# 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. - Mathematics

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,

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

Concept: Shifting of Origin
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#### APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 23 The straight lines
Exercise 23.12 | Q 17 | Page 93