# NCERT Class 11 Maths Straight Lines

# NCERT Class 11 Maths Straight Lines

## Exercise 10.1

## Exercise 10.2

## Exercise 10.3

## Miscellaneous Exercise

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Straight Lines

Exercise 10.1

## Chapter 10 Ex.10.1 Question 1

Draw a quadrilateral in the Cartesian plane, whose vertices are \(\left( { - 4,5} \right),\left( {0,7} \right),\left( {5, - 5} \right)\) and \(\left( { - 4, - 2} \right)\). Also, find its area.

**Solution**

Let \(ABCD\) be the given quadrilateral with vertices \(A\left( { - 4,5} \right),\;B\left( {0,7} \right),\;C\left( {5, - 5} \right)\) and \(D\left( { - 4, - 2} \right)\).

Then, by plotting \(A, B, C\) and \(D\) on the Cartesian plane and joining \(AB, \,BC,\, CD\) and \(DA, \)the given quadrilateral can be drawn as

To find the area of quadrilateral \(ABCD,\) we draw one diagonal, say \(AC.\)

Accordingly, \(ar\left( \square ABCD \right)=ar\left( \Delta ABC \right)+ar\left( \Delta ACD \right)\)

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

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

Therefore,

\[\begin{align}ar\left( {\Delta ABC} \right) &= \frac{1}{2}\left| { - 4\left( {7 + 5} \right) + 0\left( { - 5 + 5} \right) + 5\left( {5 - 7} \right)} \right|\\&= \frac{1}{2}\left| { - 48 + 0 - 10} \right|\\&= \frac{1}{2}\left| { - 58} \right|\\&= \frac{1}{2} \times 58\\&= 29\end{align}\]

\[\begin{align}ar\left( {\Delta ACD} \right) &= \frac{1}{2}\left| {\left( { - 4} \right)\left( { - 5 + 2} \right) + 5\left( { - 2 - 5} \right) + \left( { - 4} \right)\left( {5 + 5} \right)} \right|\\&= \frac{1}{2}\left| {12 - 35 - 40} \right|\\&= \frac{1}{2}\left| { - 63} \right|\\&= \frac{1}{2} \times 63\\&= \frac{{63}}{2}\end{align}\]

Thus,

\[\begin{align} ar\left( \square ABCD \right)&=ar\left( \Delta ABC \right)+ar\left( \Delta ACD \right) \\& =\left( 29+\frac{63}{2} \right)sq.\ unit \\& =\left( \frac{58+63}{2} \right)sq.\ unit \\& =\frac{121}{2}sq.\ unit\end{align}\]

## Chapter 10 Ex.10.1 Question 2

The base of an equilateral triangle with side \(2a\) lies along the *\(y\)*-axis such that the midpoint of the base is at the origin. Find vertices of the triangle.

**Solution**

Let \(ABC\) be the given equilateral triangle with side \(2a\).

Accordingly, \(AB = BC = CA = 2a\)

Assume that base \(BC\) lies along the *\(x\)*-axis such that the mid-point of \(BC\) is at the origin. i.e., \(BO = OC = a\), where \(O\) is the origin.

Now, it is clear that the coordinates of point \(C\left( {0,a} \right)\) , while the coordinates of point \(B\left( {0, - a} \right)\).

It is known that the line joining a vertex of an equilateral triangle with the mid-point of its opposite side is perpendicular.

Hence, vertex \(A\) lies on the *\(y\)*-axis.

On applying Pythagoras theorem to \(\Delta AOC\), we obtain

\[\begin{align} A{{C}^{2}}&=O{{A}^{2}}+O{{C}^{2}} \\ O{{A}^{2}}&=A{{C}^{2}}-O{{C}^{2}} \\ & ={{\left( 2a \right)}^{2}}+{{\left( a \right)}^{2}} \\ & =4{{a}^{2}}-{{a}^{2}} \\ & =3{{a}^{2}} \\ OA&=\pm \sqrt{3}a \end{align}\]

Therefore, coordinates of point \(A\left( { \pm \sqrt 3 a,0} \right)\)

Thus, the vertices of the given equilateral triangle are \((0,a),\;(0, - a)\) and \(\left( {\sqrt 3 a,0} \right)\) or\((0,a),\;(0, - a)\) and \(\left( { - \sqrt 3 a,0} \right)\).

## Chapter 10 Ex.10.1 Question 3

Find the distance between \(P\left( {x_1},{y_1} \right)\) and \(Q\left( {x_2},{y_2} \right)\) when: (i) \(PQ\) is parallel to the *\(y\)*-axis, (ii) \(PQ\) is parallel to the *\(x\)*-axis

**Solution**

The given points are \(P\left( {{x_1},{y_1}} \right)\) and \(Q\left( {{x_2},{y_2}} \right)\)

(i) When \(PQ\) is parallel to the *\(y\)*-axis, \({x_1} = {x_2}\).

In this case, distance between \(P\) and \( Q= \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} \)

\[\begin{align}PQ &= \sqrt {{{\left( {{y_2} - {y_1}} \right)}^2}} \\&= \left| {{y_2} - {y_1}} \right|\end{align}\]

(ii) When \(PQ\) is parallel to the *\(x\)*-axis \({y_1} = {y_2}\)

In this case, distance between \(P\) and \(Q\) \( = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} \)

\[\begin{align}PQ &= \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2}} \\&= \left| {{x_2} - {x_1}} \right|\end{align}\]

## Chapter 10 Ex.10.1 Question 4

Find a point on the \(x\)-axis, which is equidistant from the points \((7,\, 6)\) and \((3, \,4)\).

**Solution**

Let \((a,\, 0)\) be the point on the \(X\)- axis that is equidistance from the points \((7, \,6)\) and \((3,\, 4).\)

Accordingly,

\[\begin{align}&\sqrt {{{\left( {7 - a} \right)}^2} + {{\left( {6 - 0} \right)}^2}} = \sqrt {{{\left( {3 - a} \right)}^2} + {{\left( {4 - 0} \right)}^2}} \\&\Rightarrow \;\sqrt {49 + {a^2} - 14a + 36} = \sqrt {9 + {a^2} - 6a + 16} \\&\Rightarrow \;\sqrt {{a^2} - 14a + 85} = {\sqrt {{a^2} - 6a + 25} ^{}}\end{align}\]

On squaring on both sides, we obtain

\[\begin{align}&{a^2} - 14a + 85 = {a^2} - 6a + 25\\&\Rightarrow \;- 14a + 6a = 25 - 85\\&\Rightarrow\; - 8a = 60\\&\Rightarrow\; a = \frac{{60}}{8}\\&\Rightarrow \;a = \frac{{15}}{2}\end{align}\]

Thus, the required point on the *\(x\)*-axis is \(\left( {\frac{{15}}{2},0} \right)\)

## Chapter 10 Ex.10.1 Question 5

Find the slope of a line, which passes through the origin, and the mid-point of the segment joining the points \(P\left( {0, - 4} \right)\) and \(B\left( {8,0} \right)\).

**Solution**

The coordinates of the mid-point of the line segment joining the points

\(P\left( {0, - 4} \right)\) and \(B\left( {8,0} \right)\) are \(\left( {\frac{{0 + 8}}{2},\frac{{ - 4 + 0}}{2}} \right) = (4, - 2)\)

It is known that the slope (\(m\)) of a non-vertical line passing through the points \(\left( {x_1},{y_1} \right)\) and is given by \(m=\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}},{{x}_{2}}\ne {{x}_{1}}\)

Therefore, the slope of the line passing through \(\left( 0,0 \right)\) and \(\left( \text{4},-\text{2} \right)\) is

\[\frac{{ - 2 - 0}}{{4 - 0}} = - \frac{2}{4} = - \frac{1}{2}\]

Hence, the required slope of the line is \( - \frac{1}{2}\)

## Chapter 10 Ex.10.1 Question 6

Without using the Pythagoras theorem, show that the points \(\left( 4,4\right),\;\left( 3,5 \right)\) and \(\left( - 1, - 1 \right)\) are vertices of a right-angled triangle.

**Solution**

The vertices of the given triangles are \(A\left( 4,4 \right),\;B\left( 3,5 \right)\) and \(C\left( -\text{1},-\text{1} \right)\).

It is known that the slope (\(m\)) of a non-vertical line passing through the points \(\left( {x_1},{y_1} \right)\) and \(\left( {x_2},{y_2} \right)\) is given by \(m=\frac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}},{{x}_{2}}\ne {{x}_{1}}\)

Therefore, slope of AB \(\left( {{m}_{1}} \right)=\frac{5-4}{3-4}=-1\)

Slope of \(BC\) \(\left( {m_2} \right) = \frac{ - 1 - 5}{ - 1 - 3} = \frac{ - 6}{- 4} = \frac{3}{2}\)

Slope of \(CA\) \(\left( {{m_3}} \right) = \frac{{4 + 1}}{{4 + 1}} = \frac{5}{5} = 1\)

It is observed that \({m_1}{m_3} = - 1\)

This shows that line segments \(AB\) and \(CA\) are perpendicular to each other i.e., the given triangle is right-angled at \(A\left(4,\,4 \right)\).

Thus, the points \(\left( 4, 4 \right),\;\left( 3, 5 \right)\) and \(\left( - 1, - 1 \right)\) are the vertices of a right-angled triangle.

## Chapter 10 Ex.10.1 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**

If a line makes an angle of \(30^\circ \) with positive direction of the *\(y\)*-axis measured anticlockwise, then the angle made by the line with the positive direction of the *\(x\)*-axis measured anticlockwise is \(90^\circ + 30^\circ = 120^\circ \).

Thus, the slope of the given line is

\[\begin{align} \tan 120{}^\circ &=\text{ }\tan \left( 180{}^\circ -60{}^\circ \right)~ \\& =-\tan 60{}^\circ \\& =-\surd 3\end{align}\]

## Chapter 10 Ex.10.1 Question 8

Find the value of *\(x\)* for which the points \(\left( {x, - 1} \right),\;\left( {2,1} \right)\) and \(\left(4,\, 5\right)\) are collinear.

**Solution**

If points \(A\left( {x, - 1} \right),\;B\left( {2,1} \right)\) and \(C\left( 4,5\right)\) are collinear,

Then, Slope of \(AB = \) Slope of \(BC\)

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

Thus, the required value of \(x = 1\)

## Chapter 10 Ex.10.1 Question 9

Without using distance formula, show that points \(\left( - 2, - 1\right),\left( {4},0 \right),\left( 3,3 \right)\) and \(\left( - 3,2\right)\) are vertices of a parallelogram.

**Solution**

Let points \(\left( -2,-1 \right),\left( 4,0 \right),\left( 3,3 \right)\) and \(\left( -\text{3},\text{ 2} \right)\) be respectively denoted by \(A,\,B,\, C\) and \(D.\)

Slope of \(AB= \frac{{0 + 1}}{{4 + 2}} = \frac{1}{6}\)

Slope of \(CD= \frac{{2 - 3}}{{ - 3 - 3}} = \frac{{ - 1}}{{ - 6}} = \frac{1}{6}\)

Therefore, Slope of \(AB =\) Slope of \(CD\)

Hence, \(AB\) and \(CD\) are parallel to each other.

Now,

Slope of \( BC= \frac{3 - 0}{3 - 4} = \frac{3}{ - 1} = - 3\)

Slope of \( AD= \frac{2 + 1}{ - 3 + 2} = \frac{3}{ - 1} = - 3\)

Therefore, Slope of \(BC = \)Slope of \(AD\)

Hence, \(BC\) and \(AD\) are parallel to each other.

Therefore, both pairs of opposite side of quadrilateral \(ABCD\) are parallel. Hence, \(ABCD\) is a parallelogram.

Thus, points \(\left(- 2, - 1 \right),\left( 4,0 \right),\left( 3,3 \right)\) and \(\left( { - 3,2} \right)\) are the vertices of a parallelogram.

## Chapter 10 Ex.10.1 Question 10

Find the angle between the \(x\)-axis and the line joining the points \(\left(3, - 1 \right)\) and \(\left( 4, - 2\right)\)

**Solution**

The slope of the line joining the points \(\left(3, - 1\right)\) and \(\left(4, - 2 \right)\) is

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

Now, the inclination \(\left( {\rm{\theta }} \right)\) of the line joining the points \(\left(3, - 1 \right)\) and \(\left(4, - 2 \right)\) is given by

\[\begin{align}\tan \theta &= - 1\\\theta &= \left( {90^\circ + 45^\circ } \right)\\\theta & = 135^\circ \end{align}\]

Thus, the angle between the *\(x\)*-axis and the line joining the points \(\left( \text{3},-\text{1} \right)\) and \(\left( \text{4},-\text{2} \right)\) is \(\text{135}{}^\circ \).

## Chapter 10 Ex.10.1 Question 11

The slope of a line is double of the slope of another line. If tangent of the angle between them is \(\frac{1}{3}\), find the slopes of the lines.

**Solution**

We know that if \({\rm{\theta }}\) is the angle between the lines \({l_1}\) and \({l_2}\) with slopes \({m_1}\) and \({m_2}\) then

\[\tan \theta = \left| {\frac{{{m_2} - {m_1}}}{{1 + {m_1}{m_2}}}} \right|\]

It is given that the tangent of the angle between the two lines is \(\frac{1}{3}\) and slope of a line is double of the slope of another line.

Let \(m\) and \(2m\) be the slopes of the given lines

Therefore,

\(\begin{align}\frac{1}{3} &= \left| {\frac{{2m - m}}{{1 + m\left( {2m} \right)}}} \right|\\\frac{1}{3} &= \left| {\frac{m}{{1 + 2{m^2}}}} \right|\\\frac{1}{3} &= \frac{m}{{1 + 2{m^2}}}\end{align}\) or \(\begin{align}

\frac{1}{3} &= \left| {\frac{{m - 2m}}{{1 + m\left( {2m} \right)}}} \right|\\\frac{1}{3} &= \left| {\frac{{ - m}}{{1 + 2{m^2}}}} \right|\\\frac{1}{3} &= \frac{{ - m}}{{1 + 2{m^2}}}\end{align}\)

**Case I**

\[\begin{align}&\frac{1}{3} = \frac{{ - m}}{{1 + 2{m^2}}}\\&1 + 2{m^2} = - 3m\\&2{m^2} + 2m + m + 1 = 0\\&2m\left( {m + 1} \right) + 1\left( {m + 1} \right) = 0\\&\left( {m + 1} \right)\left( {2m + 1} \right) = 0 \end{align}\]

\( \Rightarrow m = - 1\) or \(m = - \frac{1}{2}\)

If \(m = - 1\), then the slopes of the lines are \( -1\) and \( - 2\).

If \(m = - \frac{1}{2}\), then the slopes of the lines are \( - \frac{1}{2}\) and \( - 1\).

**Case II**

\[\begin{align}&\frac{1}{3} = \frac{m}{{1 + 2{m^2}}}\\&2{m^2} + 1 = 3m\\&2{m^2} - 3m + 1 = 0\\& 2{m^2} - 2m - m + 1 = 0\\&2m\left( {m - 1} \right) - 1\left( {m - 1} \right) = 0\\&\left( {2m - 1} \right)\left( {m - 1} \right) = 0\end{align}\]

\( \Rightarrow m = 1\) or \(m = \frac{1}{2}\)

If \(m = 1\), then the slopes of the lines are \(1\) and \(2.\)

If \(m = \frac{1}{2}\), then the slopes of the lines are \(\frac{1}{2}\) and \(1.\)

Hence, the slopes of the lines are \( -1\) and \( -2\), or \(-\frac{1}{2}\) and \(-\text{1}\), or \(1\) and \(2,\) or \(\frac{1}{2}\) and \(1.\)

## Chapter 10 Ex.10.1 Question 12

A line passes through \(\left( {{x}_{1}},{{y}_{1}} \right)\) and \(\left( h,k \right)\). If slope of the line is \(m\), show that \(k-{{y}_{1}}=m\left( h-{{x}_{1}} \right)\)

**Solution**

The slope of the line passing through \(\left( {{x_1},{y_1}} \right)\) and \(\left( {h,k} \right)\) is \(\frac{{k - {y_1}}}{{h - {x_1}}}\)

It is given that the slope of the line is \(m\)

Therefore,

\[\begin{align}\frac{{k - {y_1}}}{{h - {x_1}}}& = m\\k - {y_1} &= m\left( {h - {x_1}} \right)\end{align}\]

Hence, \(k - {y_1} = m\left( {h - {x_1}} \right)\) proved.

## Chapter 10 Ex.10.1 Question 13

If three points \(\left( {h,0} \right),\left( {a,b} \right)\) and \(\left( {0,k} \right)\) lie on a line, show that \(\frac{a}{h} + \frac{b}{k} = 1\)

**Solution**

If three points \(A\left( {h,0} \right),B\left( {a,b} \right)\) and \(C\left( {0,k} \right)\) lie on a line, then

Slope of \(AB =\) Slope of \(BC\)

\[\begin{align}\frac{b - 0}{a - h} &= \frac{k - b}{0 - a}\\\frac{b}{a - h} &= \frac{k - b}{{ - a}}\\- ab& = (k - b)(a - h)\\- ab &= ka - kh - ab + bh\\ka + bh &= kh\end{align}\]

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

\[\begin{align}\frac{ka}{kh} + \frac{bh}{kh} &= \frac{kh}{kh}\\\frac{a}{h} + \frac{b}{k}& = 1\end{align}\]

## Chapter 10 Ex.10.1 Question 14

Consider the given population and year graph. Find the slope of the line \(AB\) and using it, find what will be the population in the year 2010?

**Solution**

Since line \(AB\) passes through points \(A\left( {1985,92} \right)\) and \(B\left( {1995,97} \right)\), its slope is

\[\begin{align} \frac{97-92}{1995-1985}&=\frac{5}{10} \\& =\frac{1}{2}\end{align}\]

Let \(y\) be the population in the year 2010.\(A\left( {1985,92} \right)\)

Then, according to the given graph, line \(AB\) must pass through point \(C\left( {2010,y} \right)\).

Therefore, Slope of \(AB =\) Slope of BC

\[\begin{align}\frac{1}{2} &= \frac{{y - 97}}{{2010 - 1995}}\\\frac{1}{2} &= \frac{{y - 97}}{{15}}\\\frac{{15}}{2} &= y - 97\\y &= 7.5 + 97\\y& = 104.5\end{align}\]

Thus, the slope of line \(AB\) is \(\frac{1}{2}\), while in the year 2010, the population will be \(104.5\) crores.