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Let $(x_{1},4),(-2,y_{1})$ lies on the line joining the points $(2,-1),(5,-3)$ then the point  $P(x_{1},y_{1})$ lies on the line:

1. $6(x+y)-25=0$
2. $2x+6y+1=0$
3. $2x+3y-6=0$
4. $6(x+y)+25=0$

### 1 comment

Equation of line joining the points ($x_{1}$, $y_{1}$) and ($x_{2}$, $y_{2}$) is $\frac{y-y_{1}}{y_{2}-y_{1}} =\frac{x-x_{1}}{x_{2}-x_{1}}$

Here, two points are (2, -1) and (5, -3), so equation of the line will be $\frac{y+1}{-3+1} =\frac{x-2}{5-2}$

2x+3y-1=0 is the equation of line. ($x_{1},4$) and $(-2,y_{1})$ lies on this line, so

2*$x_{1}$ + 3*4 -1 = 0 → On solving, we get $x_{1}$ = $-\frac{11}{2}$. Similarly,

2*-2+3*$y_{1}$ -1 = 0 → On solving, we get $y_{1}$ = $\frac{5}{3}$. So point is ($x_{1}$,$y_{1}$) = ($-\frac{11}{2}$, $\frac{5}{3}$)

Only option B 2x + 6y + 1 = 0 satisfies the point ($x_{1}$,$y_{1}$).

2*$-\frac{11}{2}$ + 6*$\frac{5}{3}$ + 1 = 0

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