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 line2x-6y-1-06-x-y-25-02x-3y-6-0no solution

Question

Toppr Admin · Accepted Answer

Equation of the line joining the points $(2, - 1)$ and $(5, - 3)$ is
$\frac{x - 2}{y + 1} = \frac{5 - 2}{- 3 + 1} => \frac{x - 2}{y + 1} = \frac{3}{- 2} => - 2 x + 4 = 3 y + 3 => 2 x + 3 y - 1 = 0.... (i)$
$(x_{1}, 4)$ lies on the line $(i)$ , thus we have
$2 x_{1} + 12 - 1 = 0 => 2 x_{1} = - 11 => x_{1} = \frac{- 11}{2}$
Also $(- 2, y_{1})$ lies on the line $(i)$ , thus we have
$- 4 + 3 y_{1} - 1 = 0 => 3 y_{1} = 5 => y_{1} = \frac{5}{3}$
Point $(\frac{- 11}{2}, \frac{5}{3})$ is not satisfying any given straight line
Therefore, no solution is the option

If each of the points (x1,4),(−2,y1) lies on the line joining the points (2,−1),(5,−3), then the point P(x1,y1) lies on line2x+6y+1=06(x+y)−25=02x+3y−6=0no solution

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
$2 x + 6 y + 1 = 0$
$6 (x + y) - 25 = 0$
$2 x + 3 y - 6 = 0$
$n o$ $s o l u t i o n$