Question

write the equation of a line perpendicular to the line that passes through the given point.$\frac{1}{2}y = 8x + 3 = 0$(-8,0)

Answer 1

Given the equation:

$\frac{1}{2}y=8x+3$

Rewrite the equation in slope intercept form:

y = mx + b

Multiply both through by 2

$\begin{gathered} \frac{1}{2}y\ast2=\text{ }8x\ast2\text{ + 3}\ast2 \\ \\ y\text{ = 16x + 6} \end{gathered}$

The slope of this line is 16.

therefore, the slope of the line perperndicular to it will be it's inverse:

$-\frac{1}{16}$

The perpendicular line has the points:

(x, y) ===> (-8, 0)

We have:

y = mx + b

$0\text{ =-}\frac{1}{16}(-8)+b$

Solve for b which is the y-intercept:

$\begin{gathered} 0\text{ = }\frac{1}{2}+b \\ \\ b\text{ = -}\frac{1}{2} \end{gathered}$

Since the y intercept is -½

slope = -1/16

The equation of the line in point slope form:

(y - y1) = m(x - x1)

$(y\text{ - 0) = -}\frac{1}{16}(x\text{ + 8)}$

Therefore, the equation of the perpendicular line in slope intercept is:

$y\text{ = -}\frac{1}{16}x\text{ - }\frac{1}{2}$