please solve Q5 part (i) and (ii)
Answers
Answer:
don't know your answer which chapter
Answer:
(i) Slope of the line perpendicular to x - y/2 + 3 = 0 is -(1/2)
(ii) Slope of the line perpendicular to x/3 - 2y = 4 is -6
Step-by-step explanation:
x - y/2 + 3 = 0
Arranging the given equation in the form of y = mx + c,
where m = slope
Multiplying throughout the equation by 2, we get,
2x - y + 6 = 0
y = 2x + 6
∴ m1 = 2
The product of the slopes of two perpendicular lines = -1
∴ m1 x m2 = -1
∴ 2 x m2 = -1
∴ m2 = -(1/2)
Similarly,
(ii) x/3 - 2y = 4
Arranging the given equation in the form of y = mx + c,
where m = slope
Multiplying throughout the equation by 3, we get,
x - 6y = 12
6y = x - 12
y = (1/6)x - 12/6
y = (1/6)x - 2
∴ m1 = 1/6
The product of the slopes of two perpendicular lines = -1
∴ m1 x m2 = -1
∴ 1/6 x m2 = -1
∴ m2 = -1 / (1/6)
∴ m2 = -6