Find the equation of the line with the given properties. Express the equation in general form or slope -intercept form.

Question

Perpendicular to the line -5x+y=-29, contains the points (5,-5)

The equation of the line is

Answer 1

The line equation is - 5x + y = - 29 and the point is (5, - 5).

To find the slope, write the line equation - 5x + y = - 29 in slope intercept form y = mx + b, where m = slope and b = y-intercept.

Add 5x to each side.

- 5x + y + 5x  = - 29 + 5x

y  = 5x - 29.

The slope of line equation y  = 5x - 29 is 5.

If the two lines are perpendicular, then their slopes are negative reciprocals to each other.

The slope of the perpendicular line equation is - 1/5.

To find the y-intercept of perpendicular line, substitute m = - 1/5 and (x, y) = (5, -5) in slope intercept form.

(- 5) = (- 1/5)(5) + b

- 5 = - 1 + b

Add 1 to each side.

- 5 + 1 = - 1 + b + 1

- 4 = b.

The slope-intercept form of line equation is y = (- 1/5)x - 4.

Write the above equation in the general form line equation is Ax + By = C. A shouldn't be negative, A and B shouldn't both be zero, and A, B and C should be integers.

Multiply each side by 5.

5y = 5[(- 1/5)x - 4]

Apply distributive property : a(b +c) = ab + ac.

5y = 5(- 1/5)x + 5(-4)

5y = - x - 20

Add x to each side.

5y + x = - x - 20 + x

5y + x = - 20

The general form of line equation is 5y + x = - 20.