A line that contains the point (-8,-1) and is perpendicular to -15x +3y = 9

1 answer:

7 0

Put the other line in point slope form...(solve for y :D )

3y=15x+9

y=5x+3, so this line has a slope of 5

For a line to be perpendicular to another the slopes have to be negative reciprocals of each other...mathematically...

s1*s2=-1

So for the other line to be perpendicular to the first:

5m=-1, m=-1/5, thus our other line (so far) is:

y=b-x/5, now we can use the point (-8,-1) to solve for the y-intercept, "b":

-1=b-(-8)/5

-1=b+8/5

-1=b+1.6

b=-2.6

So the perpendicular line that contains the point (-8,-1) is:

y=-0.2x-2.6

You might be interested in

5x^2+11x=0 quadratic equation

hammer [34]

$5x^{2} + 11x = 0
$
$5x^{2} + 11x + 0 = 0$
$x = \frac{-(11) +/- \sqrt{(11)^{2} - 4(5)(0)}}{2(5)}$

$x = \frac{-11 +/- \sqrt{121 - 0}}{10}$

$x = \frac{-11 +/- \sqrt{121}}{10}$

$x = \frac{-11 +/- 11}{10}$

$x = \frac{11 + 11}{10}$    $x = \frac{11 - 11}{10}$

$x = \frac{22}{10}$          $x = \frac{0}{10}$

$x = 2.2$       $x = 0$