NCERT Solutions for Class 11 Maths Chapter 10 Straight Lines

Ans : Let ABCD be the given quadrilateral with vertices A (–4, 5), B (0, 7), C (5, –5), and D (–4, –2).
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) = area (ΔABC) + area (ΔACD)
We know that the area of a triangle whose vertices are  \(\left(x_{1}, y_{1}\right),\left(x_{2}, y_{2}\right), \text { and }\left(x_{3}, y_{3}\right)\) 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, area of ΔABC
\(\begin{aligned} &=\frac{1}{2}|-4(7+5)+0(-5-5)+5(5-7)| \text { unit }^{2} \\ &=\frac{1}{2}|-4(12)+5(-2)| \text { unit }^{2} \\ &=\frac{1}{2}|-48-10| \text { unit }^{2} \\ &=\frac{1}{2}|-58| \text { unit }^{2} \\ &=\frac{1}{2} \times 58 \text { unit }^{2} \\ &=29 \text { unit }^{2} \end{aligned}\)
Area of ΔACD
\(\begin{aligned} &=\frac{1}{2}|-4(-5+2)+5(-2-5)+(-4)(5+5)| \text { unit }^{2} \\ &=\frac{1}{2}|-4(-3)+5(-7)-4(10)| \text { unit }^{2} \\ &=\frac{1}{2}|12-35-40| \text { unit }^{2} \end{aligned}\) 
\(\begin{array}{l}{=\frac{1}{2}|-63| \text { unit }^{2}} \\ {=\frac{63}{2} \text { unit }^{2}}\end{array}\)
Thus, area (ABCD)
\(=\left(29+\frac{63}{2}\right) \text { unit }^{2}=\frac{58+63}{2} \text { unit }^{2}=\frac{121}{2} \text { unit }^{2}\) 

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 obtain
\(\begin{array}{l}{(A C)^{2}=(O A)^{2}+(O C)^{2}} \\ {\Rightarrow(2 a)^{2}=(O A)^{2}+a^{2}} \\ {\Rightarrow 4 a^{2}-a^{2}=(O A)^{2}} \\ {\Rightarrow(O A)^{2}=3 a^{2}} \\ {\Rightarrow O A=\sqrt{3} a}\end{array}\)
\(\therefore \text { Coordinates of point } \mathrm{A}=\left( \begin{array}{c}{ \pm \sqrt{3} a, 0 )}\end{array}\right.\)
 \(\begin{array}{l}{\text { Thus, the vertices of the given equilateral triangle are }(0, a),(0,-a), \text { and }(\sqrt{3} a, 0) \text { or }} \\ {(0, \text { a), }(0,-a), \text { and }}\end{array}\) \((-\sqrt{3} a, 0)\). 

Ans : The given points are \(\mathrm{P}\left(x_{1}, y_{1}\right)_{\text { and }} \mathrm{Q}\left(x_{2}, y_{2}\right)\).
(i) When PQ is parallel to the y-axis, \(x_{1}=x_{2}\).
In this case, distance between P and Q 
\(\begin{array}{l}{=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}} \\ {=\sqrt{\left(y_{2}-y_{1}\right)^{2}}} \\ {=\left|y_{2}-y_{1}\right|}\end{array}\)
(ii) When PQ is parallel to the x-axis, \(\mathrm{y}_{1}=\mathrm{y}_{2}\) .
In this case, distance between P and Q 
\(\begin{array}{l}{=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}} \\ {=\sqrt{\left(x_{2}-x_{1}\right)^{2}}} \\ {=\left|x_{2}-x_{1}\right|}\end{array}\) 

Ans : Let (a, 0) be the point on the x axis that is equidistant from the points (7, 6) and (3, 4).
\(\begin{array}{l}{\text { Accordingly, } \sqrt{(7-a)^{2}+(6-0)^{2}}=\sqrt{(3-a)^{2}+(4-0)^{2}}} \\ {\Rightarrow \sqrt{49+a^{2}-14 a+36}=\sqrt{9+a^{2}-6 a+16}} \\ {\Rightarrow \sqrt{a^{2}-14 a+85}=\sqrt{a^{2}-6 a+25}}\end{array}\)
On squaring both sides, we obtain
\(a^{2}-14 a+85=a^{2}-6 a+25\)
⇒ –14a + 6a = 25 – 85
⇒ –8a = –60
\(\Rightarrow a=\frac{60}{8}=\frac{15}{2}\)
Thus, the required point on the x-axis is \(\left(\frac{15}{2}, 0\right)\) . 

Ans : The coordinates of the mid-point of the line segment joining the points
P (0, –4) and B (8, 0) are  
It is known that the slope (m) of a non-vertical line passing through the points (x1, y1) and (x2, y2) is given by \(m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}, x_{2} \neq x_{1}\). 
Therefore, the slope of the line passing through (0, 0) and (4, –2) is
\(\frac{-2-0}{4-0}=\frac{-2}{4}=-\frac{1}{2}\). 
Hence, the required slope of the line is \(-\frac{1}{2}\).