Determine an equation of a line in slope-intercept form that goes through a given point and is parallel or perpendicular to a given line

Question

Write an equation that describes the line that passes through the point (-3,8) and is perpendicular to the graph of y= -5/3x +2/3

Answer 1

The slope of the given line is -5/3, so a line perpendicular to it will have a slope of 3/5.

 

Using the slope intercept formula, y = mx + b, your new line will have the equation 8 = (3/5)*(-3) + b.

 

Solving for b, we get 8 = -9/5 + b

                              8 + 9/5 = b

                             40/5 + 9/5 = b

                               49/5 = b

 

Your equation is y = 3/5x + 49/5

Answer 2

First you need to identify the slope of the perpendicular line. To do this you find the negative reciprocal slope of the line given.

 

The given/referenced equation is already in y=mx+b form where m=slope.

m=-5/3

 

The negative reciprocal (simply flip the fraction and change the sign) of -5/3 is 3/5

 

The slope, or m, of the perpendicular line is:

m= 3/5.

 

Next, you have to use point slope form to get the equation of the line you're looking for.

 

Point Slope Form is y-y₁=m(x-x₁)

 

y₁ is the y coordinate of the given ordered pair and x₁ is the x coordinate of the given ordered pair. 

 

Plug in all the information you know so far

 

y-8=3/5(x-(-3))

 

Put in y=mx+b form by solving for y.

 

y-8=3/5(x-(-3))

y-8=3/5(x+3)

y-8=3/5x+9/5

y=3/5x+49/5

 

Your solution is y=3/5x+49/5