Question

Find the equation of the line that

is perpendicular to y = –
-x and
contains the point (4,-8).
y=
x + [ ]

Answer 1

y = 3x/2-14

Explanation:

We are given that the line is perpendicular to y = -2/3 and contains (4,-8).

Perpendicular Def.

`\displaystyle \large{m_1m_2 = - 1}`

Both slopes multiply each others equal to -1.

Finding another slope that is perpendicular to -2/3, substitite m1 = -2/3 in.

`\displaystyle \large{ - (2)/(3) m_2 = - 1}`

Multiply both sides by 3.

`\displaystyle \large{ - (2)/(3) m_2( 3) = - 1(3)} \\ \displaystyle \large{ - 2 m_2= - 3} \\ \displaystyle \large{ m_2= (3)/(2) }`

Therefore, another slope that is perpendicular to -2/3 is 3/2.

Then rewrite in slope-intercept form.

Slope-Intercept

`\displaystyle \large{y = mx + b}`

where m = slope and b = y-intercept; substitute m = 3/2 in.

`\displaystyle \large{y = (3)/(2) x + b}`

Since the line contains (4,-8), substitute x = 4 and y = -8 in and solve for b.

`\displaystyle \large{ - 8 = (3)/(2) (4) + b} \\ \displaystyle \large{ - 8 = 3(2)+ b} \\ \displaystyle \large{ - 8 = 6 + b} \\ \displaystyle \large{ - 8 - 6 = b} \\ \displaystyle \large{ - 14 = b}`

Therefore, b is -14; rewrite again in slope-intercept form.

Thus:-

`\displaystyle \large{y = (3)/(2) x + b} \\ \displaystyle \large{y = (3)/(2) x - 14}`