Question

What is the y-intercept of the equation of the line that is perpendicular to the line y =3/5x + 10 and passes through the point (15, –5)? The equation of the line in slope-intercept form is y = -5/3x + .

Answer 1

The answe would be y=-5/3x+20

Answer 2

Answer:

Equation of the line in the slope-intercept form will be $y=-\frac{5}{3}x+20$

Step-by-step explanation:

An equation of the line perpendicular to $y=\frac{3}{5}x+10$ will be in the form of y = mx + c

Where m = slope of the line

c = y intercept of the line

From the property of the perpendicular line

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

where $m_{1}$ and $m_{2}$ are the slopes of the perpendicular lines.

If  $m_{1}$ = $\frac{3}{5}$

then $\frac{3}{5}\times m_{2}=-1$

$m_{2}=-\frac{5}{3}$

So the equation will be $y=-\frac{5}{3}x+c$

This line passes through the point (15, -5)

$(-5)=-\frac{5}{3}(15)+c$

-5 = -25 + c

c = 25 - 5

c = 20

Finally the equation will be $y=-\frac{5}{3}x+20$