Question

Find the value of k for which the point (1, -2) lies on the graph of the linear equation $x-2y+k=0.$

Answer 1

$\because$ Point (1, -2) lies on the graph of the equation x - 2y + k = 0

$\therefore x=1,y=-2$ will satisfy the equation. Now substituting the value of x = 1, y = -2 in it, we get

$1-2(-2)+k=0\Rightarrow 1+4+k=0\Rightarrow 5+k=0\Rightarrow k=-5\therefore k=-5$