Question

I am so lost please help

Answer 1

(a) -
`y=-2x+18`

(b) -
`y=(1)/(2) x-(9)/(2)`

Explanation:

Given the equation of a line, which we'll call line 1, find the following.

(a) - The equation of a line, which we'll call line 2, that is parallel to line 1 and travels through the point (9,0)

(b) - The equation of a line, which we'll call line 3, that is perpendicular to line 1 and travels through the point (9,0)

Given:

`3y+6x=-6`

(1) - Write the equation of line 1 in slope-intercept form

`\boxed{\left\begin{array}{ccc}\text{\underline{Slope-Intercept Form:}}\\y=mx+b\\\bullet \ m \ \text{is the slope of the line}\\\bullet \ b \ \text{is the y-intercept of the line}\end{array}\right}`

`3y+6x=-6\\\\\Longrightarrow 3y=-6-6x\\\\\Longrightarrow y=-(6)/(3) -(6)/(3)x \\\\\therefore \boxed{y=-2x-2}`

Thus, we can conclude the slope of line 1 is -2.

(2) - Answering part (a)

To find a line that is parallel to line 1, the slopes must be the same, -2. Use the point-slope form for a line to find the equation for line 2.

`\boxed{\left\begin{array}{ccc}\text{\underline{Point-Slope Form:}}\\y-y_1=m(x-x_1)\\\bullet \ m \ \text{is the slope of the line}\\\bullet \ (x_1,y_1) \ \text{is a point the line passes through}\end{array}\right}`

`y-y_1=m(x-x_1); \ \text{Recall that} \ m=-2 \ \text{and} \ (x_1,y_1)=(9,0)\\\\\Longrightarrow y-0=-2(x-9)\\\\\therefore \boxed{\boxed{ y=-2x+18}}`

Thus, the equation for line 2 is found.

(2) - Answering part (b)

To find a line that is perpendicular to line 1, the slope of line 3 must be the opposite-reciprocal of line 1's. Once again, use the point-slope form of a line to find the equation of line 3.

`y-y_1=m(x-x_1); \ \text{Recall that} \ m=(1)/(2) \ \text{and} \ (x_1,y_1)=(9,0)\\\\\Longrightarrow y-0=(1)/(2)(x-9)\\\\\therefore \boxed{\boxed{ y=(1)/(2) x-(9)/(2) }}`

Thus, the equation of line 3 is found.