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.

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.`