The point $P$ is the foot of perpendicular from $A (- 5, 7)$ to the line $2 x - 3 y + 18 = 0$ . Find co-ordinates of $P$ .

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Solution

$2 x - 3 y + 18 = 0$ ...(1)

$\Rightarrow y = \frac{2}{3} x + 6$

$m_{1} = \frac{2}{3} \Rightarrow m_{2} = - \frac{3}{2}$

$(y - y_{1}) = m (x - x_{1})$

$A (- 5, 7)$

$(y - 7) = (- \frac{3}{2}) (x + 5)$

$2 y - 14 = - 3 x - 15$

$3 x + 2 y + 1 = 0$ ...(2)

$2 \times (1) + 2 \times (3)$ gives

$x = - 3$

From (1)

$2 x - 3 y + 18 = 0 \Rightarrow 2 (- 3) - 3 y + 18 = 0 \Rightarrow y = 4$

$P (- 3, 4)$

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The point P is the foot of perpendicular from A(−5,7) to the line 2x−3y+18=0. Find co-ordinates of P .

2x−3y+18=0...(1)⇒y=23x+6m1=23⇒m2=−32(y−y1)=m(x−x1)A(−5,7)(y−7)=(−32)(x+5)2y−14=−3x−153x+2y+1=0...(2)2×(1)+2×(3) givesx=−3From (1)2x−3y+18=0⇒2(−3)−3y+18=0⇒y=4P(−3,4)

The point $P$ is the foot of perpendicular from $A (- 5, 7)$ to the line $2 x - 3 y + 18 = 0$ . Find co-ordinates of $P$ .