What is the equation of the line perpendicular to $y = - \frac{3}{16} x$ $y=-3/16x$ that passes through $(- 2, 4)$ $(-2,4)$ ?

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Answer 1 · 2016-06-07T09:19:04

If lines are perpendicular then the one slope is the negative reciprocal of the other. this means that $m_{1} \times m_{2} = - 1$ $m_1 xx m_2 = -1$

In this case $m_{1} = - \frac{3}{16}$ $m_1 = -3/16$
The perpendicular slope to this is $\frac{16}{3}$ $16/3$

Now we have the slope and we also have a point (-2,4).

Use the formula $y - y_{1} = m (x - x_{1})$ $y - y_1 = m(x - x_1)$

$y - 4 = \frac{16}{3} (x - (- 2)) \Rightarrow y - 4 = \frac{16}{3} (x + 2)$ $y -4 = 16/3(x - (-2)) " " rArr y - 4 = 16/3(x + 2)$

$y = \frac{16}{3} x + \frac{32}{3} + 4$ $y = 16/3x + 32/3 +4$

$y = \frac{16}{3} x + 14 \frac{2}{3}$ $y = 16/3x + 14 2/3$

What is the equation of the line perpendicular to y=-3/16x y=−316xy=-3/16x that passes through (-2,4) (−2,4) (-2,4) ?