Question

Write the equation of a line that is perpendicular to y=75x+6y=\dfrac{7}{5}x+6y=57​x+6y, equals, start fraction, 7, divided by, 5, end fraction, x, plus, 6 and that passes through the point (2,−6)(2,-6)(2,−6)(, 2, comma, minus, 6, ).

Answer 1

Answer:

The equation of a line that is perpendicular to $y=\frac{7}{5}x+6$ and that passes through the point (2, −6) is $y=-\frac{5}{7}x-\frac{32}{7}$.

Step-by-step explanation:

If a linear equation is written in this form $y=mx+b$, m represents the slope and b represents the y-intercept.

When one line has a slope of $m$, a perpendicular line has a slope of $\frac{-1}{m}$.

We know the slope/intercept equation of the line $y=\frac{7}{5}x+6$, the slope of this equation is $\frac{7}{5}$ and the negative reciprocal of that slope is:

$m=\frac{-1}{\frac{7}{5} } \\m=-\frac{5}{7}$

The perpendicular line will have a slope of:

$y-y_1=(-\frac{5}{7})(x-x_1)$

we use the point given (2, -6) to find the equation of the perpendicular line

$y-(-6)=(-\frac{5}{7})(x-2)$ and we solve for y

$y+6-6=\left(-\frac{5}{7}\right)\left(x-2\right)-6\\\\y=-\frac{5}{7}x+\frac{10}{7}-6\\\\y=-\frac{5}{7}x-\frac{32}{7}$

The equation of a line that is perpendicular to $y=\frac{7}{5}x+6$ and that passes through the point (2, −6) is $y=-\frac{5}{7}x-\frac{32}{7}$.

We can check our work with the graph of the two lines.