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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One diagonal of a square is along the line $8x – 15y = 0$ and one of its vertex is $(1, 2)$ Then the equation of the sides of the square passing through this vertex, are

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The triangle formed by the lines $x + y – 4 = 0,\,$ $3x + y = 4,$ $x + 3y = 4$ is

easy

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$P (x, y)$ moves such that the area of the triangle formed by $P, Q (a , 2 a)$ and $R (- a, – 2 a)$ is equal to the area of the triangle formed by $P, S (a, 2 a)\,\,\, \&\,\, \,T (2 a, 3 a)$. The locus of $'P'$ is a straight line given by :

normal

The pair of straight lines $x^2 – 4xy + y^2 = 0$ together with the line $x + y + 4 = 0$ form a triangle which is :

normal

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

Solution

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