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

Video 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

Video 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 \(\frac{x}{3} + \frac{y}{4} = 1$2x - 3y = 6\) and \( - 3x + 2y = 6\)

}\) respectively.

 

 

Solution

Video Solution
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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

Video 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

Video 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

Video 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.

 

 

Solution

Video Solution
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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

Video 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

Video 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

Video 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

Video 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

Video 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

Video 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

Video 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

Video 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

Video 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

Video 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

Video 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

Video 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

Video 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

Video 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

Video 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

Video 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

Video 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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