Question

Write an equation perpendicular to y = -2x - 7 that passes through the point (4, 3).

Answer 1

We are asked to find the equation of a line that is perpendicular to the line y = -2x - 7 and passes through the point (4,3)

Recall that the standard form of the equation of a line in slope-intercept form is given by

$y=mx+b$

Where m is the slope and b is the y-intercept.

Comparing the standard form with the given equation we see that the slope is -2

$m=-2$

Since we are given that the lines are perpendicular so the slope of the other line must be negative reciprocal of the given line.

$m=\frac{1}{-(-2)}=\frac{1}{2}$

So the slope of the required equation is m = 1/2

Since we are also given that the line passes through the point (4,3)

The point-slope form of the equation of a line is given by

$y-y_1=m(x-x_1)$

Let us substitute the value of slope and the given point into the above equation.

$y-3_{}=\frac{1}{2}(x-4_{})$

Solving the equation for y.

$\begin{gathered} y-3=\frac{1}{2}x-\frac{4}{2} \\ y-3=\frac{1}{2}x-2 \\ y=\frac{1}{2}x-2+3 \\ y=\frac{1}{2}x+1 \end{gathered}$

Therefore, the equation of the line is

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