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Question
Find the equation of one of the sides of an isosceles right angled triangle whose hypotenuse is given by 3x + 4y = 4 and the opposite vertex of the hypotenuse is (2, 2).
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Solution
Given that equation of the hypotenuse is 3x + 4y = 4 and opposite vertex is (2, 2)
Slope BC = `(-3)/4`
Let slope of AC be m
∴ tan 45° = `|(m + 3/4)/(1 + ((-3)/4))|`
⇒ 1 = `|(4m + 3)/(4 - 3m)|`
⇒ `(4m + 3)/(4 - 3m)` = ± 1
Taking (+) sign,
`(4m + 3)/(4 - 3m)` = 1
⇒ 4m + 3 = 4 – 3m
⇒ 7m = 1
⇒ m = `1/7`
Taking (–) sign,
`(4m + 3)/(4 - 3m)` =– 1
⇒ 4m + 3 = – 4 + 3m
⇒ 4m – 3m = – 3 – 4
⇒ m = – 7
∴ Equation of AC with slope `(1/7)` is
y – 2 = `1/7(x - 2)`
⇒ 7y – 14 = x – 2
⇒ x – 7y + 12 = 0
Equation of AC with slope (– 7) is
y – 2 = – 7(x – 2)
⇒ y – 2 = – 7x + 14
⇒ 7x + y – 16 = 0
Hence, the required equation are x – 7y + 12 = 0 and 7x + y – 16 = 0.
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