Question

What is the equation of a line that is perpendicular to y=-5x+11 and passes through the point (3,3)?

Answer 1

Answer:

$y=\frac{1}{5}x+\frac{12}{5}$

Step-by-step explanation:

The slope of a line that is perpendicular to another will have an opposite-reciprocal slope. So, if the slope was -2, then the perpendicular slope would be $\frac{1}{2}$ .

$y=-5x+11$

This is written in slope-intercept form:

$y=mx+b$

m is the slope and b is the y-intercept. Find the opposite-reciprocal of the slope, -5:

$y=\frac{1}{5} x+b$

Now we need to find the y-intercept. For this, substitute the given points for its appropriate value:

$(3_{x},3_{y})\\\\3=\frac{1}{5}(3)+b$

Solve for b:

Simplify multiplication:

$\frac{1}{5}*\frac{3}{1}=\frac{3}{5}$

Insert:

$3=\frac{3}{5}+b$

Subtract b from both sides:

$3-b=\frac{3}{5}+b-b\\\\3-b=\frac{3}{5}$

Subtract 3 from both sides:

$3-3-b=\frac{3}{5}-3\\\\-b=\frac{3}{5}-3$

Simplify subtraction:

$\frac{3}{5}-3\\\\\frac{3}{5}-\frac{3}{1}\\\\\frac{3}{5}-\frac{15}{5}=-\frac{12}{5}$

Insert:

$-b=-\frac{12}{5}$

Multiply both sides by - 1 to simplify b (it can be seen as -1b):

$-b*(-1)=-\frac{12}{5}*(-1)\\\\b=\frac{12}{5}$

Insert:

$y=\frac{1}{5}x+\frac{12}{5}$

:Done