What is the slope of the line perpendicular to $y = \frac{7}{4} x - 3$ $y=7/4x -3$ ?

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Answer 1 · 2016-01-07T21:07:02

$s l o p e = - \frac{4}{7}$ $slope = - 4/7$

One form of the equation of a straight line is y = mx + c
where m is the slope and c the y-intercept.

the equation here fits this form and by comparison $m = \frac{7}{4}$ $m =7/4$

Consider 2 lines with slopes $m_{1} and m_{2}$ $m_1 and m_2$

when they are perpendicular to each other then $m_{1} \times m_{2} = - 1$ $m_1 xx m_2 = - 1$

$\Rightarrow \frac{7}{4} \times m_{2} = - 1$ $rArr 7/4 xx m_2 = -1$

$\Rightarrow m_{2} = - \frac{1}{\frac{7}{4}} = - \frac{4}{7}$ $rArr m_2 = -1/(7/4) = - 4/7$

What is the slope of the line perpendicular to y=74x−3 y=7/4x -3 ?