Browse Questions

# Find the equations of the lines through the point of intersection of the lines $x-y+1=0$ and $2x-3y+5 = 0$ and whose distance from the point (3, 2) is $\large\frac{7}{5}$

$\begin {array} {1 1} (A)\;4x+3y-1=0 ; 3x+4y+6=0 & \quad (B)\;4x-3y+1=0 ; 3x-4y+6=0 \\ (C)\;4x-3y-1=0 ; 3x-4y-6=0 & \quad (D)\;4x+3y-1=0 ; 3x+4y-6=0 \end {array}$

Toolbox:
• Equation of a line passing through the intersection of two lines is $a_1x+b_1y+c_1 + \lambda (a_2x+b_2y+c_2)=0$
• Distance of a line $ax+by+c=0$ from a point $(x_1, y_1)$ is $d = \bigg| \large\frac{ax_1+by_1+c}{\sqrt{a^2+b^2}} \bigg|$
Step 1 :
Let the equation of the line passing through the point of intersection be
$(x-y+1)+ \lambda (2x-3y+5)=0$
( i.e) $x(1+2\lambda )-y(1+3\lambda)+1+5\lambda=0$
Step 2 :
Its distance from the point (3,2) is given as $\large\frac{7}{5}$
$\therefore \bigg| \large\frac{3(1+2\lambda )-2(1+3\lambda)+1+5\lambda}{\sqrt{(1+2\lambda)^2+(1+3\lambda)^2}} \bigg|$$= \large\frac{7}{5} \Rightarrow \bigg| \large\frac{3+6\lambda - 2 - 6 \lambda + 1 + 5 \lambda}{\sqrt{1+4\lambda^2+4\lambda+1+6\lambda+9\lambda^2}} \bigg|$$= \large\frac{7}{5}$
$\Rightarrow \bigg| \large\frac{2+5\lambda}{\sqrt{2+10\lambda+13\lambda^2}} \bigg|$$= \large\frac{7}{5} Step 3 : Squaring on both sides we get, \large\frac{(2+5\lambda)^2}{13\lambda^2+10\lambda+2}$$=\large\frac{49}{25}$
$\large\frac{4+20\lambda+25\lambda^2}{13\lambda^2+10\lambda+2}$$=\large\frac{49}{25} 25(4+20\lambda+25\lambda^2)=49(13\lambda^2+10\lambda+2) 100+500\lambda+625\lambda^2=637\lambda^2+490\lambda+98 \Rightarrow 12\lambda^2-10\lambda-2=0 \Rightarrow 6\lambda^2-5\lambda-1=0 On factorizing we get, (6\lambda+1)(\lambda-1)=0 Hence \lambda=-\large\frac{1}{6} or \lambda=1 << Enter Text >> Step 4 : The equation of the required line is when \lambda=-\large\frac{1}{6} x\bigg(1+2\bigg(-\large\frac{1}{6} \bigg) \bigg) -y \bigg( 1 + 3\bigg(-\large\frac{1}{6} \bigg)\bigg)$$+1+5\bigg(-\large\frac{1}{6} \bigg)=0$