Find an equation of the line perpendicular to the graph of 14x-7y=8 that passing through the point at (-2,5)

Question

asked Feb 13, 2020 164k views

1 Answer

Andriy Slobodyanyk · Answer 1 · 2020-02-19T15:24:17+0000

For this case we have to by definition, if two lines are perpendicular then the product of its slopes is -1.

We have the following line:

14x-7y = 8

Rewriting:

$-7y = 8-14x\\7y = 14x-8\\y = 2x- \frac {8} {7}$

Thus, the slope is:

$m_ {1} = 2$

We have to:

$m_ {1} * m_ {2} = - 1$ (Perpendicular condition)

$2 * m_ {2} = - 1$

$m_ {2} = - \frac {1} {2}$

Thus, the equation of the line is:

$y = - \frac {1} {2} x + b$

We find "b" knowing that the line passes through the point (-2,5).

$5 = - \frac {1} {2} (- 2) + b\\5 = 1 + b\\b = 5-1 = 4$

Finally, the equation is:

$y = - \frac {1} {2} x + 4$

Answer:

$y = - \frac {1} {2} x + 4$