What is the slope of the line perpendicular to $y = - \frac{10}{3} x + 12$ $y=-10/3x +12$ ?

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Answer 1 · 2018-07-25T20:38:51

The slope perpendicular is $+ \frac{3}{10}$ $+3/10$

If two lines are perpendicular, then there is a specific relationship between their slopes:

One slope is the negative reciprocal of the other.

The product of their slopes is $- 1$ $-1$

if one is $\frac{a}{b}$ $a/b$ , then the other is $- \frac{b}{a}$ $-b/a$

For example. $\frac{2}{3}$ $2/3$ is perpendicular to $- \frac{3}{2}$ $-3/2$

$\frac{2}{3} \times - \frac{3}{2} = - 1$ $2/3 xx-3/2 =-1$

In this case the slope is $- \frac{10}{3}$ $-10/3$

The slope perpendicular is $+ \frac{3}{10}$ $+3/10$

$- \frac{10}{3} \times \frac{3}{10} = - 1$ $-10/3 xx 3/10 =-1$

What is the slope of the line perpendicular to y=-10/3x +12 y=−103x+12 y=-10/3x +12 ?