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Home  >>  CBSE XI  >>  Math  >>  Straight Lines
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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

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

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1 Answer

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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.
answered Jul 6, 2014 by thanvigandhi_1
 

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