Which is an equation of the line passing through (-2, 3) and perpendicular to the line 5x - y = 12?

Mathematics

1 answer:

Bond [772]2 years ago

5 0

Answer:

An equation of the line passing through (-2, 3) and perpendicular to the line 5x - y = 12 will be:

$y=-\frac{1}{5}x+\frac{13}{5}$

Step-by-step explanation:

Given the equation

$5x - y = 12$

converting the line into the slope-intercept form y = mx+b, where m is the slope

$-y = 12-5x$

$y = 5x-12$

The slope of the line = m = 5

We know that a line perpendicular to another line contains a slope that is the negative reciprocal of the slope of the other line, such as:

Therefore, the slope of new line = – 1/m = -1/5 = -1/5

Using the point-slope form of the line equation

$y-y_1=m\left(x-x_1\right)$

where m is the slope of the line and (x₁, y₁) is the point

substituting the slope of new line = -1/5 and (-2, 3)

$y-y_1=m\left(x-x_1\right)$

$y-3=-\frac{1}{5}\left(x-\left(-2\right)\right)$

$y-3=-\frac{1}{5}\left(x+2\right)$

Add 3 to both sides

$y-3+3=-\frac{1}{5}\left(x+2\right)+3$

$y=-\frac{1}{5}x+\frac{13}{5}$

Therefore, an equation of the line passing through (-2, 3) and perpendicular to the line 5x - y = 12 will be:

$y=-\frac{1}{5}x+\frac{13}{5}$

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