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 line

• A. $6(x+y)-25=0$
• B. $2x+6y+1=0$
• C. $2x+3y-6=0$
• D. $no$ $solution$

Answer 1

The correct option is D $no$ $solution$

Equation of the line joining the points $(2,-1)$ and $(5,-3)$ is

$\frac{x-2}{y+1}=\frac{5-2}{-3+1}=>\frac{x-2}{y+1}=\frac{3}{-2}=>-2x+4=3y+3=>2x+3y-1=\mathrm{0....}\left(i\right)$

$(x_1,4)$ lies on the line $\left(i\right)$, thus we have

$2x_1+12-1=0=>2x_1=-11=>x_1=\frac{-11}{2}$

Also $(-2,y_1)$ lies on the line $\left(i\right)$, thus we have

$-4+3y_1-1=0=>3y_1=5=>y_1=\frac{5}{3}$

Point $(\frac{-11}{2},\frac{5}{3})$ is not satisfying any given straight line

Therefore, no solution is the option