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What is an equation of the line that is perpendicular to y - 4 = 2(x - 6) and

passes through the point (-3, -5)?
• A. y -5=1/2(x-4)
• B. y - 5 = -2(x- 3)
• C. y + 5 = 2(x+ 3)
O D. y +5- -1/2(x+3)

User Jon Adams
by
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1 Answer

2 votes

Answer:

D. y + 5 = -1/2(x + 3)

Explanation:

Let k: y = a₁x + b₁ and l: y = a₂x + b₂. Then

k ║ l ⇔ a₁ = a₂

k ⊥ l ⇔ a₁ · a₂ = -1

We have the euation of a line in point slope form:

(y - y₁) = m(x - x₁)

Convert to the slope intercept form:

y =mx + b

y - 4 = 2(x - 6)

y - 4 = 2x - 12

y - 4 + 4 = 2x - 12 + 4

y = 2x - 8

The slope is m₁ = 2.

Let y = m₂x + b.

The lines is perpendicular. Therefore:

2m₂ = -1 |divide both sides by 2

m₂ = -1/2

The line passes through the point (-3, -5).

Substitute to the point slope equation:

y - (-5) = -1/2(x - (-3))

y + 5 = -1/2(x + 3)

User Essien
by
8.7k points

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