Question

Find the equation of a line perpendicular to y=-2x+4 that contains the point (0,3).

Answer 1

Slope-intercept form:  y = mx + b

(m is the slope, b is the y-intercept or the y value when x = 0 --> (0, y) or the point where the line crosses through the y-axis)

For lines to be perpendicular, their slopes have to be negative reciprocals of each other. (flip the sign +/- and the fraction(switch the numerator and the denominator))

For example:

Slope = 2 or $\frac{2}{1}$

Perpendicular line's slope = $-\frac{1}{2}$   (flip the sign from + to -, and flip the fraction)

Slope = $-\frac{2}{5}$

Perpendicular line's slope = $\frac{5}{2}$    (flip the sign from - to +, and flip the fraction)

y = -2x + 4   The slope is -2, so the perpendicular line's slope is $\frac{1}{2}$

Now that you know the slope, substitute/plug it into the equation:

y = mx + b

$y=\frac{1}{2} x+b$   Since they gave you the point (0, 3), which is the y-intercept, you can just plug 3 into "b". Another way to find b is plugging in the point on the line (0, 3) into the equation, then isolate/get the variable "b" by itself

$3=\frac{1}{2}(0)+b$

3 = b

$y=\frac{1}{2} x+3$