Question

In a $△ABC$, $D$ is a point on $BC$ such that $\frac{BD}{DC}=\frac{AB}{AC}$. The equation of the line $AD$ is $2x+3y+4=0$ & the equation of the line $AB$ is $3x+2y+1=0$. The equation of the line $AC$ is

• A. $9x+46y-83=0$
• B. $9x-46y-83=0$
• C. $9x+46y+83=0$
• D. $9x-46y+83=0$

Answer 1

Answer: (C)

$9x+46y+83=0$

Given

$AD:2x+3y+4=0$------(1)

$AB:3x+2y+1=0$------(2)

Intersection point is A SO

On solving Eq (1) and (2)

$3\left(\frac{-4-3y}{2}\right)+2y+1=0$

$-12-9y+4y+2=0\Rightarrow y=-2,x=1$

Point$(1,-2)$

Given condition $\frac{BD}{DC}=\frac{AB}{AC}$ it means $\Delta ABC$ is isosceles

Hence AD and BC are perpendicular line

Perpendicular of line BC is $m_2(-\frac{2}{3})=-1$

$m_2=\frac{3}{2}$

Slope of line AB say $m_1=-\frac{3}{2}$

Let slope of line AC is $m$ Then the angle between AB and BC is equal to angle between AC and BC

$\left|\frac{\frac{3}{2}+\frac{3}{2}}{1-\frac{9}{4}}\right|=\left|\frac{\frac{3}{2}-m}{1+\frac{3}{2}m}\right|$

$\left|\frac{3\times 4}{4-9}\right|=\frac{3-2m}{2+3m}$

$\left|\frac{12}{-5}\right|=\frac{3-2m}{2+3m}$

$\frac{12}{5}=\frac{3-2m}{2+3m}$

$24+36m=15-10m$

$46m=-9\Rightarrow m=-\frac{9}{46}$

Equation of line AC with Slope m and point A

$y+2=\frac{-9}{46}(x-1)$

$46y+92=-9x+9$

$9x+46y+83=0$