What is the slope of a line that is perpendicular to the line v=-1/3x-6

1 answer:

7 0

If you are trying to find the slope of a line that is perpendicular to another line, you would need to change the sign and change it to the fraction's reciprocal.

m = $-\frac{1}{3}$

Change the sign of the slope:

$\frac{1}{3}$

Finally, make the reciprocal of the original slope (switching the numerator and denominator):

$\frac{3}{1}$

Simplify:

Final answer:

$m=3$

You might be interested in

write a linear equation that intersects y=x^2 at two points. Then write a second linear equation that intersects y=x^2 at just o

Liula [17]

We know that $y = x^2$ is a parabola, concave up, with vertex in the origin $(0,0)$

So, we may use three horizontal lines for our purpose: any horizontal line above the x axis will intersect the parabola twice. The x axis itself intersects the parabola once on the vertex, while any horizontal line below the x axis won't intercept the parabola.

Here's the examples:

The horizontal line $y = 4$ intercepts the parabola twice: the system $y = x^2,\ y = 4$ is solved by $x^2=4 \implies x = \pm 2$
The horizontal line $y=0$ intercepts the parabola only once: the system is $y=x^2,\ y=0$ which yields $x^2=0\implies x=0$ which is a repeated solution
The horizontal line $y=-5$ intercepts the parabola only once: the system is $y=x^2,\ y=-5$ which yields $x^2=-5$ which is impossible, because a squared number can't be negative.