Browse Questions

# Choose the correct answer from the given four options. The coordinates of the foot of perpendiculars from the point (2, 3) on the line $y = 3x + 4$ is given by

$\begin {array} {1 1} (A)\;\bigg( \large\frac{37}{10}, \large\frac{-1}{10} \bigg) & \quad (B)\;\bigg( \large\frac{-1}{10}, \large\frac{37}{10} \bigg) \\ (C)\;\bigg( \large\frac{10}{37},-10 \bigg) & \quad (D)\;\bigg( \large\frac{2}{3}, -\large\frac{1}{3} \bigg) \end {array}$

Toolbox:
• Slope of a line is $m = \large\frac{y_2-y_1}{x_2-x_1}$
• If two lines are perpendicular then the product of their slopes is -1.
Step 1 :
Let (h, k) be the coordinate of the foot of the perpendicular from the given point (2,3). Equation of the given line is $y=3x+4$.
Then the slope of the perpendicular line is $\large\frac{k-3}{h-2}$
Slope of the line $3x-y+4=0$ is $-\bigg( \large\frac{3}{-1} \bigg)$$=3 Hint : Condition for the perpendicularity of lines is m_1m_2=-1 \therefore \bigg( \large\frac{k-3}{h-2} \bigg)$$ (3)=-1$
$\Rightarrow (3k-9)=-1(h-2)$
$\Rightarrow 3k-9=-h+2$
$\therefore 3k+h=11$---------(1)
Also (h,k) lies on the line $3x-y+4=0$.
$\therefore 3(h) = k + 4 = 0$
$\Rightarrow 3h-k=-4$---------(2)
Solving equation (1) and (2) we get,
$3k+h=11$
$3h-k=-4$
______________
$9k+\not{3}\not{h}=33$
$\not{3}\not{h} - k = -4$
$(-) \quad (+) \quad (+)$
______________
$\qquad 10k=37$
$\Rightarrow k = \large\frac{37}{10}$
Substituting this value of k in equation (1) we get,
$3 \bigg( \large\frac{37}{10} \bigg)$$+h=11 \Rightarrow h = 11-\large\frac{111}{10} = -\large\frac{1}{10} Hence the coordinate are \bigg(- \large\frac{1}{10}$$, \large\frac{37}{10} \bigg)$
Hence B is the correct option.