NCERT Solutions for Class 11 Maths Chapter 8 - Straight Lines Exercise 8.1

Exercise 8.1

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.

Ans: 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, area (ABCD) $ = (\Delta ABC) + $ area$(\Delta ACD)$

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 $\Delta ABC$

\[\begin{array}{l}= \frac{1}{2}\left| { - 4\left( {7 + 5} \right) + 0\left( { - 5 - 5} \right) + 5\left( {5 - 7} \right)} \right|uni{t^2}\\= \frac{1}{2}\left| { - 4\left( {12} \right) + 5\left( { - 2} \right)} \right|uni{t^2}\\= \frac{1}{2}\left| { - 48 - 10} \right|uni{t^2}\\ = \frac{1}{2}\left| { - 58} \right|uni{t^2}\\ = \frac{1}{2} \times 58uni{t^2}\\= 29uni{t^2}\end{array}\]

Area of $\Delta ACD$

\[\begin{array}{l}= \frac{1}{2}\left| { - 4\left( { - 5 + 2} \right) + 5\left( { - 2 - 5} \right) + \left( { - 4} \right)\left( {5 + 5} \right)} \right|uni{t^2}\\= \frac{1}{2}\left| { - 4\left( { - 3} \right) + 5\left( { - 7} \right) + \left( { - 4} \right)\left( {5 + 5} \right)} \right|uni{t^2}\\= \frac{1}{2}\left| { - 4\left( { - 3} \right) + 5\left( { - 7} \right) + \left( { - 4} \right)\left( {10} \right)} \right|uni{t^2}\\= \frac{1}{2}\left| {12 - 35 - 40} \right|uni{t^2}\\ = \frac{1}{2}\left| { - 63} \right|uni{t^2}\\= \frac{{63}}{2}uni{t^2}\end{array}\]

Thus, area (ABCD) $ = \left( {29 + \frac{{63}}{2}} \right)uni{t^2} = \frac{{58 + 36}}{2}uni{t^2} = \frac{{121}}{2}uni{t^2}$

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.

Ans: Let ABC be the given equilateral triangle with side 2a. 

Accordingly, AB = BC = CA = 2a 

Assume that base BC lies along the y-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 are (0, a), while the coordinates of point B are (0, -a). 

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’ll get

\[\begin{array}{l}{\left( {AC} \right)^2} = {\left( {OA} \right)^2} + {\left( {OC} \right)^2}\\ \Rightarrow {\left( {2a} \right)^2} = {\left( {OA} \right)^2} + {a^2}\\\Rightarrow 4{a^2} - {a^2} = {\left( {OA} \right)^2}\\\Rightarrow {\left( {OA} \right)^2} = 3{a^2}\\\Rightarrow \left( {OA} \right) = \sqrt 3 a\end{array}\]

\[\therefore \] Coordinates of point A \[ = \left( { \pm \sqrt 3 a,0} \right)\]

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

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.

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

1. When PQ is parallel to y-axis, $\left( {{x_1} = {x_2}} \right)$

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

\[\begin{array}{l}= \sqrt {{{({y_2} - {y_1})}^2}} \\= \left| {{y_2} - {y_1}} \right|\end{array}\]

2. When PQ is parallel to x-axis, $\left( {{y_1} = {y_2}} \right)$

In this case, distance between P and Q \[ = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{({y_2} - {y_1})}^2}} \]\[\begin{array}{l}= \sqrt {{{({x_2} - {x_1})}^2}} \\ = \left| {{x_2} - {x_1}} \right|\end{array}\]

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

Ans: Let $\left( {a,0} \right)$ be the point on the x-axis that is equidistance from the points $\left( {7,6} \right)$ and $\left( {3,4} \right).$

Accordingly, $\sqrt {{{\left( {7 - a} \right)}^2} + {{\left( {6 - 0} \right)}^2}}  = \sqrt {{{\left( {3 - a} \right)}^2} + {{\left( {4 - 0} \right)}^2}} $

\[\begin{array}{l}\Rightarrow \sqrt {49 + {a^2} - 14a + 36}  = \sqrt {9 + {a^2} - 6a + 16} \\\Rightarrow \sqrt {{a^2} - 14a + 85}  = \sqrt {{a^2} - 6a + 25} \end{array}\]

On squaring both sides, we’ll get

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

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

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 $\left( {8,0} \right)$. 

Ans: 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  $\left( {\frac{{0 + 8}}{2},\frac{{ - 4 + 0}}{2}} \right) = \left( {4, - 2} \right)$

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, the slope of the line passing through $\left( {0,0} \right)$ and $\left( {4, - 2} \right)$ is $\frac{{ - 2 - 0}}{{4 - 0}} =  - \frac{2}{4} =  - \frac{1}{2}$

Hence, the required slope of the line is$ - \frac{1}{2}$.

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.

Ans: The vertices of the given triangle are A$\left( {4,4} \right),$ B$\left( {3,5} \right)$and C$\left( { - 1, - 1} \right)$. 

It is known that the slope (m) of a non-vertical line passing through the $\left( {{x_1},{y_1}} \right)$ 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 \[\left( {{m_1}{m_3}} \right) =  - 1\] 

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 $\left( { - 1, - 1} \right)$ the vertices of a right-angled triangle.

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

Ans: 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 \[\tan 120^\circ  = \tan \left( {180^\circ  - 60^\circ } \right) =  - \tan 60^\circ  =  - \sqrt 3 \]

8. 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. 

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

Slopes of AB = $\frac{{0 + 1}}{{4 + 2}} = \frac{1}{6}$ 

Slopes of CD = $\frac{{2 - 3}}{{ - 3 - 3}} = \frac{{ - 1}}{{ - 6}} = \frac{1}{6}$

$ \Rightarrow $Slope of AB = Slope of CD 

$ \Rightarrow $AB and CD are parallel to each other. 

Now, slope of BC = $\frac{{3 - 0}}{{3 - 4}} = \frac{3}{{ - 1}} =  - 3$

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

$ \Rightarrow $Slope of BC = Slope of AD 

$ \Rightarrow $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. 

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

Ans: The slope of the line joining the points $\left( {3, - 1} \right)$and $\left( {4, - 2} \right)$is $m = \frac{{ - 2 - \left( { - 1} \right)}}{{4 - 3}} =  - 2 + 1 =  - 1$

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

\[ \Rightarrow \theta  = \left( {90^\circ  + 45^\circ } \right) = 135^\circ \] 

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

10. 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 slope of the lines. 

Ans: Let \[{m_1}\], \[m\]be the slopes of the two given lines such that \[{m_1} = 2m\]. 

We know that if \[\theta \] is the angle between the lines \[{l_1}\] and \[{l_2}\] with slopes m and 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 \[\frac{1}{3}\].

\[\begin{array}{l}\therefore \frac{1}{3} = \left| {\frac{{m - 2m}}{{1 + \left( {2m} \right).m}}} \right|\\ \Rightarrow \frac{1}{3} = \left| {\frac{{ - m}}{{1 + 2{m^2}}}} \right|\\ \Rightarrow \frac{1}{3} = \frac{m}{{1 + 2{m^2}}}\end{array}\]

Now, Case I:

\[ \Rightarrow \frac{1}{3} = \frac{{ - m}}{{1 + 2{m^2}}}\]

\[\begin{array}{l}\Rightarrow 1 + 2{m^2} =  - 3m\\ \Rightarrow 2{m^2} + 3m + 1 = 0\\ \Rightarrow 2{m^2} + 2m + m + 1 = 0\end{array}\]

\[\begin{array}{l} \Rightarrow 2m\left( {m + 1} \right) + 1\left( {m + 1} \right) = 0\\ \Rightarrow \left( {m + 1} \right)\left( {2m + 1} \right) = 0\\ \Rightarrow m =  - 1 or m =  - \frac{1}{2}\end{array}\]

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 \[ - 1\]

Now, Case II:

\[\frac{1}{3} = \frac{m}{{1 + 2{m^2}}}\]

\[\begin{array}{l}\Rightarrow 2{m^2} + 1 = 3m\\ \Rightarrow 2{m^2} - 3m + 1 = 0\\\Rightarrow 2{m^2} - 2m - m + 1 = 0\\ \Rightarrow 2m\left( {m - 1} \right) - 1\left( {m - 1} \right) = 0\\\Rightarrow \left( {m - 1} \right)\left( {2m - 1} \right) = 0\\ \Rightarrow m = 1 or m = \frac{1}{2}\end{array}\]

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 \[ - 1\] or \[1\]and \[2\]or \[\frac{1}{2}\] and \[1\]

11.  A line passes through $\left( {{x_1},{y_1}} \right)$ and $\left( {h,k} \right)$. If slope of the line is \[m\], show that $k - {y_1} = m\left( {h - {x_1}} \right)$

Ans: 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 \[m\]. 

\[\therefore \]$\frac{{k - {y_1}}}{{h - {x_1}}} = m$

$ \Rightarrow k - {y_1} = m\left( {h - {x_1}} \right)$

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

Conclusion

Ex 8.1 Class 11 Maths NCERT Solutions has provided an introduction to the basic principles of straight lines in coordinate geometry. You have learned how to determine the slope of a line from two given points and how to express the equation of a line in various forms, such as slope-intercept form, point-slope form, and general form. By practising these problems, you have built a strong foundation in understanding and working with straight lines. This knowledge is essential for solving more complex geometric problems and will be beneficial in advanced mathematical studies.

Class 11 Maths Chapter 8: Exercises Breakdown

CBSE Class 11 Maths Chapter 8 Other Study Materials

Chapter-Specific NCERT Solutions for Class 11 Maths

Given below are the chapter-wise NCERT Solutions for Class 11 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

Important Related Links for CBSE Class 11 Maths