(a) Find the gradient of line l.
(b) Find the equation of line l in the form y = mx + c.
(c) Find the equation of the line that is perpendicular to line l and passes through the point (12, -7). Give your answer in the form y = mx + c.
▶️ Answer/Explanation
(a) -3/4 or -0.75
Gradient = rise/run = -3/4
(b) y = -3/4x + 2
Using y-intercept at (0,2) and gradient from (a)
(c) y = 4/3x – 23
Perpendicular gradient = 4/3 (negative reciprocal). Substituted (12,-7) into y=4/3x+c to find c.
A is the point (-6, 5) and B is the point (-2, -3).
(a) Find the equation of the straight line, l, that passes through point A and point B. Give your answer in the form y = mx + c.
(b) Find the equation of the line that is perpendicular to l and passes through the origin.
▶️ Answer/Explanation
(a) \( y = -2x – 7 \)
(b) \( y = \frac{1}{2}x \)
Explanation:
(a) Equation of line AB:
1. Find slope (m): \( m = \frac{y_2 – y_1}{x_2 – x_1} = \frac{-3 – 5}{-2 – (-6)} = \frac{-8}{4} = -2 \).
2. Find y-intercept (c): Using point A (-6, 5), substitute into \( y = mx + c \):
\( 5 = -2(-6) + c \) → \( c = -7 \).
3. Final equation: \( y = -2x – 7 \).
(b) Perpendicular line through origin:
1. Slope of perpendicular line: Negative reciprocal of -2 → \( \frac{1}{2} \).
2. Since it passes through (0,0): \( y = \frac{1}{2}x \) (no y-intercept term needed).