What is an equation of the line that has a y-intercept of -2 and is perpendicular to the line #x-2y = 5#?

Question

Jayden Jackson · Accepted Answer

#2x+y =-2#

Write as #y_1= 1/2x -5/2#

If you have standard form of #y=mx+c# then the gradient of its normal is #-1/m#

The gradient of a line normal to this is #-1 times (1/2)^("inverted") =-2#

As it passes through y=02 at x=0 then the equation becomes:

#y_2=-2x-2#

In same form as question gives:

#2x+y =-2#

Aurora Ayala · Answer

undefined To find the equation of a line perpendicular to $x - 2y = 5$ with a y-intercept of -2, first, we need to find the slope of the given line. Rearranging $x - 2y = 5$ into slope-intercept form gives us $y = \frac{1}{2}x - \frac{5}{2}$.

The slope of this line is $m = \frac{1}{2}$.

Since the line we're looking for is perpendicular to this line, its slope will be the negative reciprocal of $\frac{1}{2}$, which is $-2$.

Given that the y-intercept is -2, we can use the point-slope form of a line, $y - y_1 = m(x - x_1)$, where $m$ is the slope and $(x_1, y_1)$ is a point on the line.

Substituting $m = -2$ and $y_1 = -2$, we get:

$y - (-2) = -2(x - 0)$

Simplify this equation to get the final form of the equation of the line:

$y + 2 = -2x$

$$y = -2x - 2$$

So, the equation of the line that has a y-intercept of -2 and is perpendicular to the line $x - 2y = 5$ is $y = -2x - 2$.