मराठी

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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प्रश्न

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).

बेरीज
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उत्तर

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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पाठ 10: Straight Lines - Exercise [पृष्ठ १७९]

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एनसीईआरटी एक्झांप्लर Mathematics Exemplar [English] Class 11
पाठ 10 Straight Lines
Exercise | Q 12 | पृष्ठ १७९

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