Write the equation of a line that is perpendicular to y=7/5x+6 and passes through the point (2 -6)

Question

asked Nov 14, 2020 141k views

1 Answer

Brandizzi · Answer 1 · 2020-11-17T23:33:42+0000

For this case we have that by definition, the equation of a line in the slope-intersection form is given by:

y = mx + b

Where:

m: Is the slope

b: Is the cut-off point with the y axis

By definition, if two lines are perpendicular then the product of their slopes is -1, that is:

$m_ {1} * m_ {2} = - 1$

We have the following equation:

$y = \frac {7} {5} x + 6$

Where:
$m_ {1} = \frac {7} {5}$

We find
$m_ {2}:$

$m_ {2} = \frac {-1} {m_ {1}}\\m_ {2} = \frac {-1} {\frac {7} {5}}\\m_ {2} = - \frac {5} {7}$

Thus, the equation of the line is of the form:

$y = - \frac {5} {7} x + b$

We substitute point
(2, -6) and find "b":

$-6 = - \frac {5} {7} (2) + b\\-6 = - \frac {10} {7} + b\\-6+ \frac {10} {7} = b\\\frac {-42 + 10} {7} = b\\\frac {-32} {7} = b$

Finally, the equation is:

$y = - \frac {5} {7} x- \frac {32} {7}$

ANswer:

$y = - \frac {5} {7} x- \frac {32} {7}$