Find an equation of the line containing the given point and perpendicular to the given line

Question

Find an equation of the line containing the given point and perpendicular to the given line 

 

(5,2) 3x+y=5

Answer 1

Let's start by finding the new line's slope. Perpendicular lines have slopes that are negative reciprocals of each other. In other words, if one line's slope is 3/4, a perpendicular line's slope would be -4/3. So, to find the new line's slope, we'll need to find the slope of 3x + y = 5. We'll do this by putting it in slope-intercept form. The slope-intercept form of a line is:

 

y = mx + b, where m is the slope and b is the y-intercept

 

Subtracting 3x from each side the line's equation will do it:

 

y = -3x + 5

 

This means that the slope (m) is -3. The negative reciprocal of -3 is 1/3, which is the new line's slope.

 

 

Next, we'll use point-slope form to find an equation for the new line. The point-slope form of a line is:

 

(y-y₁) = m(x-x₁), where m is the slope and (x₁, y₁) is a point the line passes through

 

We know that the new line has a slope of 1/3 and passes through the point (5, 2), so you can just plug those numbers in:

 

y - 2 = 1/3(x - 5)

 

 

This is an acceptable answer for your question, but it's usually preferable to put the equation into slope-intercept form:

 

y = 1/3(x - 5) + 2

y = 1/3x - 5/3 + 6/3

y = 1/3x + 1/3