Two vertices of a traingle are $B(5,-1)$ and $c(-2,3)$. If the orthocentre of the traingle is at origin, find the third vertex.

Solution: Slope of $OB =\frac{0+1}{0-5}=-\frac{1}{5}$

slope $B E=-\frac{1}{5}$

$\therefore \text { Slope f } A C=\frac{-1}{-1 / 5}=5$

as, $\quad B E \perp\text { AC }$

Slope of $C F=\frac{0-3}{0+2}=-\frac{3}{2}$

Since $C F \perp A B$

$\Rightarrow \text { slope of } A B=\frac{-1}{-3/2}=2 / 3$

Equation of $A C$ with slope $=5 \rightarrow$ passing through $c(-2,3)$

$y-3=5(x+2)$

$\Rightarrow \quad 5 x-y+13=0 \cdots (1)$

Again equation of line $A B$ with slope: $2 / 3$ and passing through $B(5,-1)$

$y+1=\frac{2}{3}(x-5)$

$\Rightarrow 3 y+3=2 x-10$

$\Rightarrow 2 x-3 y-13=0 \cdots(2)$

from (1) and (2)

$5 x-y+13=0$

$\Rightarrow y=5 x+13$

put $y=5 x+13$ in (2)

$2 x-3(5 x+13)-13=0$

$\Rightarrow-13 x-52=0$

$\Rightarrow \quad x=-4$

$\therefore y=5 x+13$

$=5 x(-4)+13=-7$

$\therefore$ This vertex $A(-4,-7)$

A straight line is drawn through the point $P(1,0)$. It intersects the straight line $y=2 x-3$ at the point $Q$. Find the slopes of the straight line $P Q$ if $P Q=\sqrt{2}$.

Solution:

Given equation of line $y=2 x-3$

Let $Q(\alpha, \beta)$

Q lies on $y=2 x-3$

$\therefore \beta=2 \alpha-3$

$\therefore \quad Q(\alpha, 2 \alpha-3)$

$A / q \quad P Q=\sqrt{2}$

$\Rightarrow \quad P Q^2=2$

$\Rightarrow \quad(\alpha-1)^2+(2 \alpha-3)^2=2$

$\Rightarrow \quad \alpha^2-2 \alpha+1+4 \alpha^2$

$=12 \alpha+9=2$

$\Rightarrow \quad 5 \alpha^2-14 \alpha+10=2$

$\Rightarrow \quad 5 \alpha^2-14 \alpha+8=0$

$\Rightarrow 5 \alpha^2-10 \alpha-4 \alpha+8=0$

$\Rightarrow \quad 5 \alpha(\alpha-2)-4(\alpha-2)=0$

$\Rightarrow \quad(5 \alpha-4)(\alpha-2)=0$

$\Rightarrow \quad \alpha=\frac{4}{5}, 2$

$\therefore \beta =2 \alpha-3$

$\Rightarrow \beta =2 \times \frac{4}{5}-3$

or $\quad\beta=2 \times 2-3$

$\beta=\frac{8-15}{5}$

$\Rightarrow \beta =-\frac{7}{5}$

or $\beta=1$

$\therefore Q = (\frac{4}{5},-\frac{7}{5})$

$Q^{\prime}=(2,1)$

Given $P(1,0)$

$\therefore \text { slope of } P Q=\frac{-7 / 5-0}{4 / 5-1}$

$=\frac{-7 / 5}{-1 / 5}=7$

Slope of $P Q^{\prime}=\frac{1-0}{2-1}=\frac{1}{1}=1$

Show that the four lines $a x \pm b y \pm c=0$ enclose a rhombus whose area is $2 c^2 / a b$.

Solution:

Given equation

$a x \pm b y \pm c =0$

i.e. $\ a x+b y+c =0-(i)$

$a x+b y-c =0-(ii)$

$a x-b y+c =0 - (iii)$

$a x-b y-c =0 - (iv)$

$a x+b y+c=0$

$\Rightarrow \frac{x}{-c / a}+\frac{y}{-c / b}=1$

$a x+b y-c=0$

$\Rightarrow \frac{x}{c / a}+\frac{y}{c / b}=1$

$a x-b y+c=0$

$\Rightarrow \frac{x}{-c / a}+\frac{y}{c / b}=1$

$a x-b y-c=0$

$\Rightarrow \frac{x}{c / a}+\frac{y}{-c / b}=1$

$B D=\sqrt{O+(\frac{c}{b}+\frac{c}{b})^2}=\frac{2 c}{b}$

$A C=\sqrt{(\frac{c}{a}+\frac{c}{a})^2+0}=\frac{2 c}{a}$

$\text { Area of rhombus } =\frac{1}{2} \times d_1 \times d_2$

$=\frac{1}{2} \times B D \times A C$

$=\frac{1}{2} \times \frac{2 c}{b} \times \frac{2 c}{a}$

$=\frac{2 c^2}{a b}$

Two sides of a square lie on the lines $x+y=1$ and $x+y=-2$. Find its area.

Solution:

Given equation of lines

$x+y=1 \cdots(i)$

$x+y=-2\cdots(ii)$

$\because$ slopes of lines (i) and (ii) are equal that is slope $=-1$

$\therefore \quad$ line are parallel distance between parallel lines

$=\frac{|c_2-c_1|}{\sqrt{a^2+b^2}}$

in the given equation $c_1=-1$ and $c_2=2$

$\therefore \text {distance}$

$(d) =\frac{|c_2-c_1|}{\sqrt{a^2+b^2}}$

$=\frac{|2+1|}{\sqrt{1^2+1^2}} =\frac{3}{\sqrt{2}}$

$\therefore$ length of sides of square.

$=$ distance between given parallel lines $=\frac{3}{\sqrt{2}}$

$\text { Area of square }$

$=d^2=\left(\frac{3}{\sqrt 2}\right)^2$

$=\frac{9}{2} \text { square units. }$

If the line joining two points $A(2,0)$ and $B(3,1)$ is rotated about $A$ in anti-clockwise direction through an angle of $15^{\circ}$, find the equation of the line in new position.

Solution:

Let- the line makes angle $\theta$ with $x$-axis.

$\text { slope of line }=\tan \theta$

$\Rightarrow \frac{1-0}{3-2}=\tan \theta$

$\Rightarrow \tan\theta =1 =\tan 45^{\circ}$

$\Rightarrow \ \theta=45^{\circ}$

In new position when line rotates about $A(2,0)$ in anticlockwise then resultant angle with $x$-axis

$=\theta+15^{\circ}$

$=45^{\circ}+15^{\circ}$

$=60^{\circ}$

$\therefore$ slope of line in new position.

$=\tan 60^{\circ}$

$=\sqrt{3}$

$\therefore$ Equation of line with slolpe $\sqrt 3$ and passing through the point $A(2,0)$

$y-0=\sqrt{3}(x-2)$

$\Rightarrow \quad y=\sqrt{3} x-2 \sqrt{3}$

$\Rightarrow \quad \sqrt{3} x-y-2 \sqrt{3}=0$

Find the equation of the bisector of $\angle A$ of $\triangle A B C$, whose vertices are $A(4,3), B(0,0)$ and $C(2,3)$

Solution:

$\frac{A B}{A C}=\frac{B D}{D C}$

$A B=\sqrt{4^2+3^2}=\sqrt{25}=5$

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

$\therefore \frac{B D}{D C}=\frac{A B}{A C}=\frac{5}{2}$

Point $D(\frac{2 \times 0+5 \times 2}{5+2}, \frac{5 \times 3+2 \times 0}{5+2})$

i.e.,

$D(\frac{10}{7}, \frac{15}{7})$

Equation of $A D$

$(y-3) =\frac{\frac{15}{7}-3}{\frac{10}{7}-4}(x-4)$

$\Rightarrow(y-3) =\frac{1}{3}(x-4)$

$\Rightarrow 3 y-9 =x-y \Rightarrow x-3 y+5=0$

A ray of light -passing through the point $P(1,2)$ reflects on $x$-axis at the point $A$ and the reflected ray passes through the point $Q(5,3)$.

Find the coordinate of $A$.

Solution:

Let, reflected ray $AQ$ makes angle $\theta$ with the $x$-axis

$\therefore$ slope of $AQ=\tan \theta$

$AP$ makes angle $(\pi-\theta)$ with $x$-axis

$\therefore \text { slope of } A P=\tan (\pi-\theta)=-\tan \theta$

$\therefore$ Slope of $A Q=-$ slope of $A P$

$\Rightarrow \quad \frac{3-0}{5-a}=-\frac{2-0}{1-a}$

$\Rightarrow 3(1-a)=-2(5-a) \Rightarrow \ 3-3 a=-10+2 a$

$\Rightarrow \quad-5 a=-13$

$\Rightarrow \quad 5 a=13 \Rightarrow \ a=13 / 5$

$\therefore \quad \text { Required Point }A(a, 0)$

$\text {i.e. }\ \ A(13 / 5,0)$

Find the equation of the stright line Which cuts off intercept on $x$-axis which is twice that on $y$-axis and is at a unit distance from the origin.

Solution:

Let the equation of line be

$\frac{x}{2 a}+\frac{y}{a}=1$

$\Rightarrow x+2 y=2 a$

$\Rightarrow x+2 y-2 a=0\cdots (i)$

According to the question, distance of line (i) from origin is 1 unit.

$\therefore |\frac{0+2 \times 0-2 a}{\sqrt{1^2+2^2}}|=1$

$\Rightarrow|\frac{-2 a}{\sqrt{5}}|=1$

$\Rightarrow\frac{2 a}{\sqrt{5}}= \pm 1$

$\Rightarrow 2 a= \pm \sqrt{5}$

$\Rightarrow a= \pm \frac{\sqrt{5}}{2}$

$\therefore \quad$ Required equation of line are

$x+2y=2 a$

$\Rightarrow x+2 y= \pm 2 \times \frac{\sqrt{5}}{2}$

$\Rightarrow \quad x+2 y \pm \sqrt{5}=0$