Question

What is an equation of the line that passes through the point (2,-3) and is perpendicular to the line 2x+5y=10?

Answer 1

An equation of the line that passes through the point (2,-3) and is perpendicular to the line is:

• 
  `y=(5)/(2)x-8`

Explanation:

The slope-intercept form of the line equation

`y = mx+b`

where m is the slope and b is the y-intercept

Given the line

`y=-(2)/(5)x+2`

Here, the slope = m = -2/5

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 = -2/5

perpendicular slope = – 1/m = -1/(-2/5) = 5/2

Using the point-slope form of the line equation

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

where m is the slope and (x₁, y₁) is the point

substituting the values m = 5/2 and the point (2,-3)

`y-\left(-3\right)=(5)/(2)\left(x-2\right)`

`y+3=(5)/(2)\left(x-2\right)`

subtract 3 from both sides

`y+3-3=(5)/(2)\left(x-2\right)-3`

`y=(5)/(2)x-8`

Therefore, an equation of the line that passes through the point (2,-3) and is perpendicular to the line is:

• 
  `y=(5)/(2)x-8`