Question

Write an equation of a line that passes through the point (4,3) and is perpendicular to the graph of the equation y=−13x+4.

Answer 1

An equation of a line that passes through the point (4,3) and is perpendicular to the graph of the equation will be:

• 
  `y=(1)/(13)x+(35)/(13)`

Explanation:

We know that the slope-intercept form of the line equation is

`y=mx+b`

where

• m is the slope

• b is the y-intercept

Given the line

y = -13x+4

comparing with the slope-intercept form of the line equation

The slope of the line = m = -13

We know that a line perpendicular to another line contains a slope that is the negative reciprocal of the slope of the other line, such as:

slope = m = -13

Thus, the slope of the the new perpendicular line = – 1/m = -1/-13 = 1/13

Using the point-slope form of the line equation

`y-y_1=m\left(x-x_1\right)`

where

• m is the slope of the line

• (x₁, y₁) is the point

substituting the values of the slope m = 1/13 and the point (4, 3)

`y-y_1=m\left(x-x_1\right)`

`y-3=(1)/(13)\left(x-4\right)`

Add 3 to both sides

`y-3+3=(1)/(13)\left(x-4\right)+3`

`y=(1)/(13)x-(4)/(13)+3`

`y=(1)/(13)x+(35)/(13)`

Therefore, an equation of a line that passes through the point (4,3) and is perpendicular to the graph of the equation will be:

• 
  `y=(1)/(13)x+(35)/(13)`