Find the foot of the perpendicular of (3, 6) on the line x - 2y + 4 = 0

(5, 2)

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(4, 4)

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(-2, 1)

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(2, 3)

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Solution

The correct option is B
(4, 4)

Let the co-ordinates of Q ( foot of the perpendicular ) be (h,k)

Slope of PQ = $\frac{- 1}{s l o p e o f t h e l i n e}$

= $\frac{\frac{- 1}{1}}{2} = - 2$

We can find the slope using P(3,6) and Q(h,k)

$\Rightarrow$ $\frac{k - 6}{h - 3}$ = -2

$\Rightarrow$ k - 6 = -2 h + 6

2h + k = 12 -(1)

(h,k) is a point on x - 2 y + 4 = 0

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