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