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What is the equation of the line perpendicular to the line 2x + 3y = -6 and having Y-intercept 3?
Answer & Solution
Correct Answer:
Option
B
$$\eqalign{
& 2x + 3y = - 6 \cr
& y = - \frac{2}{3}x - \frac{6}{3} \cr
& \therefore {\text{Slope}} = - \frac{2}{3} \cr} $$
⇒ Two lines are perpendicular if m1 × m2 = -1
Y intercept = 3
⇒ y = mx + c
⇒ m2 = $$\frac{3}{2}$$
The equation of line with slope $$\frac{3}{2}$$ is:-
y - y1 = m2(x - x1)
where, (x1 y1) = (0, 3)
y - 3 = $$\frac{3}{2}$$(x - 0)
2y - 6 = 3x
3x - 2y = -6
⇒ Two lines are perpendicular if m1 × m2 = -1
Y intercept = 3
⇒ y = mx + c
⇒ m2 = $$\frac{3}{2}$$
The equation of line with slope $$\frac{3}{2}$$ is:-
y - y1 = m2(x - x1)
where, (x1 y1) = (0, 3)
y - 3 = $$\frac{3}{2}$$(x - 0)
2y - 6 = 3x
3x - 2y = -6
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