Question

Find the equation of the line perpendicular to x−5y=15 that passes through the point (−2,5).

Answer 1

1. Solve the given equation for y.

x - 5y = 15

-5y = -x + 15

y = (-x + 15)/-5

y = (x/5) - 3

y = (1/5)(x) - 3

The slope is 1/5. See it?

The equation we are looking for has a slope which is the negative inverse of the slope in the equation we just solved for y.

The slope for the equation we want is -5 which is the negative inverse of 1/5. Undetstand?

We have the slope of the new equation and one point is given.

Plot BOTH into the point-slope formula and solve for y. To solve for a variable means to isolate the variable ALONE on one side of the equation.

y - y_1 = m(x - x_1)...This is the point-slope formula. Our given point is (5,-2)

y - 5 = -5(x - (-2))

y - 5 = -5(x + 2)

We now solve for y and that's it.

y - 5 = -5x - 10

y = -5x - 10 + 5

The equation we want is y = -5x - 5.

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Answer 2

First off, let's solve x−5y=15  for "y".

$\bf x-5y=15\implies x-15=5y\implies \cfrac{x-15}{5}=y\implies \stackrel{slope}{\cfrac{1}{5}}x-3=y$

now, notice the function in slope-intercept form, well, it has a slope of 1/5.

now, a perpendicular line to that one, will have a negative reciprocal to that, let's check what that is.

$\bf \textit{perpendicular, negative-reciprocal slope for slope}\quad \cfrac{1}{5}\\\\ slope=\cfrac{1}{{{ 5}}}\qquad negative\implies -\cfrac{1}{{{ 5}}}\qquad reciprocal\implies - \cfrac{{{ 5}}}{1}\implies -5$

so, we're looking for the equation of a line whose slope is -5 and goes through -2,5.

$\bf \begin{array}{lllll} &x_1&y_1\\ % (a,b) &({{ -2}}\quad ,&{{ 5}}) \end{array} \\\\\\ % slope = m slope = {{ m}}= \cfrac{rise}{run} \implies -5 \\\\\\ % point-slope intercept \stackrel{\textit{point-slope form}}{y-{{ y_1}}={{ m}}(x-{{ x_1}})}\implies y-5=-5[x-(-2)] \\\\\\ y-5=-5(x+2)\implies y-5=-5x-10\implies y=-5x-5$