Question

If each of the points $(x_1,4),(-2,y_1)$ lies on the line joining the points $(2,-1),(5,-3),$ then the point $P(x_1,y_1)$ ) lies on the line

• A. $6(x+y)-25=0$
• B. $2x+6y+1=0$
• C. $2x+3y-6=0$
• D. $6(x+y)+25=0$

Answer 1

Answer: (C)

$2x+3y-6=0$

$slopeofline=\left(\frac{-3-(-1)}{5-2}\right)=\left(\frac{-2}{3}\right)\therefore e{q}^{n}ofline=\left(\frac{y-(-3)}{x-5}\right)=\left(\frac{-2}{3}\right)3(y+3)=-2(x-5)3y+9=-2x+103y+2x=1now(x,4)liesonline\therefore 3\cdot 4+2x,=1\therefore x,\left(\frac{-11}{2}\right)and(-2,y)liesonline\therefore 3y+2(-2)=1\therefore y=\left(\frac{5}{3}\right)now(x,y)satistione{q}^{n}2x+3y=-6$