Answer to Question #132892 in Algebra for Everlyn

Question

Use the given conditions to write an equation for the line in​ point-slope form and in​ slope-intercept form.
Passing through ​(3​,-2​) and perpendicular to the line whose equation is y=1/2x+4. What's the point-slope form

Answer 1

The products of the gradients of perpendicular lines is $-1.$ That is to say, $mm' = -1$

Let's assume that the gradient of the line perpendicular to the line $y = \dfrac {1} {2} x + 4$ is $m.$

$=> m(\dfrac {1}{2}) = -1$

$=> \dfrac {1}{2}m = -1$

$=> m = -2$

Point-slope form

$ANSWER: \space y + 2 = -2(x - 3)$

$SOLUTION$

The general point-slope form of a linear equation is

$y - y_{1} = m(x - x_{1})$ where $m$ is the slope/gradient and $(x_{1}; y_{1})$ is the point.

In this case, $m = -2 \space and \space points \space x_{1} = 3 \space \& \space y_{1} = -2 \space from \space the \space point \space (3; -2)$

$\therefore \space y - (-2) = -2(x-3)$

$=> y + 2 = -2 (x - 3)$

Slope-intercept form

$ANSWER: y = -2x + 4$

$SOLUTION$

The general slope-intercept form of a linear equation is given by $y = mx + c$ where $m$ is the slope/gradient and $c$ is the $y-intercept.$

Now, $m = -2$ , therefore $y = -2x + c$

Substituting the point $(3; -2)$ to find $c:$ $-2 = -2(3) + c$

$=> -2 = -6 + c$

$=> c = -2 + 6$

$=> c = 4$

$\therefore \space y = -2x + 4$