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

Answer:

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

• $y=\frac{5}{2}x-8$

Step-by-step 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

$y=-\frac{2}{5}x+2$

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)

$y-\left(-3\right)=\frac{5}{2}\left(x-2\right)$

$y+3=\frac{5}{2}\left(x-2\right)$

subtract 3 from both sides

$y+3-3=\frac{5}{2}\left(x-2\right)-3$

$y=\frac{5}{2}x-8$

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

• $y=\frac{5}{2}x-8$