Question

If PM is the perpendicular from P (2,3) on to the line x+y=3 then the co-ordinates of M are

• A. (2,1)
• B. (−1,4)
• C. (1,2)
• D. (4,−1)

Answer 1

The correct option is C

(1,2)

If (x,y) is the foot of the perpendicular from $(x_1,y_1)$ to the line ax + by + c = 0 then

$\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{-(a{x}_1+b{y}_1+c)}{a^2+b^2}$

Here $(x_1,y_1)=(2,3);ax+by+c=x+y-3$

$\therefore$ $\frac{x-2}{1}=\frac{y-3}{1}=\frac{-\left(\right(1\times 2)+(1\times 3)+(-3\left)\right)}{1^2+1^2}$

$\Rightarrow$ $\frac{x-2}{1}=\frac{y-3}{1}=\frac{-\left(\right(2)+(3)+(-3\left)\right)}{2}$

$\Rightarrow$ $\frac{x-2}{1}=\frac{y-3}{1}=\frac{-2}{2}$

$\Rightarrow$ $\frac{x-2}{1}=\frac{y-3}{1}=-1$

$\therefore$ x - 2 = -1 ; y-3 = -1

$\Rightarrow$ x = -1 + 2 ; y = -1 + 3

$\Rightarrow$ x = 1 ; y = 2

$\therefore$ (x,y) = (1,2)