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.

#### Solution

The given line is y = mx + c which can be written as \[mx - y + c = 0\] ... (1)

The slope of the line perpendicular to y = mx + c is \[- \frac{1}{m}\]

So, the equation of the line with slope \[- \frac{1}{m}\] and passing through the origin is \[y = - \frac{1}{m}x\]

\[x + my = 0\] ... (2)

Solving eq (1) and eq (2) by cross multiplication, we get

\[\frac{x}{mc - 0} = \frac{y}{0 - c} = \frac{1}{- 1 - m^2}\]

\[ \Rightarrow x = - \frac{mc}{m^2 + 1}, y = \frac{c}{m^2 + 1}\]

Thus, the point of intersection of the perpendicular from the origin to the line y = mx + c is \[\left( - \frac{mc}{m^2 + 1}, \frac{c}{m^2 + 1} \right)\]

It is given that the perpendicular from the origin to the line *y* = *mx* + *c* meets it at the point (\[-\] 1,2)

\[- \frac{mc}{m^2 + 1} = - 1 \text { and } \frac{c}{m^2 + 1} = 2\]

\[ \Rightarrow m^2 + 1 = mc \text { and } m^2 + 1 = \frac{c}{2}\]

\[ \Rightarrow mc = \frac{c}{2}\]

\[ \Rightarrow m = \frac{1}{2}\]

Now, substituting the value of m in \[m^2 + 1 = mc\] we get

\[\frac{1}{4} + 1 = \frac{1}{2}c\]

\[ \Rightarrow c = \frac{5}{2}\]

Hence,

\[m = \frac{1}{2} \text { and } c = \frac{5}{2}\].