Question what is the equation of a line (in slope-intercept form) that is perpendicular to y=2x+3 and contains the point (4,5).

Answer 1

y=mx+b

First find the slope :m

The slope is PERPENDICULAR to the line: y=2x+3

A line that is perpendicular to another line has a slope that is the negative reciprocal.

In this example the negative reciprocal of 2 is -1/2 (flip the slope and change the sign)

Therefore m=-1/2

Now you need to find b in the equation: y=mx+ b To find b, use the slope that you found for m (-1/2) and substitute the given point (4,5) in the equation and solve for b

5=(-1/2) x 4 + b

5=-2+b

Add a positive 2 to each side and you have 5+2=-2+2+b

The result is 7=b.

Your final equation is y=(-1/2)x+7

Answer 2

The slope-intercept form of a line is y = mx + b. In order to write this, we need to find the slope (m) and the y-intercept (b). We are given the equation of a line that is perpendicular to the line we want; the keyword here is "perpendicular" because the slopes of lines that are perpendicular are opposite reciprocals. In this case, the perpendicular line has a slope of positive 2, so the perpendicular slope must be -1/2 ("negative one half"). Now onto the y-intercept: since we found the slope, we can plug this AND the coordinate (4, 5) for x and y into the equation to solve for b: y = (-1/2)x + b --> 5 = (-1/2)4 + b. Use algebra to solve for b: 5 = -2 + b --> 7 = b. Plug in the values you calculated for m and b back into the slope-intercept form to write your equation.