Question

Find the equation perpendicular to the line y=−3x+5 that passes through point (2, 6). Show Work

Answer 1

so hmmm, let's see, the equation above, is already in slope-intercept form, thus
`\bf y=\stackrel{slope}{-3}x+5`.

so, a perpendicular line to that, will have a negative reciprocal slope, that is, if the slope for that one is -3, then


`\bf \textit{perpendicular, negative-reciprocal slope for slope }-3\implies \cfrac{-3}{1}\\\\ slope=\cfrac{-3}{{{ 1}}}\qquad negative\implies +\cfrac{3}{{{1}}}\qquad reciprocal\implies +\cfrac{{{ 1}}}{3}`

so then, what is the equation of a line whose slope is 1/3 and goes through 2,6?

Answer 2

Hey there! :)

Perpendicular to y = -3x + 5 ; passes through (2, 6).

Using y=mx+b (where m=slope, y=y-intercept) we can find the slope from our given equation. Since -3 is in the "m" spot, we know that -3 is the slope for the given equation.

However, since the equation we're looking for is perpendicular to our given one, we must find the negative reciprocal of -3.

A negative reciprocal is simply a number flipped over and is given a negative or has a negative taken away. Using this information, we can come to the conclusion that the negative reciprocal of -3 is 1/3. So, our new slope is 1/3.

Now, to find our new equation, we must use point-slope form ( y-y₁=m(x-x₁) ). Using 1/3 as our slope and (2, 6) as our intercepts, let's plug & chug! :)

y-y₁=m(x-x₁)y - (6) = 1/3(x - 2)

Simplify.

y - 6 = 1/3x - 2/3

~Hope I helped!~