S Chand CLASS 10 Chapter 11 Coordinate Geometry Exercise 11 B

 Exercise 11 B

Question 1 

Ans: If the inclination of line to the positive direction is θ then slope of that line will be tan θ

Therefore. 

(a) $\tan 30^{\circ}=\frac{1}{\sqrt{3}} \quad$ (b) $\tan 45^{\circ}=1 \quad$ (c) $\tan 60^{\circ}=\sqrt{3}$

(d) $\tan 15^{\circ}=0.2679 \quad$ (e) $\tan 75^{\circ}=3.7231$

Question 2

Ans: (a) Points are (1,2) and (5,6)

So slope (m) = $\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{6-2}{5-1}=\frac{4}{4}=1$

Angle of inclination $=45^{\circ}$ (if $\tan 45^{\circ}=1$ )

(b) Points are $(0,0)$ and $(\sqrt{3}, 3)$

So slope $(m)=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{3-0}{\sqrt{3-0}}=\frac{3}{\sqrt{3}}$

$=\frac{3 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}=\frac{3 \sqrt{3}}{3}=\sqrt{3}$

So Angle of inclination $=60^{\circ}$

Question 3

Ans: $\begin{aligned}&\Rightarrow 1=\frac{a-3}{-2} \\&\Rightarrow-2=a-3 \\&\Rightarrow a=-2+3=1\end{aligned}$

Hence $a=1$

Question 4

Ans: $A B C D E F$ is a regular hexagon in which side $A B$ and $E D$ are parallel to $x$-axis. 

Now slope of $A B=\tan 0^{\circ}=0$

(TO BE ADDED)

Slope of $B C=\tan 60^{\circ}=\sqrt{3}$

Slope of $C O=\tan 120^{\circ}=-53$

Stope of $E D=\tan 0^{\circ}=0$

slope of $E F=\tan 60^{\circ}=\sqrt{3}$ and slope of $F A=\tan 130^{\circ}=-\sqrt3$

Question 5

Ans: we know that slope $(m)=\frac{y_{2}-y_{1}}{n_{2}-n_{1}}$

$\Rightarrow \frac{-4}{3}=\frac{y-11}{2-(-8)}$

$\Rightarrow \frac{-4}{3}=\frac{y-11}{2+8}$

$\Rightarrow \frac{-4}{3}=\frac{y-11}{10}$

$\Rightarrow 3 y=-33=-40$

$\Rightarrow 3 y=-40+33$

$\Rightarrow 3 y=-7$

So $=y=\frac{-7}{3}$

Question 6

Ans: Slope (m1) of the line passing through the Points $(-5,-8)$ and $(3,0)$

$=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{0-(-8)}{3-(-5)}$

$=\frac{8}{3+5}=\frac{8}{8}=1$

and slope $\left(\mathrm{m}_{2}\right)$ of the line passing through the Points $(6.3)$ and $(4,9)=\frac{9-3}{4-6}=\frac{9-3}{-2}$

If The two lines are parallel 

so their slopes are equal $\Rightarrow m_{1}=m_{2}$

$=\frac{-(b+k)}{-(f+h)}=\frac{b+k}{f+h}$

(d) Points are $\left(n_{1}, y_{1}\right)$ and $\left(n_{2}, y_{2}\right)$

So slope of the line Joining there points $=m$

$=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

Slope of the perpendicular to it $=\frac{-1}{m}$

$=\frac{x_{2}-x_{1}}{y_{2}-y_{1}}$

Question 7

Ans: we know that slope of a line is $m$ then Slope of the line perpendicular to the Given line will be $\frac{-1}{m}$

(a) Slope of the line ( m) =$\frac{1}{3}$

Slope of its perpendicular $=\frac{-1}{m}=\frac{-3}{1}=-3$

(b) Slope of the line (m) $=\frac{-5}{6}$

So slope of its penpendicular $=\frac{-1}{m}=-\left(\frac{-6}{5}\right)=\frac{6}{5}$

(c) slope of the line $(m)=5$

So slope of its perpendicular $=\frac{-1}{m}=\frac{-1}{5}$

(d)slope of the line $(m)=-5 \frac{1}{7}=\frac{-26}{7}$

So slope of ate penpen dicular $=\frac{-1}{m}=-\left(\frac{-7}{36}\right)$

$=\frac{7}{36}$

(e) Slope of the line (m) =0 

So slope of its perpendicular $\left(\frac{-1}{m}\right)=\frac{-1}{0}=$ infinite 

(f) slope of the line (m) = infinite 

So slope of its perpendicular = $\frac{-1}{m}=\frac{-1}{\text { infinite }}$

=0

Question 8

Ans: (a) Point are $(0,8)$ and $(-5,2)$

So slope of the line joining there points $(\mathrm{(m)}$

$\begin{aligned}&=\frac{y_{2}-y_{1}}{x_{2}-x_{1}} \\&=\frac{2-8}{-5-0}=\frac{-6}{-5}=\frac{6}{5}\end{aligned}$

So slope of the line perpendicular to it $\left(\frac{-1}{m}\right)$ $=\frac{-5}{6}$

(b) Points are $(1,-11)$ and $(5,2)$.

So slope of the line Joining there points

$=m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{2-(-11)}{5-1}=\frac{2+11}{5-1}=\frac{13}{4}$

So slope of the line perpendicular to it $=\frac{-1}{m}$ $=\frac{-4}{13}$

(c) Points are $(-k, h)$ and $(b,-f)$

So slope of the line joining these points 

$=m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=$ $\frac{-f-h}{b-(-k)$

$=\frac{-(f+h)}{b+k}$

Question 9

Ans:  In rectangle $A B C D$

slope of $A B=\frac{5}{6}$, There fore

(IMAGE TO BE ADDED)

(a) Slope of $B C=\frac{-1}{m}$

$=\frac{-6}{5}$

(b) Slope of CD $=m$

$=\frac{5}{6}$

(C) Slope of $D A=\frac{-1}{m}$

$=\frac{-6}{5}$

Question 10

Ans:  In parallelogram $A B C D, A B \| C D$ and $D A \| B C$

Slope of $A B=-2$ and slope of $B C=\frac{3}{5}$

(a) So DA|| BC

So slope of AD = Slope of BC = $\frac{3}{5}$

(b) If $C D \| A B$

So slope of $C D=$ slope of $A B=-2$

(c) Slope of altitude of $A B=-\left(\frac{1}{\text { slope of AD }}\right)$ $=\frac{-5}{3}$

(d) Slope of the altitude of $C D=\frac{-1}{\text { slope of } C D}=\left(\frac{-1}{-2}\right)=\frac{1}{2}$

Question 11

Ans: Vertices of a $\triangle A B C$ are $A(1,1) B(7,3)$ and $C(3,6)$

(IMAGE TO BE ADDED)

(a) so $\text { slope of } A B=\frac{y_{2}-y_{1}}{n_{2-x_{1}}}$

$=\frac{3-1}{7-1}=\frac{2}{6}=\frac{1}{3}$

and slope of its altitude $=\frac{-1}{\text { Slope of } A B}$

$=\frac{-3}{1}=-3$

Similarly 

(b) Slope of BC = $\frac{6-3}{3-7}=\frac{3}{-4}$

and slope of altitude of $B C=\frac{-1}{\text { slope of } B C}$

$=-\left(\frac{-4}{3}\right)=\frac{4}{3}$

(c) Slope of $A C=\frac{6-1}{3-1}=\frac{5}{2}$

So slope of altitude of $A C=\frac{-1}{\text { Slope of } A C}$

$=\frac{-2}{5}$

Question 12

Ans: Slope of line joining the points (-5,7) and (0,-2)

$=m_{1}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$=\frac{-2-1}{0-(-5)}=\frac{-9}{5}$

and slope of the line joining the points 

$(1,-3)$ and $(4, x)=m_{2}=\frac{x-(-3)}{4-1}=\frac{x+3}{3}$

If these line are perpendicular to each other 

So , $m_{1} \times m_{2}=-1$

$\Rightarrow \frac{-9}{5} \times \frac{x+3}{3}=-1$

$\Rightarrow \frac{-3(x+3)}{5}=-1$

$\Rightarrow x+3=-1 \times \frac{5}{-3}=\frac{5}{3}$

$\Rightarrow x=\frac{5}{3}-3$

$=\frac{5-9}{3}$

$=\frac{-4}{3}$

Hence x = $\frac{-4}{3}$

Question 13

Ans: Vertices of a quad PMQS are p$(0,0), m(3,2)$Q $(7,7)$ and $s(4,5)$

Slope of PM $\left(m_{1}\right)=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{2-0}{3-0}=\frac{2}{3}$

Similarly, 

slope of $m Q=\left(m_{2}\right)=\frac{7-2}{7-3}=\frac{5}{4}$

slope of QS $\left(m_{3}\right)=\frac{5-7}{4-7}=\frac{-2}{-3}=\frac{2}{3}$

and slope of $s p=\frac{5-0}{4-0}=\frac{5}{4}$

If slope of $\mathrm{pm}=$ slope of $QS=\frac{2}{3}$

So PM||QS ..........(i)

Similarly slope of $m \theta=$ slope of $s p=\frac{s}{4}$

So $M Q || S P$

so from ii and (iii)

PMQS is a parallelogram

Question 14

Ans: co - ordinater of $P=(a, b) \cdot Q(a-3, b+4)$ $R(a-1, b+7)$, S $(a-4, b+3)$

So $P Q=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

$\begin{aligned}&=\sqrt{(a+3-a)^{2}+(b+4-b)^{2}} \\&=\sqrt{(3)^{2}+(4)^{2}}=\sqrt{9+16} \\&=\sqrt{25}=5\end{aligned}$

Similarly 

$\begin{aligned} Q R &=\sqrt{(a-1-a-3)^{2}+\left(b+7-b-41^{2}\right.} \\ &=\sqrt{(-4)^{2}+(3)^{2}}=\sqrt{16+9}=\sqrt{25}=5 \\ \text { RS } &=\sqrt{(a-4-a+1)^{2}+\left(b+3-b-71^{2}\right.} \\ &=\sqrt{131^{2}+(-4)^{2}} \\ &=\sqrt{9+16} \\ &=\sqrt{25} \\ &=5 \\ \text { and sp } &=\sqrt{(a-a+4)^{2}+(b-b-3)^{2}} \end{aligned}$

$=\sqrt{(4)^{2}+(-3)^{2}}$

$=\sqrt{16+9}=\sqrt{25}$

$=5$

Diagonal PR = $\sqrt{(a-1-a)^{2}+(b+7-b)^{2}}$

$=\sqrt{(-1)^{2}+(7)^{2}}=\sqrt{1+49}=\sqrt{50}$

and diagonal

$\begin{aligned} \theta S &=\sqrt{(a-4-a-3)^{2}+(b+3-b-4)^{2}} \\ &=\sqrt{(-7)^{2}+(-1)^{2}}=\sqrt{49+1} \\ &=\sqrt{50} \end{aligned}$

IF all the sides PQ,QR, RS and SP are equal and both diagonal PR and QS are also equal 

So PQRS is a square 

Now area of square PQRS = (side)^{2}

$=(P R)^{2}=(5)^{2}=25$ square units

Question 15

Ans: Vertices of a $\triangle A B C$ are $A(0,4) \cdot B(1,2)$ and $C(3,3)$

Now slope of $A B\left(m_{1}\right)=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{2-4}{1-0}$

$=\frac{-2}{1}=-2$

Similarly 

Slope of BC $\left(m_{2}\right)=\frac{3-2}{3-1}=\frac{1}{2}$

and slope of $C A\left(m_{3}\right)=\frac{4-3}{0-3}=\frac{1}{-3}$

if $m_{1} \times m_{2}=-2 \times \frac{1}{2}=-1$

So $A D$ and $B C$ are pen perpendicular to each other So $\triangle A B C$ is a right angled triangle.

Question 16

Ans: Pointt are $A(x, 1), B(1,2)$ and $C(0, y+1)$

So

slope of $A B=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{2-1}{1-x}=\frac{1}{1-x}$

similarly slope of $B C=\frac{y+1-2}{0-1}=\frac{y-1}{-1}$

if $A, B$ and $C$ are collinear

So slope of $A B=$ slope of $B C$

$\Rightarrow \frac{1}{1-x}=\frac{y-1}{-1}$

$\Rightarrow(1-x)(y-1)=-1$

$\Rightarrow y-1-x y+x=-1$

$\Rightarrow x+y=-1+1+x y$

$\Rightarrow y+x=x y$

Dividing by x y

$\begin{aligned}\frac{y}{x y}+\frac{x}{x y} &=\frac{x y}{x y} \\\Rightarrow \frac{1}{x}+\frac{1}{y} &=1\end{aligned}$

Hence $\frac{1}{x}+\frac{1}{y}=1$