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Question
Find the equation of a line passing through (3, – 2) and perpendicular to the line.
x - 3y + 5 = 0.
Solution
x - 3y + 5 = 0
⇒ 3y = x + 5
∴ y = `x/(3) + (5)/(3)`
∴ m1 = `(1)/(3)`
Since lines are perpendicular to each other
∴ m1 x m2 = -1
`(1)/(3) xx m_2` = -1
m2 = -1 x 3
m2 = -3
Passing point is (3, -2)
∴ Equation of line
y - y1 = m(x - x1)
⇒ y + 2 = -3(x - 3)
⇒ y + 2 = -3x + 9
⇒ 3x + y + 2 - 9 = 0
⇒ 3x + y =7.
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