A line passes through(5,0) and is perpendicular to the graph of y+1=2(x-3). What equation represents the line in slop-intercept form

Answer 1

y = mx + b

The product of slopes of perpendicular lines is -1. The given equation is in point-slope form, with a slope of 2, so the slope of the perpendicular line is

m = -1/2

y = (-1/2)x + b

Get b from the given point.

0 = (-1/2)(5) + b

You can finish from here.

Answer 2

The graph of y+1=2(x-3) will have a slope of 2, since in the slope-intercept form of y = mx + b, that equation is y = 2x - 7, and m = 2.

For a line to be perpendicular to a line of m = 2, it must have a slope of -1/2 (the negative reciprocal). So this perpendicular line will have the form y = -x/2 + b. Given the point (5,0), plug in the values for x and y of 5 and 0, respectively, so b = y + x/2 = 0 + 5/2 = 5/2. Therefore, the equation is y = -x/2 + 5/2, or -(1/2)x + 5/2. The slope is -1/2, and y-intercept is 5/2.