If each of the points `(x_1,4),(-2,y_1)` lies on the line joining the points `(2,-1)a n d(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`

Options

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

Answer

B

Explanation

The equation of the line joining the points (2,-1) and (5,-3) is given by
`y+1 = (-1+3)/(2-5)(x-2)`
` "or " 2x+3y-1 = 0 " " (1)`
Since `(x_(1),4) " and " (-2,y_(1))` lie on 2x+3y-1=0, we have
`2x_(1)+12-1=0" or"x_(1) = -(11)/(2)`
`"and " -4+3y_(1)-1 = 0 " or " y_(1) = (5)/(3)`
`"Thus, "(x_(1),y_(1)) " satisfies " 2x+6y+1=0.`