What is the equation of a line that satisfies the given conditions: perpendicular to $y = - 2 x + 5$ $y= -2x + 5$ and passing through (4, -10)?

Question

20484 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-12-06T19:55:24

$y = \frac{1}{2} x - 12$ $y=1/2x-12$

Perpendicular lines have opposite reciprocal slopes.

The slope of the line $y = - 2 x + 5$ $y=-2x+5$ is $- 2$ $-2$ .

The opposite reciprocal of $- 2$ $-2$ is $\frac{1}{2}$ $1/2$ , which is the slope of the perpendicular line.

We also know that the line passes through the point $(4, - 10)$ $(4,-10)$ .

We can use the $y = m x + b$ $y=mx+b$ form of the line to determine the equation.

We know that:
$x = 4$ $x=4$
$y = - 10$ $y=-10$
$m = \frac{1}{2}$ $m=1/2$
And we want to solve for $b$ $b$ .

$- 10 = \frac{1}{2} (4) + b$ $-10=1/2(4)+b$

$- 10 = 2 + b$ $-10=2+b$

$- 12 = b$ $-12=b$

The equation of the line is $y = \frac{1}{2} x - 12$ $y=1/2x-12$ .

What is the equation of a line that satisfies the given conditions: perpendicular to y= -2x + 5y=−2x+5y= -2x + 5 and passing through (4, -10)?