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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