What is the an equation of the line that goes through (−1, −3) and is perpendicular to the line 2x + 7y + 5 = 02x+7y+5=0?

1 Answer
Lotusbluete
Dec 3, 2015

y = 7/2 x + 1/2y=72x+12

Explanation:

Let's write the equation of the line 2x + 7y + 5 = 02x+7y+5=0 in standard form first.
To do so, solve the equation for yy:

color(white)(xxx) 2x + 7y + 5 = 0×x2x+7y+5=0

<=> 7y = -2x - 5⇔7y=−2x−5

<=> y = -2/7 x - 5/7⇔y=−27x−57

Now, we know that m_1 = -2/7m1=−27 is the slope of this line.

To make the line that you need to construct perpendicular to the given line, its slope mm must fulfill the condition m = -1/m_1m=−1m1, so the slope of the line that you need to create is

m = 7/2m=72.

Now that you have the slope mm and your point (-1, -3)(−1,−3), you can compute the equation of the line.

The standard form is

y = mx + ny=mx+n.

Since you know that the point (-1, -3)(−1,−3) is on your line, x = -1x=−1 and y = -3y=−3 must fulfill this equation.
Thus, plug xx, yy and mm and solve for nn:

-3 = 7/2 * (-1) + n−3=72⋅(−1)+n

<=> n = 1/2⇔n=12

The equation of your line is y = 7/2 x + 1/2y=72x+12.

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