Question

Write a slope intercept equation for a line passing through the point (5,-3) that is parallel to the line 5x+7y=8. Then write a second equation for a line passing through the point (5,-3) that is perpendicular to the line 5x+7y=8. Thanks so much!!

Answer 1

We're starting with the line

$5x + 7y = 8$

All the lines parallel to it will be of the form

$5x + 7y = \textrm{constant}$

and when our intersection point is (5,-3) the constant must be

$5x + 7y = 5(5)+7(-3) = 25 - 21 = 4$

That's the answer but we have to turn this into slope intercept form so we solve for y:

$7y = -5x + 4$

$y = - \frac 5 7 x + \frac 4 7$

Answer: y = (-5/7)x + 4/7

For perpendicular lines, we swap the coefficients on x and y, negating one:

$7x - 5y = \textrm{constant}$

Again the point gives the constant:

$7x - 5y = 7(5) - 5(-3) = 35 + 15 = 50$

Solve for y for slope/intercept:

$5y = 7x - 50$

[tex]y = \frac 7 5  x - 10

Answer: y = (7/5) x - 10

Let's check with a nice plot:

Answer 2

The equation will be equal to y = (7/5) x - 10

What is an equation?

The equation in mathematics is the relationship between the variables and the number and establishes the relationship between the two or more variables.

We're starting with the line

5x + 7y = 8

All the lines parallel to it will be of the form

5x + 7y = constant

and when our intersection point is (5,-3) the constant must be

5x+7y=5(5)+7(-3)=25-21=4

That's the answer but we have to turn this into the slope-intercept form so we solve for y:

7y = -5x +4

y= (-5/7)x +(4/7)

Answer: y = (-5/7)x + 4/7

For perpendicular lines, we swap the coefficients on x and y, negating one:

7x - 5y = constant

Again the point gives the constant:

7x-5y=7(5)-5(-3)=35+15=50

Solve for y for slope/intercept:

5y=7x-50

Answer: y = (7/5) x - 10

Therefore the equation will be equal to y = (7/5) x - 10

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