Question

Give the equation of the line perpendicular to y +7= -2(x - 6) through the point (6,-3)

Answer 1

Note: this answer assumes the equation of the line can be put in slope-intercept form.

Answer:

$y + 3 = \frac{1}{2}(x-6)$

Step-by-step explanation:

1) First, find the slope of y + 7 = -2 (x - 6). We can see that it's already in point-slope form, or $y-y_1 = m (x-x_1)$ format. Remember that the number in place of the $m$ is the slope. Therefore, -2 is the slope of that equation.

What we need is the slope that is perpendicular to that, though. So, find the opposite reciprocal of -2. To do this, change its sign, convert it into a fraction ($-\frac{2}{1}$), and flip its numerators and denominators. Therefore, the perpendicular slope would be $\frac{1}{2}$.

2) Now that we have a slope and a point the line passes through, we can write an equation using the point-slope formula, $y-y_1 = m (x-x_1)$. In order to write an equation, the $x_1$, $y_1$, and $m$ have to be substitute for with real values.

The $m$ represents the slope. We already calculated that in the last step, so put $\frac{1}{2}$ in place of the $m$. The $x_1$ and $y_1$ represent the x and y values of a point the line passes through. We know that the line has to pass through (6, -3), so substitute 6 for $x_1$ and -3 for $y_1$:

$y - (-3) = \frac{1}{2}(x -(6))\\y + 3 = \frac{1}{2}(x - 6)$

Therefore, (again, assuming that the line can be put in point-slope form) the answer is $y + 3 = \frac{1}{2}(x-6)$.