Discussion

What is the equation of the line passing through (-3, 5) and perpendicular to the line through the points (2, 5) and (-3, 6)?

Answer: A

First, find the slope (m₁) of the line through (2, 5) and (-3, 6).

m₁ = \(\frac{6-5}{-3-2} = \frac{1}{-5} = -1/5\).

The slope (m₂) of the required perpendicular line is the negative reciprocal of m₁. m₂ = 5.

Now use the point-slope form with point (-3, 5) and slope m₂=5.

y - 5 = 5(x - (-3))

y - 5 = 5(x + 3)

y - 5 = 5x + 15

5x - y + 20 = 0.

No comment is present. Be the first to comment.

Loading…