# NCERT Solutions For Class 11 Maths Chapter 10 Exercise 10.2

Straight Lines

Exercise 10.2

## Chapter 10 Ex.10.2 Question 1

Write the equation for the *\(x\)* and *\(y\)*-axes.

**Solution**

The *\(y\)*-coordinate of every point on the *\(x\)*-axis is \(0.\)

Therefore, the equation of the *\(x\)*-axis is \(y = 0\).

The *\(x\)*-coordinate of every point on the *\(y\)*-axis is \(0.\)

Therefore, the equation of the *\(y\)*-axis is \(x = 0\).

## Chapter 10 Ex.10.2 Question 2

Find the equation of the line which passes through the point \(\left( { - {\rm{4}},{\rm{3}}} \right)\) with slope.

**Solution**

We know that the equation of the line passing through point \(\left( {{x_0},{y_0}} \right)\), whose slope is \(m\), is \(\left( {y - {y_0}} \right) = m\left( {x - {x_0}} \right)\)

Thus, the equation of the line passing through point \(\left( { - {\rm{4}},{\rm{3}}} \right)\), whose slope is \(\frac{1}{2}\),

\[\begin{align}\left( {y - 3} \right)& = \frac{1}{2}\left( {x + 4} \right)\\2\left( {y - 3} \right) &= x + 4\\2y - 6 &= x + 4\\x - 2y + 10 &= 0\end{align}\]

Hence the equation is \(x - 2y + 10 = 0\).

## Chapter 10 Ex.10.2 Question 3

Find the equation of the line which passes through \(\left( 0,0 \right)\) with slope \(m\).

**Solution**

We know that the equation of the line passing through point \(\left( {{x_0},{y_0}} \right)\), whose slope is \(m,\)

\[\left( {y - {y_0}} \right) = m\left( {x - {x_0}} \right)\]

Thus, the equation of the line passing through point \(\left( {0,0} \right)\), whose slope is \(m\), is

\[\begin{align}\left( {y - 0} \right) &= m\left( {x - 0} \right)\\y &= mx\end{align}\]

Hence the equation is \(y = mx\).

## Chapter 10 Ex.10.2 Question 4

Find the equation of the line which passes through \(\left( {2,2\sqrt 3 } \right)\)and is inclined with the *\(x\)*-axis at an angle of \(75^\circ \).

**Solution**

The slope of the line that inclines with the *x*-axis at an angle of \(75^\circ \) is \(m = \tan 75^\circ \)

\[\begin{align}m &= \tan \left( {45^\circ + 30^\circ } \right)\\&= \frac{{\tan 45^\circ + \tan 30^\circ }}{{1 - \tan 45^\circ \tan 30^\circ )}}\\&= \frac{{1 + \frac{1}{{\sqrt 3 }}}}{{1 - 1 \times \frac{1}{{\sqrt 3 }}}}\\&= \frac{{\frac{{\sqrt 3 + 1}}{{\sqrt 3 }}}}{{\frac{{\sqrt 3 - 1}}{{\sqrt 3 }}}}\\&= \frac{{\sqrt 3 + 1}}{{\sqrt 3 - 1}}\end{align}\]

We know that the equation of the line passing through point \(\left( {{x_0},{y_0}} \right)\), whose slope is \(m,\)

\[\left( {y - {y_0}} \right) = m\left( {x - {x_0}} \right)\]

Thus, if a line passes through \(\left( {2,2\sqrt 3 } \right)\) and inclines with the *x*-axis at an angle of \(75^\circ \), then the equation of the line is given as

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

Hence the equation is \(\left( {\sqrt 3 + 1} \right)x - \left( {\sqrt 3 - 1} \right)y = 4\left( {\sqrt 3 - 1} \right)\).

## Chapter 10 Ex.10.2 Question 5

Find the equation of the line which intersects the *\(x\)*-axis at a distance of \(3\) units to the left of origin with slope \(-2\).

**Solution**

It is known that if a line with slope \(m\) makes *\(x\)*-intercept \(d\), then the equation of the time is given as \(y = m\left( {x - d} \right)\)

For the line intersecting the *\(x\)*-axis at a distance of \(3\) units to the left of the origin, \(d = - 3\).

The slope of the line is given as \(m = - 2\)

Thus, the required equation of the given line is

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

Hence the equation is \(2x + y + 6 = 0\)

## Chapter 10 Ex.10.2 Question 6

Find the equation of the line which intersects the *\(y\)*-axis at a distance of \(2\) units above the origin and makes an angle of \(30^\circ \) with the positive direction of the *\(x\)*-axis.

**Solution**

It is known that if a line with slope \(m\) makes *\(y\)*-intercept \(c\), then the equation of the line is given as \(y = mx + c\)

Here, \(c = 2\) and \(m = \tan 30^\circ = \frac{1}{{\sqrt 3 }}\)

Thus, the required equations of the given line is

\[\begin{align}y &= \frac{1}{{\sqrt 3 }}x + 2\\y &= \frac{{x + 2\sqrt 3 }}{{\sqrt 3 }}\\\sqrt 3 y &= x + 2\sqrt 3 \\x - \sqrt 3 y + 2\sqrt 3 &= 0\end{align}\]

Hence, the equation of the line is \(x - \sqrt 3 y + 2\sqrt 3 = 0\)

## Chapter 10 Ex.10.2 Question 7

Find the slope of the line, which makes an angle of \(30^\circ \) with the positive direction of *\(y\)*-axis measured anticlockwise.

**Solution**

It is known that the equation of the line passes through points \(\left( {{x_1},{y_1}} \right)\) and \(\left( {{x_2},{y_2}} \right)\) is

\[y - {y_1} = \frac{{y_2 - y_1}}{{x_2 - x_1}}\left( {x - {x_1}} \right)\]

Therefore, the equation of the line passing through the points \(\left( - 1,1 \right)\) and \(\left( 2, - 4 \right)\) is

\[\begin{align}\left( {y - 1} \right) &= \frac{{ - 4 - 1}}{{2 + 1}}\left( {x + 1} \right)\\\left( {y - 1} \right) &= - \frac{5}{3}\left( {x + 1} \right)\\3\left( {y - 1} \right) &= - 5\left( {x + 1} \right)\\3y - 3 &= - 5x - 5\\5x + 3y + 2 &= 0\end{align}\]

Hence, the equation of the line is \(5x + 3y + 2 = 0\)

## Chapter 10 Ex.10.2 Question 8

Find the equation of the line which is at a perpendicular distance of \(5\) units from the origin and the angle made by the perpendicular with the positive *\(x\)*-axis is \(30^\circ \).

**Solution**

If \(p\) is the length of the normal from the origin to a line and \(\omega \) is the angle made by the normal with the positive direction of the *\(x\)*-axis, then the equation of the line given by

\[x\cos \omega + y\sin \omega = p\]

Here, \(p = 5\) units and \(\omega = 30^\circ \)

Thus, the required equation of the given line is

\[\begin{align}x\cos 30^\circ + y\sin 30^\circ &= 5\\x.\frac{{\sqrt 3 }}{2} + y.\frac{1}{2} &= 5\\\sqrt 3 x + y - 10 &= 0\end{align}\]

Hence, the equation of the line is \(\sqrt 3 x + y - 10 = 0\)

## Chapter 10 Ex.10.2 Question 9

The vertices of \(\Delta PQR\) are \(P\left( {2,1} \right),\;Q\left( { - 2,3} \right)\) and \(R\left( {4,5} \right)\). Find equation of the median through the vertex \(R.\)

**Solution**

It is given that the vertices of \(\Delta PQR\) are \(P\left( {2,1} \right),\;Q\left( { - 2,3} \right)\) and \(R\left( {4,5} \right)\). Let \(RL\) be the median through vertex \(R.\)

Accordingly, \(L\) be the mid-point of \(PQ.\)

By mid-point formula, the coordinates of point L are given by \(\left( {\frac{{2 - 2}}{2},\frac{{1 + 3}}{2}} \right) = \left( {0,2} \right)\)

Therefore, the equation of the line passing through points \(\left( {4,5} \right)\) and \(\left( {0,2} \right)\) is

\[\begin{align}y - 5 &= \frac{{2 - 5}}{{0 - 4}}\left( {x - 4} \right)\\y - 5 &= \frac{{ - 3}}{{ - 4}}\left( {x - 4} \right)\\4\left( {y - 5} \right) &= 3\left( {x - 4} \right)\\ 4y - 20 &= 3x - 12\\3x - 4y + 8 &= 0\end{align}\]

Thus, the equation of the median through vertex \(R\) is \(3x - 4y + 8 = 0\).

## Chapter 10 Ex.10.2 Question 10

Find the equation of the line passing through \(\left(- 3,5 \right)\) and perpendicular to the line through the points \(\left( 2,5\right)\) and \(\left( -3,6 \right)\).

**Solution**

The slope of the line joining the points \(\left( 2,5 \right)\) and \(\left( -3,6 \right)\) is \(m=\frac{6-5}{-3-2}=-\frac{1}{5}\)

We know that two non-vertical lines are perpendicular to each other if and only if their slopes are negative reciprocals of each other.

Therefore, slope of the line perpendicular to the line through the points \(\left( 2,5 \right)\) and \(\left( { - 3,6} \right)\) is \( - \frac{1}{m} = - \frac{1}{{\left( { - \frac{1}{5}} \right)}} = 5\)

Now, the equation of the line passing through point \(\left( - 3,5 \right)\), whose slope is \(5,\) is

\[\begin{align}\left( {y - 5} \right) &= 5\left( {x + 3} \right)\\y - 5 &= 5x + 15\\5x - y + 20 &= 0\end{align}\]

Hence, the equation of the line is \(5x - y + 20 = 0\)

## Chapter 10 Ex.10.2 Question 11

A line perpendicular to the line segment joining the points \(\left( {1,0} \right)\) and \(\left( {{\rm{2}},{\rm{3}}} \right)\) divides it in the ratio \(1:n\). Find the equation of the line.

**Solution**

According to the section formula, the coordinates of the point that divides the line segment joining the points \(\left( {1,0} \right)\) and \(\left( {{\rm{2}},{\rm{3}}} \right)\) in the ratio \(1:n\) is given by

\[\left( {\frac{{n\left( 1 \right) + 1\left( 2 \right)}}{{1 + n}},\frac{{n\left( 0 \right) + 1\left( 3 \right)}}{{1 + n}}} \right) = \left( {\frac{{n + 2}}{{n + 1}},\frac{3}{{n + 1}}} \right)\]

The slope of the line joining the points \(\left( {{\rm{1}},0} \right)\) and \(\left( {{\rm{2}},{\rm{3}}} \right)\) is \(m = \frac{{3 - 0}}{{2 - 1}} = 3\)

We know that two non-vertical lines are perpendicular to each other if and only if their slopes are negative reciprocals of each other.

Therefore, slope of the line that is perpendicular to the line joining the points \(\left( {{\rm{1}},0} \right)\) and \(\left( {{\rm{2}},{\rm{3}}} \right)\) is \( - \frac{1}{m} = - \frac{1}{3}\)

Now the equation of the line passing through \(\left( {\frac{{n + 2}}{{n + 1}},\frac{3}{{n + 1}}} \right)\) and whose slope is \( - \frac{1}{3}\), given by

\[\begin{align}\left( {y - \frac{3}{{n + 1}}} \right) &= - \frac{1}{3}\left( {x - \frac{{\left( {n + 2} \right)}}{{\left( {n + 1} \right)}}} \right)\\3\left[ {\left( {n + 1} \right)y - 3} \right] &= - \left[ {x\left( {n + 1} \right) - \left( {n - 2} \right)} \right]\\3\left( {n + 1} \right)y - 9 &= - \left( {n + 1} \right)x + n + 2\\\left( {1 + n} \right)x + 3\left( {1 + n} \right)y &= n + 11\end{align}\]

Hence, the equation of the line is \(\left( {1 + n} \right)x + 3\left( {1 + n} \right)y = n + 11\)

## Chapter 10 Ex.10.2 Question 12

Find the equation of a line that cuts off equal intercepts on the coordinate axes and passes through the points \(\left( {2,3} \right)\).

**Solution**

The equation of the line in the intercept form is

\[\frac{x}{a} + \frac{y}{b} = 1 \qquad \quad \ldots \left( 1 \right)\]

Here, *\(a\)* and *\(b\)* are the intercepts on *\(x\)* and *\(y\)* axes respectively.

It is given that the line cuts off equal intercepts on both the axes. This means that \(a = b\).

Accordingly, equation (1) reduces to

\[x + y = a \qquad \quad \ldots \left( 2 \right)\]

Since the given line passes through point \(\left( {2,3} \right)\), equation (2) reduces to

\[\begin{align}2 + 3 &= a\\a &= 5\end{align}\]

On substituting the value of *\(a\)* in equation (2), we obtain

\(x + y = 5\), which is the required equation of the line.

## Chapter 10 Ex.10.2 Question 13

Find the equation of the line passing through the points \(\left( 2,2\right)\) and cutting off intercepts on the axis whose sum is \(9.\)

**Solution**

The equation of a line in the intercept form is

\[\frac{x}{a} + \frac{y}{b} = 1 \qquad \quad \ldots \left( 1 \right)\]

Here, \(a\) and \(b\) are the intercepts on \(x\) and \(y\) axis respectively.

It is given that

\[\begin{align}a + b &= 9\\b &= 9 - a \qquad \quad \ldots \left( 2 \right)\end{align}\]

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

\[\frac{x}{a} + \frac{y}{{9 - a}} = 1 \qquad \quad \ldots \left( 3 \right)\]

It is given that the line passes through point \(\left( {{\rm{2}},{\rm{2}}} \right)\).

Therefore, equation (3) reduces to

\(\begin{align}&\Rightarrow \;\frac{2}{a} + \frac{2}{{9 - a}} = 1\\&\Rightarrow \;\frac{{2(9 - a) + 2a}}{{a(9 - a)}} = 1\\&\Rightarrow \;\frac{{18 - 2a + 2a}}{{(9a - {a^2})}} = 1\\&\Rightarrow \;\frac{{18}}{{9a - {a^2}}} = 1\\&\Rightarrow \;18 = 9a - {a^2}\\&\Rightarrow\; {a^2} - 9a + 18 = 0\\&\Rightarrow\; {a^2} - 6a - 3a + 18 = 0\\&\Rightarrow \;a(a - 6) - 3(a - 6) = 0\\&\Rightarrow \;(a - 6)(a - 3) = 0\end{align}\)

\( \Rightarrow a = 6\) or \(a = 3\)

If \(a = 6\) then \(b = 9 - 6 = 3\),

Hence, the equation of the line is

\[\begin{align}&\Rightarrow \;\frac{x}{6} + \frac{y}{3} = 1\\&\Rightarrow\; x + 2y - 6 = 0\end{align}\]

If \(a = 3\) then \(b = 9 - 3 = 6\)

Hence, the equation of the line is

\[\begin{align}&\Rightarrow \; \frac{x}{3} + \frac{y}{6} = 1\\&\Rightarrow \; 2x + y - 6 = 0\end{align}\]

Thus, the equation of the line is \(x + 2y - 6 = 0\) or \(2x + y - 6 = 0\).

## Chapter 10 Ex.10.2 Question 14

Find equation of the line through the points \(\left( {0,{\rm{2}}} \right)\) making an angle \(\frac{{2\pi }}{3}\) with the positive *\(x\)*-axis. Also, find the equation of the line parallel to it and crossing the *\(y\)*-axis at a distance of \(2\) units below the origin.

**Solution**

The slope of the line making an angle \(\frac{{2\pi }}{3}\) with the positive *\(x\)*-axis is \(m = \tan \left( {\frac{{2\pi }}{3}} \right) = - \sqrt 3 \)

Now, the equation of the line passing through points \(\left( {0,{\rm{2}}} \right)\) and having a slope \( - \sqrt 3 \) is

\[\begin{align}\left( {y - 2} \right) &= - \sqrt 3 \left( {x - 0} \right)\\\sqrt 3 x + y - 2 &= 0\end{align}\]

The slope of line parallel to line \(\sqrt 3 x + y - 2 = 0\) is \( - \sqrt 3 \)

It is given that the line parallel to line \(\sqrt 3 x + y - 2 = 0\) crosses the *\(y\)*-axis \(2\) units below the origin, i.e., it passes through point \(\left( {0,{\rm{2}}} \right)\).

Hence, the equation of the line passing through points \(\left( {0,{\rm{2}}} \right)\) and having a slope \( - \sqrt 3 \) is

\[\begin{align}y - \left( { - 2} \right) &= - \sqrt 3 \left( {x - 0} \right)\\ y + 2 &= - \sqrt 3 x\\\sqrt 3 x + y + 2 &= 0\end{align}\]

Thus, the equation of the line is \(\sqrt 3 x + y + 2 = 0\)

## Chapter 10 Ex.10.2 Question 15

The perpendicular from the origin to a line meets it at the point \(\left( { - 2,9} \right)\), find the equation of the line.

**Solution**

The slope of the line joining the origin \(\left( 0,0 \right)\) and point \(\left( - 2,9 \right)\), \(m = \frac{9 - 0}{ - 2 - 0} = - \frac{9}{2}\)

Accordingly, the slope of the line perpendicular to the line joining the origin and point \(\left( { - 2,9} \right)\) is

\[{m_2} = \frac{1}{{{m_1}}} = - \frac{1}{{\left( { - \frac{9}{2}} \right)}} = \frac{2}{9}\]

Now, the equation of the line passing through point \(\left( { - 2,9} \right)\) and having a slope \({m_2}\) is

\[\begin{align}\left( {y - 9} \right) &= \frac{2}{9}\left( {x + 2} \right)\\9y - 81 &= 2x + 4\\2x - 9y + 85 &= 0\end{align}\]

Thus, the equation of the line is \(2x - 9y + 85 = 0\)

## Chapter 10 Ex.10.2 Question 16

The length \(L\) (in centimeter) of a copper rod is a linear function of its Celsius temperature \(C.\) In an experiment, if \(L = 124.942\) when \(C = 20\) and \(L = 125.134\) when \(C = 110\), express \(L\) in terms of \(C.\)

**Solution**

It is given that when \(C = 20\), the value of \(L = 124.942\) , whereas when \(C = 110\), the value of \(L = 125.134\).

Accordingly, points \(\left( {20,124.942} \right)\) and \(\left( {110,125.134} \right)\) satisfy the linear relation between \(L\) and \(C.\)

Now, assuming \(C\) along the *\(x\)*-axis and \(L\) along the *\(y\)*-axis, we have two points i.e., \(\left( {20,124.942} \right)\) and \(\left( {110,125.134} \right)\) in the \(XY\) plane.

Therefore, the linear relation between \(L\) and \(C\) is the equation of the line passing through points \(\left( {20,124.942} \right)\) and \(\left( {110,125.134} \right)\).

\[\begin{align}\left( {L - 124.942} \right) &= \frac{{125.134 - 124.942}}{{110 - 20}}\left( {C - 20} \right)\\\left( {L - 124.942} \right) &= \frac{{0.192}}{{90}}\left( {C - 20} \right)\\L &= \frac{{0.192}}{{90}}\left( {C - 20} \right) + 124.942\end{align}\]

Thus, the required linear relation is \(L = \frac{{0.192}}{{90}}\left( {C - 20} \right) + 124.942\)

## Chapter 10 Ex.10.2 Question 17

The owner of a milk store finds that, he can sell \(980\) liters of milk each week at \(₹ 14/\rm{litre}\) and 1220 liters of milk each week at \(₹ 16/\rm{litre}\). Assuming a linear relationship between selling price and demand, how many liters could he sell weekly at \(₹\,17/\rm{litre}\)?

**Solution**

The relationship between selling price and demand is linear.

Assuming selling price per liter along the *\(x\)*-axis and demand along the *\(y\)*-axis, we have two points i.e., \(\left( {14,980} \right)\) and \(\left( {16,1220} \right)\) in the \(XY\) plane that satisfy the linear relationship between selling price and demand.

Therefore, the line passing through points \(\left( {14,980} \right)\) and \(\left( 16,1220 \right)\).

\[\begin{align} y-980&=\frac{1220-980}{16-14}\left( x-14 \right) \\ y-980&=\frac{240}{2}\left( x-14 \right) \\ y-980&=120\left( x-14 \right) \\ y&=120\left( x-14 \right)+980 \\ \end{align}\]

When \(x = 17\),

\[\begin{align}y &= 120\left( {17 - 14} \right) + 980\\&= 120 \times 3 + 980\\&= 360 + 980\\&= 1340\end{align}\]

Thus, the owner of the milk store could sell \(1340\) litres of milk weekly at ₹ \(17/\rm{litre.}\)

## Chapter 10 Ex.10.2 Question 18

\(P\left( {a,b} \right)\) is the mod-point of a line segment between axes. Show that the equation of the line is \(\frac{x}{a}+\frac{y}{b}=2\).

**Solution**

Let \(AB\) be the line segment between the axis and let \(P\left( {a,b} \right)\) be its mid-point.

Let the coordinates of \(A\) and \(B\) be \(\left( {0,y} \right)\) and \(\left( {x,0} \right)\) respectively.

Since \(P\left( {a,b} \right)\) is the mid-point of \(AB\),

\(\begin{align} &\left( {\frac{{0 + x}}{2},\frac{{y + 0}}{2}} \right) = (a,b)\\ &\Rightarrow \;\left( {\frac{x}{2},\frac{y}{2}} \right) = (a,b)\end{align}\)

\(\Rightarrow \frac{x}{2} = a\) and \(\frac{y}{2} = b\)

\( \Rightarrow x = 2a\) and \(y = 2b\)

Thus, the respective coordinates of \(A\) and \(B\) are \(\left( {0,2b} \right)\) and \(\left( {2a,0} \right)\). The equation of the line passing through points \(\left( {0,2b} \right)\)and \(\left( {2a,0} \right)\) is

\[\begin{align}\left( {y - 2b} \right) &= \frac{{\left( {0 - 2b} \right)}}{{\left( {2a - 0} \right)}}\left( {x - 0} \right)\\\left( {y - 2b} \right) &= - \frac{{2b}}{{2a}}\left( x \right)\\a\left( {y - 2b} \right) &= - bx\\ay - 2ab &= - bx\\bx + ay &= 2ab\end{align}\]

On dividing both sides by \(ab\), we obtain

\[\begin{align} & \frac{bx}{ab}+\frac{ay}{ab}=\frac{2ab}{ab} \\ & \Rightarrow \frac{x}{a}+\frac{y}{b}=2 \\ \end{align}\]

Hence, the equation of the line is \(\frac{x}{a} + \frac{y}{b} = 2\) proved.

## Chapter 10 Ex.10.2 Question 19

Point \(R\left( {h,k} \right)\) divides a line segment between the axes in the ratio \(1:2\). Find equation of the line.

**Solution**

Let AB be the line segment between the axes such that point \(R\left( {h,k} \right)\) divides \(AB\) in the ratio \(1:2\).

Let the respective coordinates of \(A\) and \(B\) be and \(\left( {0,y} \right)\).

Since point \(R\left( {h,k} \right)\) divides \(AB\) in the ratio \(1:2\), according to the section formula,

\(\begin{align}\left( {h,k} \right) &= \left( {\frac{{1\left( 0 \right) + 2\left( x \right)}}{{1 + 2}},\frac{{1\left( y \right) + 2\left( 0 \right)}}{{1 + 2}}} \right)\\\Rightarrow \left( {h,k} \right) &= \left( {\frac{{2x}}{3},\frac{y}{3}} \right)\end{align}\)

\( \Rightarrow h = \frac{{2x}}{3}\) and \(k = \frac{y}{3}\)

\( \Rightarrow x = \frac{{3h}}{2}\) and \(y = 3k\)

Therefore, the respective coordinates of \(A\) and \(B\) \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\) are \(\left( {\frac{{3h}}{2},0} \right)\) and \(\left( {0,3k} \right)\)

Now, the equation of the line \(AB\) passing through points \(\left( {\frac{{3h}}{2},0} \right)\) and \(\left( {0,3k} \right)\)is

\[\begin{align}\left( {y - 0} \right) &= \frac{{3k - 0}}{{0 - \frac{{3h}}{2}}}\left( {x - \frac{{3h}}{2}} \right)\\y &= - \frac{{2k}}{h}\left( {x - \frac{{3h}}{2}} \right)\\hy &= - 2k\left( {\frac{{2x - 3h}}{2}} \right)\\hy &= - k\left( {2x - 3h} \right)\\hy &= - 2kx + 3kh\\2kx + hy &= 3kh\end{align}\]

Thus, the required equation of a line is \(2kx + hy = 3kh\)

## Chapter 10 Ex.10.2 Question 20

By using the concept of equation of a line, prove that the three points \(\left(3,0 \right),\left( - 2, - 2 \right)\) and \(\left( 8,2 \right)\) are collinear.

**Solution**

In order to show that the points \(\left(3,0 \right),\left( - 2, - 2 \right)\) and \(\left( 8,2 \right)\) are collinear, it suffices to show that the line passing through points \(\left( 3,0 \right)\) and \(\left( - 2, - 2 \right)\) also passes through point \(\left( {8,2} \right)\).

The equation of the line passing through points \(\left(3,0 \right)\) and \(\left( - 2, - 2 \right)\) is

\[\begin{align}\left( {y - 0} \right) &= \frac{{\left( { - 2 - 0} \right)}}{{\left( { - 2 - 3} \right)}}\left( {x - 3} \right)\\y& = \frac{{ - 2}}{{ - 5}}\left( {x - 3} \right)\\5y &= 2x - 6\\2x - 5y - 6 &= 0\end{align}\]

It is observed that at \(x = 8,\;y = 2\)

\[\begin{align}LHS &= 2 \times 8 - 5 \times 2 - 6\\&= 16 - 10 - 6\\&= 0\\&= RHS\end{align}\]

Therefore, the line passing through points \(\left( 3,0 \right)\) and \(\left( - 2, - 2 \right)\) also passes through point \(\left( {8,2} \right)\).

Hence, points \(\left( 3,0 \right),\left( - 2, - 2 \right)\) and \(\left( {8,2} \right)\) are collinear, proved.