Question

The equation of line c is y = -6x - 3. Line d includes the point (3, -3) and is perpendicular to line c. What is the equation of line d?

Answer 1

Answer::

`y=(1)/(6)x-(7)/(2)\text{ or }x-6y=21`

Step-by-step explanation:

Two lines are perpendicular if the product of their slopes is -1.

First, find the slope of line c (by comparing it with the slope-intercept form).

`\begin{gathered} y=mx+b \\ y=-6x-3 \\ \implies\text{Slope, m=-6} \end{gathered}`

Let the slope of line d = n

Since Lines c and d are perpendicular lines, thus:

`n*-6=-1\implies n=(1)/(6)`

So, we have that line d has the following properties:

• Slope = 1/6

,

• Point = (3,-3)

We find the equation of line d using the point-slope form:

`y-y_1=m(x-x_1)\text{ where }\begin{cases}m=(1)/(6) \\ (x_1,y_1)=(3,-3)\end{cases}`

Substitute the given values:

`\begin{gathered} y-(-3)=(1)/(6)(x-3)\text{ } \\ y+3=(1)/(6)x-(3)/(6) \\ y=(1)/(6)x-(1)/(2)-3 \\ y=(1)/(6)x-(7)/(2) \\ \text{Multiply all through by 6} \\ 6y=x-21 \\ x-6y=21 \end{gathered}`

The equation of line d is:

`y=(1)/(6)x-(7)/(2)\text{ or }x-6y=21`