Question

The equation of a the line whose slope is $\frac{3}{2}$ and which passes through point P, where P divides the line segment joining $A(-2,6)$ and $B(3,-4)$ in the ratio $2:3$ is

• A. $3x-2y-4=0$
• B. $3x-2y+4=0$
• C. $3x+2y+4=0$
• D. $3x+2y-4=0$

Answer 1

Answer: (B)

$3x-2y+4=0$

At first, evaluate $P$ by using the section formula ie, let $P$ =$(x,y)$

$\therefore (x,y)=(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n})$

Here $m:n=2:3$

$\therefore (x,y)=(\frac{2\left(3\right)+3(-2)}{2+3},\frac{2(-4)+3\left(6\right)}{3+2})$

$(x,y)$=$(0,2)$ , $m=\frac{3}{2}$

Now, equation of the line would be

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

$\therefore 2y=3x+4$