What is the standard form of a line that goes through $(5, - 4)$ $(5, -4)$ and is perpendicular to $y = \frac{5}{4} x - 5$ $y = 5/4x -5$ ?

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Answer 1 · 2017-04-20T12:56:54

$5 y + 4 x = 0$ $5y+4x=0$

Since the line is perpendicular to another line with slope $\frac{5}{4}$ $5/4$ , its slope will be the negative reciprocal of the other line's slope.

Hence the slope of the line is $- \frac{4}{5}$ $-4/5$ . We also know it passes through $(5, - 4)$ $(5, -4)$ .

Using

$y = m x + c$ $y=mx+c$

we know

$m (slope) = - \frac{4}{5}$ $"m (slope) =" -4/5$

therefore

$y = - \frac{4}{5} x + c$ $y = -4/5x+c$

Substituting $(5, - 4)$ $(5, -4)$ gives you

$- 4 = - \frac{4}{5} (5) + c$ $-4=-4/5 (5)+c$

$c = 0$ $c=0$

Therefore

$y = - \frac{4}{5} x$ $y=-4/5x$

$5 y = - 4 x$ $5y=-4x$

$5 y + 4 x = 0$ $5y+4x=0$

What is the standard form of a line that goes through (5,−4)(5, -4) and is perpendicular to y=54x−5y = 5/4x -5?