# NCERT Solutions For Class 11 Maths Chapter 10 Exercise 10.3

Sets

Ex.10.1

## 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**

**Video 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**

**Video 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**

**Video 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**

**Video 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**

**Video 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**

**Video 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**

**Video 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**

**Video 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**

**Video 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**

**Video 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**

**Video 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**

**Video 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**

**Video 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**

**Video 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**

**Video 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**

**Video 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**

**Video 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**

**Video 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 \(\frac{1}{{{p^2}}} = \frac{1}{{{a^2}}} + \frac{1}{{{b^2}}}\)