# Miscellaneous Exercise Straight Lines - NCERT Class 11

## Chapter 10 Ex.10.ME Question 1

Find the value of $$k$$ for which the line $$\left( {k - 3} \right)x - \left( {4 - {k^2}} \right)y + {k^2} - 7k + 6 = 0$$ is

(a) Parallel to $$x$$-axis.

(b) Parallel to $$y$$-axis.

(c) Passing through the origin.

### Solution

The given equation of the line is

$\left( {k - 3} \right)x - \left( {4 - {k^2}} \right)y + {k^2} - 7k + 6 = 0 \quad \qquad \ldots \left( 1 \right)$

(a) If the given line is parallel to the $$x$$-axis then,

Slope of the given line $$=$$ Slope of the x-axis

Then given line can be written as

\begin{align}&\left( {k - 3} \right)x + {k^2} - 7k + 6 = \left( {4 - {k^2}} \right)y\\&y = \frac{{\left( {k - 3} \right)}}{{\left( {4 - {k^2}} \right)}}x + \frac{{{k^2} - 7k + 6}}{{\left( {4 - {k^2}} \right)}}\end{align}

Which is of the form $$y = mx + c$$

Slope of the given line $$= \frac{{\left( {k - 3} \right)}}{{\left( {4 - {k^2}} \right)}}$$

Slope of the $$x$$-axis = 0

\begin{align}&\Rightarrow\; \frac{{k - 4}}{{\left( {4 - {k^2}} \right)}} = 0\\&\Rightarrow \; k - 3 = 0\\&\Rightarrow \; k = 3\end{align}

Thus, the given line is parallel to $$x$$-axis, then the value of $$k = 3$$.

(b) If the given line is parallel to the $$y$$-axis, it is vertical.

Hence, its slope will be undefined.

The slope of the given line is $$= \frac{{\left( {k - 3} \right)}}{{\left( {4 - {k^2}} \right)}}$$

Now, $$\frac{{\left( {k - 3} \right)}}{{\left( {4 - {k^2}} \right)}}$$is defined at $${k^2} = 4$$

\begin{align}&\Rightarrow \; {k^2} = 4\\&\Rightarrow \; k = \pm 2\end{align}

Thus, if the given line is parallel to the $$y$$-axis, then the value of $$k = \pm 2$$.

(c) if the given line is passing through the origin, then point $$\left( {0,0} \right)$$ satisfies the given equation of the line.

\begin{align}&\left( {k - 3} \right)\left( 0 \right) - \left( {4 - {k^2}} \right)\left( 0 \right) + {k^2} - 7k + 6 = 0\\&{k^2} - 7k + 6 = 0\\&{k^2} - 6k - k + 6 = 0\\&\left( {k - 6} \right)\left( {k - 1} \right) = 0\end{align}

$$\Rightarrow k = 6$$ or $$k = 1$$

Thus, if the given line is passing through the origin, then the values of $$k$$ is either $$1$$ or 6.

## Chapter 10 Ex.10.ME Question 2

Find the values of $$\theta$$ and $$p$$, if the equation $$x\cos \theta + y\sin \theta = p$$ is the normal form of the line$$\sqrt 3 x + y + 2 = 0$$

### Solution

The equation of the given line is $$\sqrt 3 x + y + 2 = 0$$

This equation can be reduced as

\begin{align}&\Rightarrow\; \sqrt 3 x + y + 2 = 0\\&\Rightarrow\; - \sqrt 3 x - y = 2\end{align}

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

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

On comparing equation ($$1$$) to $$x\cos \theta + y\sin \theta = p$$, we obtain

$$\cos \theta = - \frac{{\sqrt 3 }}{2},\sin \theta = - \frac{1}{2}$$ and $$p = 1$$

Since the value of $$\sin \theta$$and $$\cos \theta$$ are negative $$\theta = \pi + \frac{\pi }{6} = \frac{{7\pi }}{6}$$

Thus, the respective values of $${\rm{\theta }}$$ and p are $$\frac{{7\pi }}{6}$$ and 1.

## Chapter 10 Ex.10.ME Question 3

Find the equation of the line, which cut-off intercepts on the axis whose sum and product are $$1$$ and $$- {\rm Let the intercepts cut by the given lines on the axis be \(a$$ and $$b$$.

It is given that

\begin{align}a + b = 1 \quad \qquad \ldots \left( 1 \right)\\ab = - 6 \quad \qquad \ldots \left( 2 \right)\end{align}

On solving equation ($$1$$) and ($$2$$), we obtain

$$a = 3$$ and $$b = - 2$$ or $$a = - 3$$ and $$b = 3$$

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

$$\frac{x}{a} + \frac{y}{b} = 1$$ or $$bx + ay - ab = 0$$

Case I: $$a = 3$$ and $$b = - 2$$

In this case, the equation of the line is $$- 2x + 3y + 6 = 0 \Rightarrow 2x - 3y = 6$$

Case II: $$a = - 3$$ and $$b = 3$$

In this case, the equation of the line is $$3x - 2y + 6 = 0 \Rightarrow - 3x + 2y = 6$$

Thus, the required equations of the lines are $$\frac{x}{3} + \frac{y}{4} = 12x - 3y = 6$$ and $$- 3x + 2y = 6$$

}\) respectively.

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## Chapter 10 Ex.10.ME Question 4

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

### Solution

Let $$\left( {0,b} \right)$$ be the point on $$y$$-axis whose distance from line $$\frac{x}{3} + \frac{y}{4} = 1$$ is $$4$$ units.

The given line can be written as

$4x + 3y - 12 = 0 \qquad \ldots \left( 1 \right)$

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

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

Therefore, if $$\left( {0,b} \right)$$ is the point on the $$y-$$axis whose distance from line $$\frac{x}{3} + \frac{y}{4} = 1$$ is 4 units,

Then,

\begin{align}&\Rightarrow\; 4 = \frac{{\left| {4\left( 0 \right) + 3\left( b \right) - 12} \right|}}{{\sqrt {{4^2} + {3^2}} }}\\&\Rightarrow \;4 = \frac{{\left| {3b - 12} \right|}}{5}\\&\Rightarrow \;20 = \left| {3b - 12} \right|\\&\Rightarrow \; 20 = \pm \left( {3b - 12} \right)\end{align}

$$\Rightarrow \left( {3b - 12} \right) = 20$$ or $$\left( {3b - 12} \right) = - 20$$

$$\Rightarrow 3b = 20 + 12$$ or $$3b = - 20 + 12$$

$$\Rightarrow b = \frac{{32}}{3}$$ or $$b = \frac{{ - 8}}{3}$$

Thus, the required points are $$\left( {0,\frac{{32}}{3}} \right)$$ and $$\left( {0, - \frac{8}{3}} \right)$$

## Chapter 10 Ex.10.ME Question 5

Find the perpendicular distance from the origin to the line joining the points $$\left( {\cos {\rm{\theta }},\sin {\rm{\theta }}} \right)$$ and $$\left( {\cos \phi ,\sin \phi } \right)$$.

### Solution

The equation of the line joining the points $$(\cos \theta ,\sin \theta )$$ and $$(\cos \phi ,\sin \phi )$$is given by

\begin{align}\frac{{y - \sin \theta }}{{x - \cos \theta }}& = \frac{{\sin \phi - \sin \theta }}{{\cos \phi - \cos \theta }}\\y - \sin \theta &= \frac{{\sin \phi - \sin \theta }}{{\cos \phi - \cos \theta }}(x - \cos \theta )\end{align}

\begin{align} &y\left( {\cos \phi - \cos \theta } \right) - \sin \theta \left( {\cos \phi - \cos \theta } \right)= x\left( {\sin \phi - \sin \theta } \right) - \cos \theta \left( {\sin \phi - \sin \theta } \right)\\&x\left( {\sin \phi - \sin \theta } \right) + y\left( {\cos \phi - \cos \theta } \right) + \cos \theta \sin \phi - \cos \theta \sin \theta - \sin \theta \cos \phi + \sin \theta \cos \theta = 0\\&x\left( {\sin \phi - \sin \theta } \right) + y\left( {\cos \phi - \cos \theta } \right) + \sin \left( {\phi - \theta } \right) = 0\end{align}

$$Ax + By + C = 0$$, where $$A = \sin \theta - \sin \phi ,B = \cos \phi - \cos \theta$$ and $$C = \sin (\phi - \theta )$$ 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 $$d = \frac{{\left| {A{x_1} + B{y_1} + C} \right|}}{{\sqrt {{A^2} + {B^2}} }}$$

Therefore, the perpendicular distance (d) of the given line from point $$\left( {{x_1},{y_1}} \right) = \left( {0,0} \right)$$ is

\begin{align}d &= \frac{{|\left( {\sin \phi - \sin \theta } \right)(0) + \left( {\cos \phi - \cos \theta } \right)(0) + \sin \left( {\phi - \theta } \right)|}}{{\sqrt {{{\left( {\sin \phi - \sin \theta } \right)}^2} + {{\left( {\cos \phi - \cos \theta } \right)}^2}} }}\\&= \frac{{\left| {\sin \left( {\phi - \theta } \right)} \right|}}{{\sqrt {{{\sin }^2}\phi + {{\sin }^2}\theta - 2\sin \theta \sin \phi + {{\cos }^2}\phi + {{\cos }^2}\theta - 2\cos \phi \cos \theta } }}\\&= \frac{{\left| {\sin \left( {\phi - \theta } \right)} \right|}}{{\sqrt {({{\sin }^2}\phi + {{\cos }^2}\phi ) + ({{\sin }^2}\theta + {{\cos }^2}\theta ) - 2(\sin \theta \sin \phi + \cos \phi \cos \theta } )}}\\&= \frac{{\left| {\sin \left( {\phi - \theta } \right)} \right|}}{{\sqrt {1 + 1 - 2(\cos (\phi - \theta )} )}}\\&= \frac{{\left| {\sin \left( {\phi - \theta } \right)} \right|}}{{\sqrt {2\left( {1 - \cos (\phi - \theta )} \right)} }}\\&= \frac{{\left| {\sin \left( {\phi - \theta } \right)} \right|}}{{\sqrt {2\left( {2{{\sin }^2}\left( {\frac{{\phi - \theta }}{2}} \right)} \right)} }}\\&= \frac{{\left| {\sin \left( {\phi - \theta } \right)} \right|}}{{\left| {2\sin \left( {\frac{{\phi - \theta }}{2}} \right)} \right|}}\end{align}

Where we know that $$\sin {\rm{\theta }} = 2\left( {\cos \frac{{\rm{\theta }}}{2}\sin \frac{{\rm{\theta }}}{2}} \right)$$

Then,

$\frac{{\left| {2\cos \left( {\frac{{\phi - \theta }}{2}} \right)\sin \left( {\frac{{\phi - \theta }}{2}} \right)} \right|}}{{\left| {2\sin \left( {\frac{{\phi - \theta }}{2}} \right)} \right|}} = \left| {\cos \left( {\frac{{\phi - \theta }}{2}} \right)} \right|$

## Chapter 10 Ex.10.ME Question 6

Find the equation of the line parallel to $$y$$-axis and draw through the point of intersection of the lines $$x - 7y + 5 = 0$$ and $$3x + y = 0$$

### Solution

The equation of any line parallel to the $$y-$$axis is of the form

$x = a \quad \qquad \ldots \left( 1 \right)$

The two given lines are

\begin{align}x - 7y + 5 &= 0 \quad \qquad \ldots \left( 2 \right)\\3x + y &= 0 \quad \qquad \ldots \left( 3 \right)\end{align}

On solving equation ($$2$$) and ($$3$$), we obtain $$x = - \frac{5}{{22}}$$ and $$x = - \frac{{15}}{{22}}$$

Therefore, $$\left( { - \frac{5}{{22}}, - \frac{{15}}{{22}}} \right)$$ is the point of intersection of lines ($$2$$) and ($$3$$).

Since, line $$x = a$$ passes through point $$\left( { - \frac{5}{{22}}, - \frac{{15}}{{22}}} \right)$$, $$a = - \frac{5}{{22}}$$

Thus, the required equation of the line is $$x = - \frac{5}{{22}}$$.

## Chapter 10 Ex.10.ME Question 7

Find the equation of a line drawn perpendicular to the line $$\frac{x}{4} + \frac{y} The equation of the given line is \(\frac{x}{4} + \frac{y}{6} = 1$$

This equation can also be written as $$3x + 2y - 12 = 0$$

$$y = - \frac{3}{2}x + 6$$, which is of the form $$y = mx + c$$

Hence, the slope of the given line is $$- \frac{3}{2}$$

Slope of the line perpendicular to the given line is $$- \frac{1}{{\left( { - \frac{3}{2}} \right)}} = \frac{2}{3}$$

Let the given line intersect the y-axis at $$\left( {0,y} \right)$$.

On substituting $$x = 0$$ in the equation of the given line, we obtain $$\frac{y}{6} = 1 \Rightarrow y = 6$$

The given line intersects the $$y-$$axis at $$\left( {0,6} \right)$$.

The equation of the line that has a slope of $$\frac{2}{3}$$ and passes through point $$\left( {0,6} \right)$$ is

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

Thus, the required equation of the line is $$2x - 3y + 18 = 0$$.

= 1\) through the point, where it meets the $$y-$$axis.

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## Chapter 10 Ex.10.ME Question 8

Find the area of the triangle formed by the line $$y - x = 0,\;x + y = 0$$ and $$x - k = 0$$.

### Solution

The equations of the given lines are:

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

The point of interaction of lines ($$1$$) and ($$2$$) is given by

$$x = 0$$ and $$y = 0$$.

The point of interaction of lines ($$2$$) and ($$3$$) is given by

$$x = k$$ and $$y = - k$$.

The point of interaction of lines ($$3$$) and ($$1$$) is given by

$$x = k$$ and $$y = k$$.

Thus, the vertices of the triangle formed by the three given lines are $$\left( {0,0} \right),\left( {k, - k} \right)$$ and$$\left( {k, - k} \right)$$.

We know that the area of a triangle whose vertices are $$\left( {{x_1},{y_1}} \right),\;\left( {{x_2},{y_2}} \right)$$ and $$\left( {{x_3},{y_3}} \right)$$ is

$\frac{1}{2}\left| {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right|$

Therefore, area of the triangle formed by three given lines

\begin{align}&= \frac{1}{2}\left| {0\left( { - k - k} \right) + k\left( {k - 0} \right) + k\left( {0 + k} \right)} \right|\\&= \frac{1}{2}\left| {{k^2} + {k^2}} \right|\\&= \frac{1}{2}2{k^2}\\&= {k^2}\end{align}

Hence, the area of the triangle is $${k^2}$$ square units.

## Chapter 10 Ex.10.ME Question 9

​​Find the value of p so that the three lines $$3x + y - 2 = 0,\;px + 2y - 3 = 0$$ and $$2x - y - 3 = 0$$ may intersect at one point.

### Solution

The equation of the given line are

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

On solving equations ($$1$$) and ($$3$$), we obtain

$$x = 1$$ and $$y = - 1$$

Since these three lines may intersect at one point, the point of intersection of lines ($$1$$) and ($$3$$) will also satisfy line ($$2$$)

\begin{align}p\left( 1 \right) + 2\left( { - 1} \right) - 3 &= 0\\p - 2 - 3 &= 0\\p &= 5\end{align}

Thus, the required value of $$p = 5$$.

## Chapter 10 Ex.10.ME Question 10

If three lines whose equations are $${y_1} = {m_1}x + {c_1},\;y = {m_2}x + {c_2}$$ and $$y = {m_3}x + {c_3}$$ are concurrent, then show that $${m_1}\left( {{c_2} - {c_3}} \right) + {m_2}\left( {{c_3} - {c_1}} \right) + {m_3}\left( {{c_1} - {c_2}} \right) = 0$$.

### Solution

The equations of the given lines are:

\begin{align}y &= {m_1}x + {c_1} \qquad \ldots \left( 1 \right)\\y &= {m_2}x + {c_2} \qquad \ldots \left( 2 \right)\\y &= {m_3}x + {c_3} \qquad \ldots \left( 3 \right)\end{align}

On subtracting equation ($$1$$) from ($$2$$) we obtain

\begin{align}\left( {{m_2} - {m_1}} \right)x + \left( {{c_2} - {c_1}} \right) &= 0\\\left( {{m_1} - {m_2}} \right)x &= \left( {{c_2} - {c_1}} \right)\\x &= \frac{{\left( {{c_2} - {c_1}} \right)}}{{\left( {{m_1} - {m_2}} \right)}}\end{align}

On substituting this value of x in (1), we obtain

\begin{align}y &= {m_1}\left( {\frac{{{c_2} - {c_1}}}{{{m_1} - {m_2}}}} \right) + {c_1}\\&= \frac{{{m_1}{c_2} - {m_1}{c_1}}}{{{m_1} - {m_2}}} + {c_1}\\&= \frac{{{m_1}{c_2} - {m_1}{c_1} + {c_1}\left( {{m_1} - {m_2}} \right)}}{{{m_1} - {m_2}}}\\&= \frac{{{m_1}{c_2} - {m_1}{c_1} + {m_1}{c_1} - {m_2}{c_1}}}{{{m_1} - {m_2}}}\\y &= \frac{{{m_1}{c_2} - {m_2}{c_1}}}{{{m_1} - {m_2}}}\end{align}

Therefore,

$$\left( {\frac{{{c_2} - {c_1}}}{{{m_1} - {m_2}}},\frac{{{m_1}{c_2} - {m_2}{c_1}}}{{{m_1} - {m_2}}}} \right)$$ is the point of intersection of line (1) and (2)

It is given that lines ($$1$$), ($$2$$) and ($$3$$) are concurrent.

Hence the point of intersection of lines ($$1$$) and ($$2$$) will also satisfy equation ($$3$$).

On substituting this value of $$x$$ and $$y$$ in ($$3$$), we obtain

\begin{align}&\frac{{{m_1}{c_2} - {m_2}{c_1}}}{{{m_1} - {m_2}}} = {m_3}\left( {\frac{{{c_2} - {c_1}}}{{{m_1} - {m_2}}}} \right) + {c_3}\\&\frac{{{m_1}{c_2} - {m_2}{c_1}}}{{{m_1} - {m_2}}} = \frac{{{m_3}{c_2} - {m_3}{c_1} + {c_3}{m_1} - {c_3}{m_2}}}{{{m_1} - {m_2}}}\\&{m_1}{c_2} - {m_2}{c_1} - {m_3}{c_2} + {m_3}{c_1} - {m_1}{c_3} + {m_2}{c_3} = 0\\&{m_1}{c_2} - {m_1}{c_3} - {m_2}{c_1} + {m_2}{c_3} - {m_3}{c_2} + {m_3}{c_1} = 0\\&{m_1}\left( {{c_2} - {c_3}} \right) + {m_2}\left( {{c_3} - {c_1}} \right) + {m_3}\left( {{c_1} - {c_2}} \right) = 0\end{align}

Hence, $${m_1}\left( {{c_2} - {c_3}} \right) + {m_2}\left( {{c_3} - {c_1}} \right) + {m_3}\left( {{c_1} - {c_2}} \right) = 0$$ proved.

## Chapter 10 Ex.10.ME Question 11

Find the equation of the line through the points $$\left( {3,2} \right)$$ which make an angle of $$45^\circ$$ with the line $$x-2y=3$$.

### Solution

Let the slope of the required line be

The given line can be represented as $$y = \frac{1}{2}x - \frac{3}{2}$$, which is of the form $$y = mx + c$$

Slope of the given line $$= {m_2} = \frac{1}{2}$$

It is given that the angle between the required line and line $$x - 2y = 3$$ is $$45^\circ$$.

We know that if $$θ$$ is the acute angle between lines $${l_1}$$ and $${l_2}$$ with the slopes $${m_1}$$ and $${m_2}$$ respectively,

Then,

\begin{align}\tan \theta &= \left| {\frac{{{m_2} - {m_1}}}{{1 + {m_1}{m_2}}}} \right|\\\tan 45^\circ &= \left| {\frac{{{m_2} - {m_1}}}{{1 + {m_1}{m_2}}}} \right|\\1 &= \left| {\frac{{\frac{1}{2} - {m_1}}}{{1 + \frac{{{m_1}}}{2}}}} \right|\\1 &= \left| {\frac{{\frac{{1 - 2{m_1}}}{2}}}{{\frac{{2 + {m_1}}}{2}}}} \right|\\1& = \left| {\frac{{1 - 2{m_1}}}{{2 + {m_1}}}} \right|\\1 &= \pm \left( {\frac{{1 - 2{m_1}}}{{2 + {m_1}}}} \right)\end{align}

\begin{align}\Rightarrow 1 = \left( {\frac{{1 - 2{m_1}}}{{2 + {m_1}}}} \right)\\\Rightarrow 2 + {m_1} = 1 - 2{m_1}\\\Rightarrow {m_1} = - \frac{1}{3}\end{align} or \begin{align}{c}\Rightarrow 1 = - \left( {\frac{{1 - 2{m_1}}}{{2 + {m_1}}}} \right)\\\Rightarrow 2 + {m_1} = - 1 + 2{m_1}\\\Rightarrow {m_1} = 3\end{align}

Case I: $${m_1} = 3$$

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

\begin{align}y - 2 &= 3\left( {x - 3} \right)\\y - 2 &= 3x - 9\\3x - y &= 7\end{align}

Case I: $${m_1} = - \frac{1}{3}$$

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

\begin{align}y - 2 &= - \frac{1}{3}\left( {x - 2} \right)\\3y - 6 &= - x + 3\\x + 3y &= 9\end{align}

\begin{align}\frac{x}{a} + \frac{y}{b} &= 1\\\Rightarrow \;x + y &= ab \qquad \ldots \left( 1 \right)\end{align}

Thus, the equations of the line are $$3x - y = 7$$ and $$x + 3y = 9$$.

## Chapter 10 Ex.10.ME Question 12

Find the equation of the line passing through the point of intersection of the line $$4x + 7y - 3 = 0$$ and $$2x - 3y + 1 = 0$$ that has equal intercepts on the axes.

### Solution

Let the equation of the line having equal intercepts on the axes be

On solving equations $$4x + 7y - 3 = 0$$ and $$2x - 3y + 1 = 0$$, we obtain $$x = \frac{1}{{13}}$$ and $$y = \frac{5}{{13}}$$

Therefore,

$$\left( {\frac{1}{{13}},\frac{5}{{13}}} \right)$$ is the point of the intersection of the two given lines.

Since equation ($$1$$) passes through point $$\left( {\frac{1}{{13}},\frac{5}{{13}}} \right)$$

\begin{align}&\frac{1}{{13}} + \frac{5}{{13}} = a\\&\Rightarrow\; a = \frac{6}{{13}}\end{align}

Equation ($$1$$) becomes

\begin{align}&\Rightarrow \;x + y = \frac{6}{{13}}\\&\Rightarrow \;13x + 13y = 6\end{align}

Thus, the required equation of the line $$13x + 13y = 6$$

## Chapter 10 Ex.10.ME Question 13

Show that the equation of the line passing through the origin and making an angle $$θ$$ with the line $$y = mx + c$$, is $$\frac{y}{x} = \frac{{m \pm \tan \theta }}{{1 \mp m\tan \theta }}$$

### Solution

Let the equation of the line passing through the origin be $$y = {m_1}x$$

If this line makes an angle of $${\rm{\theta }}$$ with line $$y = mx + c$$, then angle $${\rm{\theta }}$$ is given by

\begin{align}\tan \theta &= \left| {\frac{{{m_1} - m}}{{1 + {m_1}m}}} \right|\\&= \left| {\frac{{\frac{y}{x} - m}}{{1 + \frac{y}{x}m}}} \right|\\\tan \theta &= \pm \left( {\frac{{\frac{y}{x} - m}}{{1 + \frac{y}{x}m}}} \right)\end{align}

$$\Rightarrow \tan \theta = \frac{{\frac{y}{x} - m}}{{1 + \frac{y}{x}m}}$$ or $$\tan \theta = - \left( {\frac{{\frac{y}{x} - m}}{{1 + \frac{y}{x}m}}} \right)$$

Case I: $$\tan \theta = \frac{{\frac{y}{x} - m}}{{1 + \frac{y}{x}m}}$$

\begin{align}\tan \theta &= \frac{{\frac{y}{x} - m}}{{1 + \frac{y}{x}m}}\\\tan \theta + \frac{y}{x}m\tan \theta &= \frac{y}{x} - m\\m + \tan \theta &= \frac{y}{x}\left( {1 - m\tan \theta } \right)\\\frac{y}{x} &= \frac{{m + \tan \theta }}{{1 - m\tan \theta }}\end{align}

Case II: $$\tan \theta = - \left( {\frac{{\frac{y}{x} - m}}{{1 + \frac{y}{x}m}}} \right)$$

\begin{align}\tan \theta &= - \left( {\frac{{\frac{y}{x} - m}}{{1 + \frac{y}{x}m}}} \right)\\\tan \theta + \frac{y}{x}m\tan \theta &= - \frac{y}{x} + m\\m - \tan \theta &= \frac{y}{x}\left( {1 + m\tan \theta } \right)\\\frac{y}{x} &= \frac{{m - \tan \theta }}{{1 + m\tan \theta }}\end{align}

Thus, the required line is given by $$\frac{y}{x} = \frac{{m \pm \tan \theta }}{{1 \mp m\tan \theta }}$$

## Chapter 10 Ex.10.ME Question 14

In what ratio, the joining $$\left( { - 1,1} \right)$$ and $$\left( {5,7} \right)$$ is divisible by the line $$x + y = 4$$?

### Solution

The solution of the line joining the points $$\left( { - 1,1} \right)$$ and $$\left( {5,7} \right)$$ is given by

\begin{align}y - 1 &= \frac{{7 - 1}}{{5 + 1}}\left( {x + 1} \right)\\y - 1 &= \frac{6}{6}\left( {x + 1} \right)\\x - y + 2 &= 0 \qquad \quad \ldots \left( 1 \right)\end{align}

The equation of the line is

$x + y = 4 \qquad \ldots \left( 2 \right)$

The points of intersection of line (1) and (2) is given by

$$x = 1$$ and $$y = 3$$

Let point $$\left( {1,3} \right)$$ divides the line segment joining $$\left( { - 1,1} \right)$$ and $$\left( {5,7} \right)$$ in the ratio $$1:k$$.

Accordingly, by section formula

\begin{align}&\left( {1,3} \right) = \left( {\frac{{k\left( { - 1} \right) + 1\left( 5 \right)}}{{1 + k}},\frac{{k\left( 1 \right) + 1\left( 7 \right)}}{{1 + k}}} \right)\\&\Rightarrow \; \left( {1,3} \right) = \left( {\frac{{ - k + 5}}{{1 + k}},\frac{{k + 7}}{{1 + k}}} \right)\\&\Rightarrow \; \frac{{ - k + 5}}{{1 + k}} = 1,\;\;\frac{{k + 7}}{{1 + k}} = 3\end{align}

Therefore,

\begin{align}&\Rightarrow\; \frac{{ - k + 5}}{{1 + k}} = 1\\&\Rightarrow \;- k + 5 = 1 + k\\&\Rightarrow \; 2k = 4\\&\Rightarrow\; k = 2\end{align}

Thus, the line joining the points $$\left( { - 1,1} \right)$$ and $$\left( {5,7} \right)$$ is divided by line $$x + y = 4$$ in the ratio$$1:2$$.

## Chapter 10 Ex.10.ME Question 15

Find the distance of the line $$4x + 7y + 5 = 0$$ from the point $$\left( {1,2} \right)$$ along the line $$2x - y = 0$$.

### Solution

The given lines are

\begin{align}2x - y &= 0 \qquad \ldots \left( 1 \right)\\4x + 7y + 5 &= 0 \qquad \ldots \left( 2 \right)\end{align}

Let $$A\left( {1,2} \right)$$ is a point on the line ($$1$$) and B be the point intersection of line ($$1$$) and ($$2$$).

On solving equations ($$1$$) and ($$2$$), we obtain $$x = - \frac{5}{{18}}$$ and $$y = - \frac{5}{9}$$

Coordinates of point B are $$\left( { - \frac{5}{{18}}, - \frac{5}{9}} \right)$$

By using distance formula, the distance between points A and B can be obtained as

\begin{align}AB &= \sqrt {{{\left( {1 + \frac{5}{{18}}} \right)}^2} + {{\left( {2 + \frac{5}{9}} \right)}^2}} \\&= \sqrt {{{\left( {\frac{{23}}{{18}}} \right)}^2} + {{\left( {\frac{{23}}{9}} \right)}^2}} \\&= \sqrt {{{\left( {\frac{{23}}{{9 \times 2}}} \right)}^2} + {{\left( {\frac{{23}}{9}} \right)}^2}} \\&= \sqrt {{{\left( {\frac{{23}}{9}} \right)}^2}{{\left( {\frac{1}{2}} \right)}^2} + {{\left( {\frac{{23}}{9}} \right)}^2}} \\&= \sqrt {{{\left( {\frac{{23}}{9}} \right)}^2}\left( {\frac{1}{4} + 1} \right)} \\&= \frac{{23}}{9}\sqrt {\left( {\frac{5}{4}} \right)} \\&= \frac{{23}}{9} \times \frac{{\sqrt 5 }}{2}\\&= \frac{{23\sqrt 5 }}{{18}}\end{align}

Thus, the required distance is $$\frac{{23\sqrt 5 }}{{18}}$$ units.

## Chapter 10 Ex.10.ME Question 16

Find the direction in which a straight line must be drawn through the points $$\left( { - 1,2} \right)$$ so that its point of intersection with line $$x - y = 4$$ may be at a distance of $$3$$ units from this point.

### Solution

Let $$y = mx + c$$ be the line through point $$\left( { - 1,2} \right)$$

Accordingly,

\begin{align}&\Rightarrow \; 2 = m\left( { - 1} \right) + c\\&\Rightarrow \; 2 = - m + c\\&\Rightarrow \; c = m + 2\\&\Rightarrow\; y = mx + m + 2 \qquad \ldots \left( 1 \right)\end{align}

The given line is

$x - y = 4 \qquad \ldots \left( 2 \right)$

On solving equation ($$1$$) and ($$2$$), we obtain

$$x = \frac{{2 - m}}{{m + 1}}$$ and $$y = \frac{{5m + 2}}{{m + 1}}$$

Therefore,

$$\left( {\frac{{2 - m}}{{m + 1}},\frac{{5m + 2}}{{m + 1}}} \right)$$ is the point of intersection of line ($$1$$) and ($$2$$)

Since this point is at a distance of 3 units from point $$\left( { - 1,2} \right)$$, accordingly to distance formula,

\begin{align}&\Rightarrow \;\sqrt {{{\left( {\frac{{2 - m}}{{m + 1}} + 1} \right)}^2} + {{\left( {\frac{{5m + 2}}{{m + 1}} - 2} \right)}^2}} = 3\\&\Rightarrow \; {\left( {\frac{{2 - m + m + 1}}{{m + 1}}} \right)^2} + {\left( {\frac{{5m + 2 - 2m - 2}}{{m + 1}}} \right)^2} = {3^2}\\&\Rightarrow\; \frac{9}{{{{\left( {m + 1} \right)}^2}}} + \frac{{9{m^2}}}{{{{\left( {m + 1} \right)}^2}}} = 9\\&\Rightarrow\; \frac{{1 + {m^2}}}{{{{\left( {m + 1} \right)}^2}}} = 1\\&\Rightarrow\; 1 + {m^2} = {m^2} + 1 + 2m\\&\Rightarrow 2m = 0\\&\Rightarrow \;m = 0\end{align}

Thus, the slope of the required line must be zero i.e., the line must be parallel to the $$x-$$axis.

## Chapter 10 Ex.10.ME Question 17

The hypotenuse of a right-angled triangle has its end at the points $$\left( {1,3} \right)$$ and $$\left( { - {\rm{4}},{\rm{1}}} \right)$$. Find the equation of the legs (perpendicular sides) of the triangle.

### Solution

Let ABC be the right-angled triangle, where $$\angle C = 90^\circ$$

Let the slope of $$AC = m$$

Hence, the slope of $$BC = - \frac{1}{m}$$

Equation of AC:

\begin{align}&\Rightarrow\; y - 3 = m\left( {x - 1} \right)\\&\Rightarrow\; \left( {x - 1} \right) = \frac{{y - 3}}{m}\end{align}

Equation of BC:

$x + 4 = - m\left( {y - 1} \right)$

For a given value of m, we can get these equations

For, $$m = 0,\;y - 3 = 0;\;x + 4 = 0$$

For $$m \to \infty ,\;x - 1 = 0;\;y - 1 = 0$$

## Chapter 10 Ex.10.ME Question 18

Find the image of the point $$\left( {3,8} \right)$$ with respect to the line $$x + 3y = 7$$ assuming the line to be a plane mirror.

### Solution

The equation of the given line is

$x + 3y = 7 \qquad \ldots \left( 1 \right)$

Let point $$B\left( {a,b} \right)$$ be the image of point $$A\left( {3,8} \right)$$

Accordingly, line ($$1$$) is the perpendicular bisector of AB

Slope of $$AB = \frac{{b - 8}}{{a - 3}}$$, while the slope of the line (1) is $$- \frac{1}{3}$$

Since line ($$1$$) is perpendicular to AB

\begin{align}&\Rightarrow\; \left( {\frac{{b - 8}}{{a - 3}}} \right)\left( { - \frac{1}{3}} \right) = 1\\&\Rightarrow \;\frac{{b - 8}}{{3a - 9}} = 1\\&\Rightarrow\; b - 8 = 3a - 9\\&\Rightarrow\; 3a - b - 1\end{align}

Mid-point of $$AB = \left( {\frac{{a + 3}}{2},\frac{{b + 8}}{2}} \right)$$

The mid-point of the line segments AB will also satisfy line (1).

Hence, from equation ($$1$$), we have

\begin{align}&\Rightarrow \;\left( {\frac{{a + 3}}{2}} \right) + 3\left( {\frac{{b + 8}}{2}} \right) = 7\\&\Rightarrow \;a + 3 + 3b + 24 = 14\\&\Rightarrow \;a + 3b = - 13 \qquad \qquad \dots(3)\end{align}

On solving equations ($$2$$) and ($$3$$), we obtain

$$a = - 1$$ and $$b = - 4$$.

Thus, the image of the given point with respect to the given line is $$\left( { - 1, - 4} \right)$$.

## Chapter 10 Ex.10.ME Question 19

If the lines $$y = 3x + 1$$ and $$2y = x + 3$$ are equally indicated to the line $$y = mx + 4$$, find the value of m.

### Solution

The equations of the given lines are:

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

Slope of line ($$1$$), $${m_1} = 3$$

Slope of line ($$2$$), $${m_2} = \frac{1}{2}$$

Slope of line ($$3$$), $${m_3} = m$$

It is given that lines ($$1$$) and ($$2$$) are equally inclined to line ($$3$$). This means that the given angle between lines ($$1$$) and ($$3$$) equals the angle between lines ($$2$$) and ($$3$$).

Therefore,

\begin{align}&\Rightarrow \; \left| {\frac{{{m_1} - {m_3}}}{{1 + {m_1}{m_3}}}} \right| = \left| {\frac{{{m_2} - {m_3}}}{{1 + {m_2}{m_3}}}} \right|\\&\Rightarrow\; \left| {\frac{{3 - m}}{{1 + 3m}}} \right| = \left| {\frac{{\frac{1}{2} - m}}{{1 + \frac{1}{2}m}}} \right|\\&\Rightarrow \;\left| {\frac{{3 - m}}{{1 + 3m}}} \right| = \left| {\frac{{1 - 2m}}{{m + 2}}} \right|\\&\Rightarrow\; \left| {\frac{{3 - m}}{{1 + 3m}}} \right| = \pm \left( {\frac{{1 - 2m}}{{m + 2}}} \right)\\&\Rightarrow\; \frac{{3 - m}}{{1 + 3m}} = \frac{{1 - 2m}}{{m + 2}}or\frac{{3 - m}}{{1 + 3m}} = - \left( {\frac{{1 - 2m}}{{m + 2}}} \right)\end{align}

Case I: If $$\frac{{3 - m}}{{1 + 3m}} = \frac{{1 - 2m}}{{m + 2}}$$

Then,

\begin{align}&\Rightarrow \;\left( {3 - m} \right)\left( {m + 2} \right) = \left( {1 - 2m} \right)\left( {1 + 3m} \right)\\&\Rightarrow \;- {m^2} + m - 6 = 1 + m - 6{m^2}\\&\Rightarrow \; 5{m^2} + 5 = 0\\&\Rightarrow \;{m^2} + 1 = 0\\&\Rightarrow \;{m^2} = - 1\\&\Rightarrow \;m = \sqrt { - 1} \end{align}

Here, $$m = \sqrt { - 1}$$, which is not real

Hence, this case is not possible

Case II: If $$\frac{{3 - m}}{{1 + 3m}} = - \left( {\frac{{1 - 2m}}{{m + 2}}} \right)$$

Then,

\begin{align}&\Rightarrow \;\left( {3 - m} \right)\left( {m + 2} \right) = - \left( {1 - 2m} \right)\left( {1 + 3m} \right)\\&\Rightarrow \; - {m^2} + m - 6 = - \left( {1 + m - 6{m^2}} \right)\\&\Rightarrow\; 7{m^2} - 2m - 7 = 0\\&\Rightarrow \;m = \frac{{ - 2 \pm \sqrt {4 - 4\left( 7 \right)\left( { - 7} \right)} }}{{2\left( 7 \right)}}\\&\Rightarrow \; m = \frac{{2 \pm 2\sqrt {1 + 49} }}{{14}}\\&\Rightarrow \; m = \frac{{1 \pm 5\sqrt 2 }}{7}\end{align}

Thus, the required value of $$m = \frac{{1 \pm 5\sqrt 2 }}{7}$$.

## Chapter 10 Ex.10.ME Question 20

If sum of the perpendicular distance of a variable point $$P\left( {x,y} \right)$$ from the lines $$x + y - 5 = 0$$ and $$3x - 2y + 7 = 0$$ is always $$10.$$ Show that $$P$$ must move on a line.

### Solution

The equations of the lines are

\begin{align}x + y - 5 &= 0 \qquad \ldots \left( 1 \right)\\3x - 2y + 7 &= 0 \qquad \ldots \left( 2 \right)\end{align}

The perpendicular distance of $$P\left( {x,y} \right)$$ from lines (1) and (2) are respectively given by

$${d_1} = \frac{{\left| {x + y - 5} \right|}}{{\sqrt {{{\left( 1 \right)}^2} + {{\left( 1 \right)}^2}} }}$$ and $${d_2} = \frac{{\left| {3x - 2y + 7} \right|}}{{\sqrt {{{\left( 3 \right)}^2} + {{\left( { - 2} \right)}^2}} }}$$

i.e., $${d_1} = \frac{{\left| {x + y - 5} \right|}}{{\sqrt 2 }}$$ and $${d_2} = \frac{{\left| {3x - 2y + 7} \right|}}{{\sqrt {13} }}$$

It is given that $${d_1} + {d_2} = 10$$

Therefore,

\begin{align}& \Rightarrow \; \frac{{|x + y - 5|}}{{\sqrt 2 }} + \frac{{|3x - 2y + 7|}}{{\sqrt {13} }} = 10\\& \Rightarrow \;\sqrt {13} |x + y - 5| + \sqrt 2 |3x - 2y + 7| - 10\sqrt {26} = 0\\&\Rightarrow \;\sqrt {13} \left( {x + y - 5} \right) + \sqrt 2 \left( {3x - 2y + 7} \right) - 10\sqrt {26} = 0\end{align}

Assuming $$x + y - 5 = 0$$ and $$3x - 2y + 7 = 0$$ are positive.

\begin{align}&\Rightarrow \;\sqrt {13} x + \sqrt {13} y - 5\sqrt {13} + 3\sqrt 2 x - 2\sqrt 2 y + 7\sqrt 2 - 10\sqrt {26} = 0\\&\Rightarrow \; \left( {\sqrt {13} + 3\sqrt 2 } \right)x + \left( {\sqrt {13} - 2\sqrt 2 } \right)y + \left( {7\sqrt 2 - 5\sqrt {13} - 10\sqrt {26} } \right) = 0\end{align}

Since, $$\left( {\sqrt {13} + 3\sqrt 2 } \right)x + \left( {\sqrt {13} - 2\sqrt 2 } \right)y + \left( {7\sqrt 2 - 5\sqrt {13} - 10\sqrt {26} } \right) = 0$$ is the equation of a line.

Similarly, we can obtain the equation of line for any signs of $$x + y - 5 = 0$$ and $$3x - 2y + 7 = 0$$.

Thus, point $$P$$ must move on a line.

## Chapter 10 Ex.10.ME Question 21

Find equation of the line which is equidistant from parallel lines $$9x + 6y - 7 = 0$$ and $$3x + 2y + 6 = 0$$.

### Solution

The equations of the given lines are

\begin{align}9x + 6y - 7 &= 0 \qquad \ldots \left( 1 \right)\\3x + 2y + 6 &= 0 \qquad \ldots \left( 2 \right)\end{align}

Let $$P\left( {h,k} \right)$$ be the arbitrary point, is equidistant from lines (1) and (2).

Then the perpendicular distance of $$P\left( {h,k} \right)$$ from the line (1) is given by

\begin{align}{d_1} &= \frac{{\left| {9h + 6k - 7} \right|}}{{\sqrt {{{\left( 9 \right)}^2} + {{\left( 6 \right)}^2}} }}\\&= \frac{{\left| {9h + 6k - 7} \right|}}{{\sqrt {117} }}\\&= \frac{{\left| {9h + 6k - 7} \right|}}{{3\sqrt {13} }}\end{align}

And the perpendicular distance of $$P\left( {h,k} \right)$$ from line (2) is given by

\begin{align}{d_2} &= \frac{{\left| {3h + 2k + 6} \right|}}{{\sqrt {{{\left( 3 \right)}^2} + {{\left( 2 \right)}^2}} }}\\&= \frac{{\left| {3h + 2k + 6} \right|}}{{\sqrt {13} }}\end{align}

Since $$P\left( {h,k} \right)$$ is equidistant from lines (1) and (2), $${d_1} = {d_2}$$

Therefore,

\begin{align}&\Rightarrow \; \frac{{\left| {9h + 6k - 7} \right|}}{{3\sqrt {13} }} = \frac{{\left| {3h + 2k + 6} \right|}}{{\sqrt {13} }}\\&\Rightarrow \; \left| {9h + 6k - 7} \right| = 3\left| {3h + 2k + 6} \right|\\&\Rightarrow \;9h + 6k - 7 = \pm 3\left( {3h + 2k + 6} \right)\end{align}

Case I: $$9h + 6k - 7 = 3\left( {3h + 2k + 6} \right)$$

\begin{align}&\Rightarrow \; 9h + 6k - 7 = 3\left( {3h + 2k + 6} \right)\\&\Rightarrow \;9h + 6k - 7 = 9h + 6k + 18\\&\Rightarrow \;9h + 6k - 7 - 9h - 6k - 18 = 0\\&\Rightarrow \;25 = 0\end{align}

Which is an absurd, hence this case is not possible.

Case II: $$9h + 6k - 7 = - 3\left( {3h + 2k + 6} \right)$$

\begin{align}&\Rightarrow\; 9h + 6k - 7 = - 3\left( {3h + 2k + 6} \right)\\&\Rightarrow \;9h + 6k - 7 = - 9h - 6k - 18\\&\Rightarrow\; 18h + 12k + 11 = 0\end{align}

Thus, the required equation of the line is $$18x + 12y + 11 = 0$$.

## Chapter 10 Ex.10.ME Question 22

A ray of light passing through the point $$\left( {1,2} \right)$$ reflects on the $$x$$-axis at point $$A$$ and the reflected ray passes through the point $$\left( {5,3} \right)$$. Find the coordinates of $$A.$$

### Solution Let the coordinates of point $$A$$ be $$\left( {a,0} \right)$$.

Draw a line, $$AL$$ perpendicular to the $$x$$-axis

We know that angle of incidence is equal to angle of reflection.

Hence, let $$\angle BAL = \angle CAL = \phi$$ and $$\angle CAX = \theta$$

Now,

\begin{align}\angle OAB &= 180^\circ - \left( {\theta + 2\phi } \right)\\&= 180^\circ - \left[ {\theta + 2\left( {90^\circ - \theta } \right)} \right]\\&= 180^\circ - \left[ {\theta + 180^\circ - 2\theta } \right]\\&= 180^\circ - 180^\circ + \theta \\&= \theta \end{align}

Therefore,

$\angle BAX = 180^\circ - \theta$

Now,

Slope of line $$AC = \frac{{3 - 0}}{{5 - a}}$$

$\Rightarrow \tan \theta = \frac{{3 - 0}}{{5 - a}} \qquad \ldots \left( 1 \right)$

Slope of line$$AC = \frac{{2 - 0}}{{1 - a}}$$

\begin{align}&\Rightarrow \; \tan \left( {180^\circ - \theta } \right) = \frac{2}{{1 - a}}\\&\Rightarrow \;- \tan \theta = \frac{2}{{1 - a}}\\&\Rightarrow \; \tan \theta = \frac{2}{{a - 1}} \qquad \ldots \left( 2 \right)\end{align}

From equations (1) and (2), we obtain

\begin{align}&\Rightarrow \;\frac{3}{{5 - a}} = \frac{2}{{a - 1}}\\&\Rightarrow\; 3a - 3 = 10 - 2a\\&\Rightarrow\; a = \frac{{13}}{5}\end{align}

Thus, the coordinates of point $$A$$ are $$\left( {\frac{{13}}{5},0} \right)$$.

## Chapter 10 Ex.10.ME Question 23

Prove that the product of the lengths of the perpendiculars drawn from the points $$\left( {\sqrt {{a^2} - {b^2}} ,0} \right)$$ and $$\left( { - \sqrt {{a^2} + {b^2}} ,0} \right)$$ to the line $$\frac{x}{a}\cos \theta + \frac{y}{b}\sin \theta = 1$$ is $${b^2}$$.

### Solution

The equation of the given line is

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

Length of the perpendicular from point $$\left( {\sqrt {{a^2} + {b^2}} ,0} \right)$$ to the line (1) is

\begin{align}{p_1} &= \frac{{\left| {b\cos \theta \left( {\sqrt {{a^2} - {b^2}} } \right) + a\sin \theta \left( 0 \right) - ab} \right|}}{{\sqrt {{b^2}{{\cos }^2}\theta + {a^2}{{\sin }^2}\theta } }}\\&= \frac{{\left| {b\cos \theta \left( {\sqrt {{a^2} - {b^2}} } \right) - ab} \right|}}{{\sqrt {{b^2}{{\cos }^2}\theta + {a^2}{{\sin }^2}\theta } }} \qquad \ldots \left( 2 \right)\end{align}

Length of the perpendicular from point $$\left( { - \sqrt {{a^2} + {b^2}} ,0} \right)$$ to the line (2) is

\begin{align}{p_2} &= \frac{{\left| {b\cos \theta \left( { - \sqrt {{a^2} - {b^2}} } \right) + a\sin \theta \left( 0 \right) - ab} \right|}}{{\sqrt {{b^2}{{\cos }^2}\theta + {a^2}{{\sin }^2}\theta } }}\\&= \frac{{\left| {b\cos \theta \left( {\sqrt {{a^2} - {b^2}} } \right) - ab} \right|}}{{\sqrt {{b^2}{{\cos }^2}\theta + {a^2}{{\sin }^2}\theta } }} \qquad \ldots \left( 3 \right)\end{align}

On multiplying equations (2) and (3), we obtain

\begin{align} {{p}_{1}}{{p}_{2}}& =\frac{\left| b\cos \theta \left( \sqrt{{{a}^{2}}-{{b}^{2}}} \right)-ab \right|\left| b\cos \theta \left( \sqrt{{{a}^{2}}-{{b}^{2}}} \right)-ab \right|}{{{\left( \sqrt{{{b}^{2}}{{\cos }^{2}}\theta +{{a}^{2}}{{\sin }^{2}}\theta } \right)}^{2}}} \\ & =\frac{\left| \left( b\cos \theta \left( \sqrt{{{a}^{2}}-{{b}^{2}}} \right)-ab \right)\left( b\cos \theta \left( \sqrt{{{a}^{2}}-{{b}^{2}}} \right)-ab \right) \right|}{\left( {{b}^{2}}{{\cos }^{2}}\theta +{{a}^{2}}{{\sin }^{2}}\theta \right)} \\ & =\frac{\left| {{\left( b\cos \theta \sqrt{{{a}^{2}}-{{b}^{2}}} \right)}^{2}}-{{\left( ab \right)}^{2}} \right|}{\left( {{b}^{2}}{{\cos }^{2}}\theta +{{a}^{2}}{{\sin }^{2}}\theta \right)} \\ & =\frac{\left| {{b}^{2}}{{\cos }^{2}}\theta \left( {{a}^{2}}-{{b}^{2}} \right)-{{a}^{2}}{{b}^{2}} \right|}{\left( {{b}^{2}}{{\cos }^{2}}\theta +{{a}^{2}}{{\sin }^{2}}\theta \right)} \\ & =\frac{\left| {{a}^{2}}{{b}^{2}}{{\cos }^{2}}\theta -{{b}^{4}}{{\cos }^{2}}\theta -{{a}^{2}}{{b}^{2}} \right|}{\left( {{b}^{2}}{{\cos }^{2}}\theta +{{a}^{2}}{{\sin }^{2}}\theta \right)} \\ & =\frac{{{b}^{2}}\left| {{a}^{2}}{{\cos }^{2}}\theta -{{b}^{2}}{{\cos }^{2}}\theta -{{a}^{2}} \right|}{\left( {{b}^{2}}{{\cos }^{2}}\theta +{{a}^{2}}{{\sin }^{2}}\theta \right)} \\ & =\frac{{{b}^{2}}\left| {{a}^{2}}{{\cos }^{2}}\theta -{{b}^{2}}{{\cos }^{2}}\theta -{{a}^{2}}{{\sin }^{2}}\theta -{{a}^{2}}{{\cos }^{2}}\theta \right|}{\left( {{b}^{2}}{{\cos }^{2}}\theta +{{a}^{2}}{{\sin }^{2}}\theta \right)}\\& \quad \left[ \because {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 \right] \\\\ & =\frac{{{b}^{2}}\left| -\left( {{b}^{2}}{{\cos }^{2}}\theta +{{a}^{2}}{{\sin }^{2}}\theta \right) \right|}{\left( {{b}^{2}}{{\cos }^{2}}\theta +{{a}^{2}}{{\sin }^{2}}\theta \right)} \\ & =\frac{{{b}^{2}}\left( {{b}^{2}}{{\cos }^{2}}\theta +{{a}^{2}}{{\sin }^{2}}\theta \right)}{\left( {{b}^{2}}{{\cos }^{2}}\theta +{{a}^{2}}{{\sin }^{2}}\theta \right)} \\ & ={{b}^{2}} \end{align}

Hence, proved

## Chapter 10 Ex.10.ME Question 24

A person standing at the junction (crossing) of two straight paths represented by the equation $$2x - 3y + 4 = 0$$ and $$3x + 4y - 5 = 0$$ wants to reach the path whose equation is $$6x - 7y + 8 = 0$$ in the least time. Find equation of the path that he should follow.

### Solution

The equations of the given lines are

\begin{align}2x - 3y + 4 = 0 \qquad \ldots \left( 1 \right)\\3x + 4y - 5 = 0 \qquad \ldots \left( 2 \right)\\6x - 7y + 8 = 0 \qquad \ldots \left( 3 \right)\end{align}

The person is standing at the junction of the paths represented by lines (1) and (2).

On solving equations (1) and (2), we obtain $$x = - \frac{1}{{17}}$$ and $$y = \frac{{22}}{{17}}$$

Thus, the person is standing at point $$\left( { - \frac{1}{{17}},\frac{{22}}{{17}}} \right)$$

The person can reach path (3) in the least time if he walks along the perpendicular line to (3) from point $$\left( { - \frac{1}{{17}},\frac{{22}}{{17}}} \right)$$

Now,

Slope of the line (3) $$= \frac{6}{7}$$

Slope of the line perpendicular to line (3) $$= - \frac{1}{{\left( {\frac{6}{7}} \right)}} = - \frac{7}{6}$$

The equation of the line passing through $$\left( { - \frac{1}{{17}},\frac{{22}}{{17}}} \right)$$ and having a slope of $$- \frac{7}{6}$$ is given by

\begin{align}&\Rightarrow \; \left( {y - \frac{{22}}{{17}}} \right) = - \frac{7}{6}\left( {x + \frac{1}{{17}}} \right)\\&\Rightarrow \; 6\left( {17y - 22} \right) = - 7\left( {17x + 1} \right)\\&\Rightarrow \;102y - 132 = - 119x - 7\\&\Rightarrow \;119x + 102y = 125\end{align}

Hence, the path that the person should follow is $$119x + 102y = 125$$.

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