Perpendicular to the line x-3y=7;containing the point (5,5)

Question

Find the slope-intercept equation of the line

Answer 1

Given line x-3y = 7

Subtract x to each side.

x-x -3y = 7-x

-3y = -x+7

Divide to each side by negitive 3.

-3y/-3 = -x/-3+7/-3

y = x/3-7/3

Slope of the line is say m1 = 1/3.

Perpendicular line slope say m2 = -3.

Given point say(x1,y1) = (5,5)

y-5 = -3(x-5)

y-5 = -3x+15

Add 5 to each side.

y-5+5 = -3x+15+5

y = -3x+20

Required line equation is y = -3x+20.

Answer 2

Slope - intercept form line equation is y = mx + b, where m is slope and b is y - intercept.

The line equation is x - 3y = 7 and the point is (5, 5).

Write the equation in slope - intercept form.

- 3y = - x + 7

y = (1/3)x - (7/3).

Compare the equation with slope - intercept form y = mx + b.

slope = 1/3.

Because the slopes of perpendicular lines are negative reciprocals, the slope of perpendicular line through (5, 5) is - 3.

Now, the perpendecular line equation is y = - 3x + b.

Find the y - intercept by substituting the point in the line equation say (x, y) = (5, 5).

5 = (- 3)(5) + b

b = 5 + 15

⇒ b = 20.

The perpendecular line equation is y = - 3x + 20.