.3 Question 9

Find the angles between the lines $$\sqrt 3 x + y = 1$$ and $$x + \sqrt 3 y = 1$$

### Solution

The given lines are $$\sqrt 3 x + y = 1$$ and $$x + \sqrt 3 y = 1$$

$$y = - \sqrt 3 x + 1 \qquad \ldots \left( 1 \right)$$ and $$y = - \frac{1}{{\sqrt 3 }}x + \frac{1}{{\sqrt 3 }} \qquad \ldots \left( 2 \right)$$

The slope of line (1) is $${m_1} = - 3$$, while the slope of the line (2) is $${m_2} = - \frac{1}{{\sqrt 3 }}$$

The actual angle i.e., $${\rm{\theta }}$$ between the two lines is given by

\begin{align}\tan {\rm{\theta }} &= \left| {\frac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}} \right|\\&= \left| {\frac{{ - \sqrt 3 + \frac{1}{{\sqrt 3 }}}}{{1 + \left( { - \sqrt 3 \left( { - \frac{1}{{\sqrt 3 }}} \right)} \right)}}} \right|\\&= \left| {\frac{{\frac{{ - 3 + 1}}{{\sqrt 3 }}}}{{1 + 1}}} \right| = \left| {\frac{{ - 2}}{{2 \times \sqrt 3 }}} \right|\\\tan {\rm{\theta }} &= \frac{1}{{\sqrt 3 }}\\{\rm{\theta }} &= 30^\circ \end{align}

Thus, the angle between the given lines is either $$30^\circ$$ or $$180^\circ - 30^\circ = 150^\circ$$.

## Chapter 10 Ex.10.3 Question 10

The line through the points $$\left( {h,3} \right)$$ and $$\left( {4,1} \right)$$ intersects the line $$7x-9y-19 = 0$$. At right angle. Find the value of $$h.$$

### Solution

The slope of the line passing through points $$\left( {h,3} \right)$$ and $$\left( {4,1} \right)$$ is

${m_1} = \frac{{1 - 3}}{{4 - h}} = \frac{{ - 2}}{{4 - h}}$

The slope of the line $$7x-9y-19 = 0$$ or $$y = \frac{7}{9}x - \frac{{19}}{9}$$ is $${m_2} = \frac{7}{9}$$

It is given that the two lines are perpendicular

Therefore,

\begin{align}&\Rightarrow\; {m_1} \times {m_2} = - 1\\&\Rightarrow \;\frac{{ - 14}}{{36 - 9h}} = - 1\\&\Rightarrow \;14 = 36 - 9h\\&\Rightarrow\; 9h = 36 - 14\\&\Rightarrow\;h = \frac{{22}}{9}\end{align}

Thus, the value of $$h = \frac{{22}}{9}$$.

## Chapter 10 Ex.10.3 Question 11

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

### Solution

The slope of line $$Ax + By + C = 0$$ or $$y = \left( { - \frac{A}{B}} \right)x + \left( { - \frac{C}{B}} \right)$$ is $$m = - \frac{A}{B}$$

It is known that parallel lines have the same slope.

Therefore, slope of the other line $$= m = - \frac{A}{B}$$

The equation of the line passing through point $$({x_1},{y_1})$$and having slope $$m = - \frac{A}{B}$$is

\begin{align}&y - {y_1} = m\left( {x - {x_1}} \right)\\&y - {y_1} = - \frac{A}{B}\left( {x - {x_1}} \right)\\ &B\left( {y - {y_1}} \right) = - A\left( {x - {x_1}} \right)\\&A\left( {x - {x_1}} \right) + B\left( {y - {y_1}} \right) = 0\end{align}

Hence, the line through point $$\left( {{x_1},{y_1}} \right)$$ and parallel to line $$Ax + By + C = 0$$ is $$A\left( {x - {x_1}} \right) + B\left( {y - {y_1}} \right) = 0$$

## Chapter 10 Ex.10.3 Question 12

Two lines passing through the points $$\left( {2,3} \right)$$ intersects each other at an angle of $$60^\circ$$. If slope of one line is $$2$$, find equation of the other line.

### Solution

It is given that the slope of the first line, $${m_1} = 2$$.

Let the slope of the other line be $${m_2}$$.

The angle between the two lines is $$60^\circ$$.

\begin{align} &\tan {\rm{\theta }} = \left| {\frac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}} \right|\\ &\tan 60^\circ = \left| {\frac{{2 - {m_2}}}{{1 + 2{m_2}}}} \right|\\ &\sqrt 3 = \pm \left( {\frac{{2 - {m_2}}}{{1 + 2{m_2}}}} \right) \\&\sqrt 3 \left( {1 + 2{m_2}} \right) = 2 - {m_2}\\&\sqrt 3 + 2\sqrt 3 {m_2} + {m_2} = 2\\&\sqrt 3 + \left( {2\sqrt 3 + 1} \right){m_2} = 2\\&{m_2} = \frac{{2 - \sqrt 3 }}{{\left( {2\sqrt 3 + 1} \right)}} \end{align}     or     \begin{align}&\sqrt 3 = - \left( {\frac{{2 - {m_2}}}{{1 + 2{m_2}}}} \right)\\&\sqrt 3 \left( {1 + 2{m_2}} \right) = - \left( {2 - {m_2}} \right)\\&\sqrt 3 + 2\sqrt 3 {m_2} - {m_2} = - 2\\&\sqrt 3 + \left( {2\sqrt 3 - 1} \right){m_2} = - 2\\&{m_2} = \frac{{ - \left( {2 + \sqrt 3 } \right)}}{{\left( {2\sqrt 3 - 1} \right)}}\end{align}

Case 1:

$${m_2} = \frac{{2 - \sqrt 3 }}{{\left( {2\sqrt 3 + 1} \right)}}$$

The equation of the line passing through the point $$\left( {2,3} \right)$$ and having a slope of $$\frac{{2 - \sqrt 3 }}{{\left( {2\sqrt 3 + 1} \right)}}$$ is

\begin{align}&\left( {y - 3} \right) = \frac{{2 - \sqrt 3 }}{{(2\sqrt 3 + 1)}}\left( {x - 2} \right)\\&\left( {2\sqrt 3 + 1} \right)y - 3\left( {2\sqrt 3 + 1} \right) = \left( {2 - \sqrt 3 } \right)x - 2\left( {2 - \sqrt 3 } \right)\\&\left( {\sqrt 3 - 2} \right)x + \left( {2\sqrt 3 + 1} \right)y = - 1 + 8\sqrt 3 \end{align}

In this case, the equation of the other line is $$\left( {\sqrt 3 - 2} \right)x + \left( {2\sqrt 3 + 1} \right)y = - 1 + 8\sqrt 3$$

Case 2:

$${m_2} = \frac{{ - \left( {2 + \sqrt 3 } \right)}}{{\left( {2\sqrt 3 - 1} \right)}}$$

The equation of the line passing through the point $$\left( {2,3} \right)$$ and having a slope of $$\frac{{ - \left( {2 + \sqrt 3 } \right)}}{{\left( {2\sqrt 3 - 1} \right)}}$$ is

\begin{align}&\left( {y - 3} \right) = \frac{{ - \left( {2 + \sqrt 3 } \right)}}{{\left( {2\sqrt 3 - 1} \right)}}\left( {x - 2} \right)\\&\left( {2\sqrt 3 - 1} \right)y - 3\left( {2\sqrt 3 - 1} \right) = - \left( {2 - \sqrt 3 } \right)x + 2\left( {2 - \sqrt 3 } \right)\\&\left( {2 - \sqrt 3 } \right)x + \left( {2\sqrt 3 - 1} \right)y = 4 - 2\sqrt 3 + 6\sqrt 3 - 3\\&\left( {2 - \sqrt 3 } \right)x + \left( {2\sqrt 3 - 1} \right)y = 1 + 8\sqrt 3 \end{align}

If the case of the equation of the other line is $$\left( {2 - \sqrt 3 } \right)x + \left( {2\sqrt 3 - 1} \right)y = 1 + 8\sqrt 3$$

Thus, the required equation of the other line is $$\left( {\sqrt 3 - 2} \right)x + \left( {2\sqrt 3 + 1} \right)y = - 1 + 8\sqrt 3$$ or $$\left( {2 - \sqrt 3 } \right)x + \left( {2\sqrt 3 - 1} \right)y = 1 + 8\sqrt 3$$.

## Chapter 10 Ex.10.3 Question 13

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

### Solution

The right bisector of a line segment bisects the line segment at $$90^\circ$$.

The end points of the line segment are given as $$A\left( {3,4} \right)$$ and $$B\left( { - 1,2} \right)$$.

Accordingly, mid-point of $$AB = \left( {\frac{{3 - 1}}{2},\frac{{4 + 2}}{0}} \right) = \left( {1,3} \right)$$

Slope of $$AB = \frac{{2 - 4}}{{ - 1 - 3}} = \frac{{ - 2}}{{ - 4}} = \frac{1}{2}$$

Slope of the line perpendicular to $$AB = - \frac{1}{{\left( {\frac{1}{2}} \right)}} = - 2$$

The equation of the line passing through $$\left( {1,3} \right)$$ and having a slope of $$- {\rm{2}}$$ is

\begin{align}&\left( {y - 3} \right) = - 2\left( {x - 1} \right)\\&y - 3 = - 2x + 2\\&2x + y = 5\end{align}

Thus, the required equation of the line is $$2x + y = 5$$.

## Chapter 10 Ex.10.3 Question 14

Find the coordinates of the foot of perpendicular from the points $$\left( { - 1,3} \right)$$ to the line $$3x - 4y - 16 = 0$$.

### Solution

Let $$\left( {a,b} \right)$$ be the coordinates of the foot of the perpendicular from the points $$\left( { - 1,3} \right)$$ to the line $$3x - 4y - 16 = 0$$. Slope of the line joining and $$\left( {a,b} \right)$$, $${m_1} = \frac{{b - 3}}{{a + 1}}$$

Slope of the line $$3x - 4y - 16 = 0$$ or $$y = \frac{3}{4}x - 4,\;{m_2} = \frac{3}{4}$$

Since these two lines are perpendicular, $${m_1} \times {m_2} = - 1$$

Therefore,

\begin{align}&\Rightarrow \;\left( {\frac{{b - 3}}{{a + 1}}} \right) \times \left( {\frac{3}{4}} \right) = - 1\\&\Rightarrow\; \frac{{3b - 9}}{{4a + 4}} = - 1\\&\Rightarrow \;3b - 9 = - 4a - 4\\&\Rightarrow\; 4a + 3b = 5 \qquad \quad \ldots \left( 1 \right)\end{align}

Point $$\left( {a,b} \right)$$ lies on the line $$3x - 4y - 16 = 0$$

Therefore,

$\Rightarrow 3a - 4b = 16 \qquad \quad \ldots \left( 2 \right)$

On solving equations (1) and (2), we obtain

$$a = \frac{{68}}{{25}}$$ and $$b = - \frac{{49}}{{25}}$$

Thus, the required coordinates of the foot of the perpendicular are $$\left( {\frac{{68}}{{25}},\frac{{49}}{{25}}} \right)$$

## Chapter 10 Ex.10.3 Question 15

The perpendicular from the origin to the line $$y = mx + c$$ meets it at the a point $$\left( { - 1,2} \right)$$. Find the values of $$m$$ and $$c.$$

### Solution

The given equation of line is $$y = mx + c$$

It is given that the perpendicular from the origin meets the given line at $$\left( { - 1,2} \right)$$.

Therefore, the line joining the points $$\left( {0,0} \right)$$ and $$\left( { - 1,2} \right)$$ is perpendicular to the given line

Slope of the line joining $$\left( {0,0} \right)$$ and $$\left( { - 1,2} \right)$$ is $$\frac{2}{{ - 1}} = - 2$$

The slope of the given line is $$m$$

Therefore,

$$m \times \left( { - 2} \right) = - 1$$ [The two lines are perpendicular]

$$\Rightarrow m = \frac{1}{2}$$

Since points $$\left( { - 1,2} \right)$$ lies on the given line, it satisfies the equation $$y = mx + c$$

Therefore,

\begin{align}\Rightarrow\; 2 &= m\left( { - 1} \right) + c\\\Rightarrow \;2 &= 2 + \frac{1}{2}\left( { - 1} \right) + c\\\Rightarrow\; c& = 2 + \frac{1}{2} = \frac{5}{2}\end{align}

Thus, the respective values of $$m$$ and $$c$$ are $$\frac{1}{2}$$ and $$\frac{5}{2}$$

## Chapter 10 Ex.10.3 Question 16

If $$p$$ and $$q$$ are the lengths of perpendicular from the origin to the lines $$x\cos \theta - y\sin \theta = k\cos 2\theta$$ and $$x\sec \theta + y\,{\rm{cosec}}\theta = k$$, respectively, prove that $${p^2} + 4{q^2} = {k^2}$$

### Solution

The equations of given lines are

$x\cos \theta - y\sin \theta = k\cos 2\theta \qquad \quad \ldots \left( 1 \right)$

$x\sec \theta + y\cos ec\theta = k \qquad \quad \ldots \left( 2 \right)$

The perpendicular distance $$(d)$$ of a line $$Ax + By + C = 0$$from a point $$\left( {{x_1},{x_2}} \right)$$ is given by

$d = \frac{{\left| {A{x_1} + B{y_1} + C} \right|}}{{\sqrt {{A^2} + {B^2}} }}$

On comparing equation(1) to the general equation of a line i.e., $$Ax + By + C = 0$$, we obtain $$A = \cos \theta ,B = - \sin \theta$$ and $$C = - k\cos 2\theta$$

It is given that $$p$$ is the length of the perpendicular from $$\left( {0,0} \right)$$ to line (1).

Therefore, $$p = \frac{{\left| {A(0) + B(0) + C} \right|}}{{\sqrt {{A^2} + {B^2}} }} = \frac{{\left| C \right|}}{{\sqrt {{A^2} + {B^2}} }} = \frac{{\left| { - k\cos 2\theta } \right|}}{{\sqrt {{{\cos }^2}\theta + {{\sin }^2}\theta } }} = \left| { - k\cos 2\theta } \right|\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 3 \right)$$

On comparing equation (2) to the general equation of line i.e., $$Ax + By + C = 0$$, we obtain $$A = \sec \theta ,B = {\rm{cosec}}\,\theta$$ and $$C = - k$$

It is given that q is the length of the perpendicular from $$\left( {0,0} \right)$$ to line (2)

Therefore, $$p = \frac{{\left| {A\left( 0 \right) + B\left( 0 \right) + C} \right|}}{{\sqrt {{A^2} + {B^2}} }} = \frac{{\left| C \right|}}{{\sqrt {{A^2} + {B^2}} }} = \frac{{\left| { - k} \right|}}{{\sqrt {{{\sec }^2}\theta + \cos e{c^2}\theta } }}\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 4 \right)$$

From (3) and (4), we have

\begin{align}{p^2} + 4{q^2} &= {\left( {\left| { - k\cos 2\theta } \right|} \right)^2} + 4{\left( {\frac{{| - k|}}{{\sqrt {{{\sec }^2}\theta + \cos e{c^2}\theta } }}} \right)^2}\\&= {k^2}{\cos ^2}2\theta + \frac{{4{k^2}}}{{\left( {{{\sec }^2}\theta + \cos e{c^2}\theta } \right)}}\\&= {k^2}{\cos ^2}2\theta + \frac{{4{k^2}}}{{\left( {\frac{1}{{{{\cos }^2}\theta }} + \frac{1}{{{{\sin }^2}\theta }}} \right)}}\\&= {k^2}{\cos ^2}2\theta + \left( {\frac{{4{k^2}}}{{\frac{{{{\sin }^2}\theta + {{\cos }^2}\theta }}{{{{\sin }^2}\theta {{\cos }^2}\theta }}}}} \right)\\&= {k^2}{\cos ^2}2\theta + \left( {\frac{{4{k^2}}}{{\frac{1}{{{{\sin }^2}\theta {{\cos }^2}\theta }}}}} \right)\\&= {k^2}{\cos ^2}2\theta + 4{k^2}{\sin ^2}\theta {\cos ^2}\theta \\&= {k^2}{\cos ^2}2\theta + {k^2}{(2\sin \theta \cos \theta )^2}\\&= {k^2}{\cos ^2}2\theta + {k^2}{\sin ^2}2\theta \\&= {k^2}\left( {{{\cos }^2}2\theta + {{\sin }^2}2\theta } \right)\\&= {k^2}\end{align}

Hence, we proved that $${p^2} + 4{q^2} = {k^2}$$

## Chapter 10 Ex.10.3 Question 17

In the triangle $$ABC$$ with vertices $$A\left( {2,3} \right),{\rm{ }}B\left( {4, - 1} \right)$$ and $$C\left( {1,2} \right)$$, find the equation and length of altitude from the vertex $$A.$$

### Solution

Let $$AD$$ be the altitude of triangle $$ABC$$ from vertex $$A.$$

Accordingly, $$AD \bot BC$$

The equation of the line passing through point $$\left( {2,3} \right)$$ and having a slope of $$1$$ is

\begin{align}&\Rightarrow \;\left( {y - 3} \right) = 1\left( {x - 2} \right)\\&\Rightarrow \;x - y + 1 = 0\\&\Rightarrow\; y - x = 1\end{align}

Therefore, equation of the altitude from vertex $$A = y - x = 1$$

Length of $$AD =$$ Length of the perpendicular from $$A\left( {2,3} \right)$$ to $$BC$$

The equation of $$BC$$ is

\begin{align}&\Rightarrow\; \left( {y + 1} \right) = \frac{{2 + 1}}{{1 - 4}}\left( {x - 4} \right)\\&\Rightarrow\; \left( {y + 1} \right) = - 1\left( {x - 4} \right)\\&\Rightarrow \;y + 1 = - x + 4\\&\Rightarrow\; x + y - 3 = 0 \qquad \quad\ldots \left( 1 \right)\end{align}

The perpendicular distance $$(d)$$ of a line $$Ax + By + C = 0$$ from a point $$\left( {{x_1},{y_1}} \right)$$ is given by

$d = \frac{{\left| {A{x_1} + B{y_1} + C} \right|}}{{\sqrt {{A^2} + {B^2}} }}$

On comparing equation (1) to the general equation of line $$Ax + By + C = 0$$, we obtain $$A = 1,\;B = 1$$ and $$C = 3$$.

Length of $$AD = \frac{{|1 \times 2 + 1 \times 3 - 3|}}{{\sqrt {{1^2} + {1^2}} }} = \frac{2}{{\sqrt 2 }} = \sqrt 2\; units$$

Thus, the equation and length of the altitude from vertex $$A$$ are $$y - x = 1$$ and $$\sqrt 2$$ units.

## Chapter 10 Ex.10.3 Question 18

If $$p$$ is the length of perpendicular from the origin to the line whose intercepts on the $$x$$-axis are $$a$$ and $$b$$, then show that $$\frac{1}{{{p^2}}} = \frac{1}{{{a^2}}} + \frac{1}{{{b^2}}}$$

### Solution

It is known that the equation of a line whose intercepts on the axis $$a$$ and $$b$$ is

\begin{align}\frac{x}{a} + \frac{y}{b} &= 1\\bx + ay &= ab\\bx + ay - ab &= 0 \qquad \quad \ldots \left( 1 \right)\end{align}

The perpendicular distance $$(d)$$ of a line $$Ax + By + C = 0$$ from a point $$\left( {{x_1},{y_1}} \right)$$ is given by

$d = \frac{{\left| {A{x_1} + B{y_1} + C} \right|}}{{\sqrt {{A^2} + {B^2}} }}$

On comparing equation (1) to the general equation of line $$Ax + By + C = 0$$, we obtain $$A = b,{\rm{ }}B = a$$ and $$C = - ab$$.

Therefore, if $$p$$ is the length of the perpendicular from point $$\left( {{x_1},{y_1}} \right) = \left( {0,0} \right)$$ to line (1),

We obtain

\begin{align}p &= \frac{{\left| {A(0) + B(0) - ab} \right|}}{{\sqrt {{a^2} + {b^2}} }}\\&= \frac{{\left| { - ab} \right|}}{{\sqrt {{a^2} + {b^2}} }}\end{align}

On squaring both sides, we obtain

\begin{align}&\Rightarrow \;{p^2} = \frac{{{{\left( { - ab} \right)}^2}}}{{{a^2} + {b^2}}}\\&\Rightarrow\; {p^2}\left( {{a^2} + {b^2}} \right) = {a^2}{b^2}\\&\Rightarrow\; \frac{{{a^2} + {b^2}}}{{{a^2}{b^2}}} = \frac{1}{{{p^2}}}\\&\Rightarrow\; \frac{1}{{{p^2}}} = \frac{1}{{{a^2}}} + \frac{1}{{{b^2}}}\end{align}

Hence, we showed $$\frac{1}{{{p^2}}} = \frac{1}{{{a^2}}} + \frac{1}{{{b^2}}}$$

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