Find the equation of the line which passes through the point (-4,3) and the portion of the line intercepted between the axes is divided internally in the ratio 5:3 by this point.

Solution:

$-4 = \frac{3 \times a + 5 \times 0}{5+3} = \frac{3a}{8}$

$\Rightarrow 3a = -32$

$\Rightarrow a = \frac{-32}{3}$

$3 = \frac{5 \times b + 3 \times 0}{5+3}$

$\Rightarrow 5b = 24$

$\Rightarrow b = \frac{24}{5}$

A($\frac{-32}{3}$ ,0) & B(0, $\frac{24}{5}$)

$\therefore$ equation of line AB

$\frac{x}{\frac{-32}{3}} + \frac{y}{\frac{24}{5}} = 1$

$\Rightarrow \frac{3 x}{-32}+\frac{5 y}{2 y}=1$

$\Rightarrow \frac{3 x}{4}-\frac{5 y}{3}=-8$

$\Rightarrow \frac{9 x-20 y}{12}=-8$

$\Rightarrow 9 x-20 y=-96$

$\Rightarrow 9 x-20 y+96=0$

Find the values of k for which the line

$(k-3)x - (4-k^2)y + k^2 - 7k + 6 = 0$ is

(i). Parrallel to x-axis (ii). Parrallel to y-axis (iii). Passes through origin.

Solution:

(i) Parallel to x-axis

$\Rightarrow$ co-efficient of x=0

$\Rightarrow k-3 = 0$

$\Rightarrow k = 3$

(ii) Parallel to y-axis

$\Rightarrow$ co-efficient of y = 0

$\Rightarrow 4 - k^2 = 0$

$\Rightarrow k^2 = 4$

$\Rightarrow k = \pm 2$

(iii) Passes through origin y = mx

y = mx + c $\Rightarrow$ c = 0

c = 0

$\Rightarrow k^2 - 7k + 6 = 0$

$\Rightarrow k^2 - 6k -k + 6 = 0$

$\Rightarrow k(K-6) -1(k-6) = 0$

$\Rightarrow (k-1)(k-6) = 0$

$\Rightarrow k = 1 or 6$

Find the equation of one of the sides of an isosceler right angled traingle whose hypotenuse is given by 3x+4y=4 and the opposite vertex of the hypotenuse is (2,2).

Solution:

Slope of AB = $- \frac{3}{4}$

Let Slope of AC = m

$\tan 45^0 = |\frac{m-m_1}{1+mm_1}|$

$\Rightarrow 1 = |\frac{m-(\frac{-3}{4})} {1 + m(\frac{-3}{4})}|$

$\Rightarrow 1 = |\frac{\frac{4m+3}{4}}{\frac{4-3m}{4}}|$

$\Rightarrow 1 = |\frac{4m+3}{4-3m}|$

$\Rightarrow \frac{4m+3}{4-3m} = \pm 1$

$\Rightarrow 4 m + 3 = 4-3m$ or $4 m + 3 = - 4 + 3 m$

$\Rightarrow 7m = 1$ or $m = -7$

$\Rightarrow m - \frac{1}{7}$ or $m = -7$

Equation of line with slope $m=\frac{1}{7}$

$y-2=\frac{1}{7}(x-2)$

$\Rightarrow 7 y-14=x-2$

$\Rightarrow x-7 y+12=0$

Again, Equation of line with slope $m=-7$

$y-2=-7(x-2)$

$\Rightarrow y-2=-7 x + 14$

$\Rightarrow 7x + y - 16=0$

If one diagonal of a square is along the line $8x-15y=0$ and one of its vertex is at (1,2), then find the equations of sides of the square passing this vertex.

Solution:

Slope of diagonal = $\frac{8}{15} (m_1)$

Let the slope of side AB = m

$\tan 45^0 = |\frac{m-m_1}{1+mm_1}|$

$\Rightarrow 1 =| \frac{m- \frac{8}{15}}{1 + m \times \frac{8}{15}}|$

$\Rightarrow 1=|\frac{15 m-8}{15+8 m}|$

$\Rightarrow \frac{15 m-8}{15+8 m}= \pm 1$

$\Rightarrow 15 m-8=15+8 m$

$15 m-8=-15-8 m$

$\Rightarrow 7 m=23 \text { or } 23 m=-7$

$\Rightarrow m=\frac{23}{7} \text { or } m= \frac{-7}{23}$

Equation of side of square with slope

$m=\frac{23}{7}$ and passing through A(1,2)

$y-2=\frac{23}{7}(x-1)$

$\Rightarrow 7 y-14=23 x-23$

$\Rightarrow 23 x-7 y-9=0$

Again,equation of side with slope $m=-\frac{7}{23}$

and passing thin $A(1,2)$ is

$y-2=-\frac{7}{23}(x-1)$

$\Rightarrow 23 y - 46 = -7 x + 7$

$\Rightarrow 7 x + 23 y - 53 = 0$

Find the image of the point (3,8) with respect to the $x+3y=7$, assuming the line to be a plane mirror.

Solution:

Slope of l is $-\frac{1}{3}$

$\therefore$ slope of PQ $\perp$ l

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

Equation of line PQ with slope 3 and passing through P(3,8)

$y-8=3(x-3) \ \Rightarrow 3x-y-1=0$

Given equation of line $x+3 y=7$

$\Rightarrow x=7-3 y$

put $x=7-3 y$ in (i)

$3(7-3 y)-y-1=0$

$\Rightarrow 21-9 y-y-1=0 \Rightarrow -10 y+20=0$

$\Rightarrow y=2$

$\therefore x=7-3 y=7-3 \times 2=7-6=1$

$\therefore$ Co-ordinate of foot of $\perp$ N is (1,2)

Since, $N$ is the mid point of PQ

$\therefore \frac{\alpha+3}{2} =1 \Rightarrow \alpha+3=2 \Rightarrow \alpha=-1$

$\frac{\beta+8}{2} =2 \Rightarrow \beta+8=4 \Rightarrow \beta=-4$

$\therefore$ Image of point P(3,8) with respect to a line

$x+3y=7$ is Q(-1,-4)

If the area of the traingle formed by a line with co-ordinate axes is $54\sqrt{3}$ square unit and the perpendicular drawn from the origin to the line makes an angle of $60^0$ with x-axis, find the equation of the line.

Solution:

Given area ($\Delta OAB$) = $54\sqrt{3}$

Let the equation of line be

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

Equation of line in normal for

$x \cos \alpha + y \sin \alpha = p$

$\Rightarrow x \cos 60^0 + y \sin 60^0 = p$

$\Rightarrow x \times \frac{1}{2}+y \times \frac{\sqrt{3}}{2}=p$

$\Rightarrow \frac{x}{2 p}+\frac{y}{\frac{2 p}{\sqrt{3}}}=1$–(ii)

on Comparring (i) & (ii), we will have

$a=2 p, b=\frac{2 p}{\sqrt{3}}$

Given, area $(\triangle O A B)=54 \sqrt{3}$ sq.unit

$\Rightarrow \frac{1}{2} \times a \times b=54 \sqrt{3}$

$\Rightarrow \frac{1}{2} \times 2 p \times \frac{2 b}{\sqrt{3}}=54 \sqrt{3}$

$\Rightarrow 2p^2 = 54 \times 3$

$\Rightarrow p^2 = 81$

$p = \pm 9$

Equation of line with $\alpha$ = $60^0$ & $p= \pm3$

$x \cos \alpha+y \sin \alpha=p$

$\Rightarrow x \cos 60^0+y \sin 60^0=3$

or

$x \cos 60^0+y \sin 60^0=-3 \Rightarrow x \times \frac{1}{2}+y \times \frac{\sqrt{3}}{2}=3$

$\Rightarrow x \times \frac{1}{2}+y \times \frac{\sqrt{3}}{2}=-3 \Rightarrow x+\sqrt{3} y-6=0$

or

$x+\sqrt{3} y+6=0$

Find the distance of the point P(4,1) from the line $4x-y=0$ measured along the line making an angle of $135^0$ with the positive direction of x-axis.

Solution:

Given equation of line l.

$4x-y=0$ (eqn-1)

Also given line $l_1$ makes angle $135^0$ with x-axis

$\therefore$ slope of $l_1 = \tan 135^0 = -1$

$y-1=-1(x-4)$

$\Rightarrow y-1=-x+4 \Rightarrow x+y-5=0$ (eqn-2)

$4 x-y=0$ (eqn-1)

$y=4 x$

put $y=4 x$ in (eqn-2)

$x+4 x-5=0$

$\Rightarrow 5x=5 \Rightarrow x=1$

$\therefore y=4$

Distance of line l from P(4,1) along the line $l_1$ is PQ

$\therefore PQ = \sqrt{(4-1)^2+ (1-4)^2}$

= $\sqrt{3^2 + 3^2}$

= $3\sqrt{2}$ unit.

8.If the line 3x+y-2=0, px + 2y - 3 = 0 and 2x-y-3 = 0 are concurrent, find the value of p

If $a_1 x + b_1 y +c_1=0$

$a_2 x + b_2 y + c_2 = 0$

$a_3 x + b_3 y + c_3 = 0$

lines are concurrent, then

$\left|\begin{array}{ccc}a_1 & b_1 & c_1 \\a_2 & b_2 & c_3 \\a_3 & b_3 & c_3\end{array}\right|$ $=0$

$a_1(b_2 c_3-b_3 c_2)-b_1(a_2 c_3-a_3 c_2) + c_1(a_1 b_3-a_3 b_2)=0$

Given equation of lines are

$3 x+y-2=0$ (i)

$p x+2 y-3=0$ (i)

$2 x-y-3=0$ (iii)

Since Given line (i), (ii) & (iii) are concurrent

$\therefore$ $\left|\begin{array}{ccc}3 & 1 & -2 \\p & 2 & -3 \\2 & -1 & -3\end{array}\right|$ $=0$

$\Rightarrow 3(-6-3)-1(-3 p+6)-2(-p-4)=0$

$\Rightarrow-27+3 p-6+2 p+8=0$

$\Rightarrow 5 p-33+8=0$

$\Rightarrow 5 p-25=0$

$\Rightarrow p=5$