# Q : 15      The perpendicular from the origin to the line  $y=mx+c$  meets it at the point  $(-1,2)$. Find the values of $m$ and $c$.

G Gautam harsolia

We can say that line passing through point $(0,0) \ and \ (-1,2)$  is perpendicular to line $y=mx+c$
Now,
The slope of the line  passing through the point $(0,0) \ and \ (-1,2)$ is , $m = \frac{2-0}{-1-0}= -2$
lines are perpendicular
Therefore,
$m = -\frac{1}{m'} = \frac{1}{2}$               - (i)
Now, the point $(-1,2)$ also lies on the line $y=mx+c$
Therefore,
$2=\frac{1}{2}.(-1)+C\\ C = \frac{5}{2} \ \ \ \ \ \ \ \ \ \ \ -(ii)$
Therefore, the value of m and C is $\frac{1}{2} \ and \ \frac{5}{2}$   respectively

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