The foot of the perpendicular drawn from the point $(7, 8)$ to the line $2 x + 3 y - 4 = 0$ is -

(\frac{23}{13}, \frac{2}{13})

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(13, \frac{23}{13})

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(- \frac{23}{13}, \frac{2}{13})

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(- \frac{2}{13}, \frac{23}{13})

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Solution

The correct option is A $(\frac{23}{13}, \frac{2}{13})$
the foot will be, $(a, \frac{4 - 2 a}{3})$
slope of the perpendicular line:
$= \frac{\frac{4 - 2 a}{3} - 8}{a - 8}$

slope of the line: $= \frac{- 2}{3}$

since both lines are perpendicular so,
$m_{1} m_{2} = - 1$

$a = \frac{23}{13}$

the foot will be: $(\frac{23}{13}, \frac{2}{13})$

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Similar questions

What are the coordinates of the foot of the perpendicular drawn from the point (3, 5) to the line, 2x + 3y + 7 = 0?

\frac{\frac{1}{2} \cdot \frac{2}{2}}{1^{3}} + \frac{\frac{2}{2} \cdot \frac{3}{2}}{1^{3} + 2^{3}} + \frac{\frac{3}{4} \cdot \frac{4}{2}}{1^{3} + 2^{3} + 3^{3}} + \dots

n

1^{3} + \frac{1^{3} + 2^{3}}{2} + \frac{1^{3} + 2^{3} + 3^{3}}{3} + . . .

(\frac{37}{10}, - \frac{1}{10})

(- \frac{1}{10}, \frac{37}{10})

(\frac{10}{37}, - 10)

(\frac{2}{3}, - \frac{1}{3})

\frac{1^{3}}{1} + \frac{1^{3} + 2^{3}}{2} + \frac{1^{3} + 2^{3} + 3^{3}}{3} + . . . n

=

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