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.