Equation of one of sides of an isosceles right angled triangle whose hypotenuse is 3x + 4y = 4 and o

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9.Straight Line

hard

Equation of one of the sides of an isosceles right angled triangle whose hypotenuse is $3x + 4y = 4$ and the opposite vertex of the hypotenuse is $(2, 2)$, will be

$x - 7y + 12 = 0$

$7x + y - 12 = 0$

$x - 7y + 16 = 0$

$y - (3 - 2\sqrt 2 )x = 0$

Solution

(a) Since $\angle A = \angle C = {45^o}$. We have to find equation of $AB$. Here let gradient of $AB$ be $m$, then equation of $AB$ is $y – 2 = m(x – 2)$……(i)
But angle between $3x + 4y = 4$ and (i) is ${45^o}$
So, $\tan {45^o} = \frac{{m + \frac{3}{4}}}{{1 – \frac{{3m}}{4}}} \Rightarrow m = \frac{1}{7}$
Hence, the required equation is $x – 7y + 12 = 0$
{By putting the value of m in (i)}.

Standard 11

Mathematics

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medium

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easy

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normal

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