Question

The two lines y = 2x + 8 and y = 2x - 12 intersect the x-axis at the P and Q.

Work out the distance PQ.

Answer 1

Answer:
`\Huge\boxed{Distance=10~units}`

Explanation:

Find the point P

Given expression

y = 2x + 8

Substitute 0 for the y value to find the x value

This is the definition of x-intercepts

(0) = 2x + 8

Subtract 8 on both sides

0 - 8 = 2x + 8 - 8

-8 = 2x

Divide 2 on both sides

-8 / 2 = 2x / 2

x = -4

`\large\boxed{P~(-4,0)}`

Find the point Q

Given expression

y = 2x - 12

Substitute 0 for the y value to find the x value

(0) = 2x - 12

Add 12 on both sides

0 + 12 = 2x - 12 + 12

12 = 2x

Divide 2 on both sides

12 / 2 = 2x / 2

x = 6

`\large\boxed{Q~(6,0)}`

Find the distance between PQ

Given information

`(x_1,~y_1)=(-4,~0)`

`(x_2,~y_2)=(6,~0)`

Substitute values into the distance formula

`Distance=√((x_2-x_1)^2+(y_2-y_1)^2)`

`Distance=√((6-(-4))^2+(0-0)^2)`

Simplify values in the parenthesis

`Distance=√((10)^2+(0)^2)`

Simplify values in the radical sign

`Distance=√(100)`

`\Huge\boxed{Distance=10~units}`

Hope this helps!! :)

Please let me know if you have any questions

Answer 2

PQ = 10 units

Explanation:

to find where the lines cross the x- axis let y = 0 and solve for x , that is

2x + 8 = 0 ( subtract 8 from both sides )

2x = - 8 ( divide both sides by 2 )

x = - 4 ← point P

and

2x - 12 = 0 ( add 12 to both sides )

2x = 12 ( divide both sides by 2 )

x = 6 ← point Q

the lines cross the x- axis at x = - 4 and x = 6

using the absolute value of the difference , then

PQ = | - 4 - 6 | = | - 10 | = 10 units

or

PQ = | 6 - (- 4) | = | 6 + 4 | = | 10 | = 10 units