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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).

$\begin {array} {1 1} (A)\;x-7y-12=0 & \quad (B)\;x+7y-12=0 \\ (C)\;x-7y+12=0 & \quad (D)\;x+7y+12=0 \end {array}$

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  • Slope of a line is $m = -\large\frac{coeff\: of \: x}{coeff\: of \: y}$
  • $ \tan \theta = \bigg| \large\frac{m_1-m_2}{1+m_1m_2} \bigg|$
Step 1 :
The equation of the hypotenuse is $3x+4y=4$ .
Hence the slope of the line is $ -\large\frac{3}{4}$
ASince it is an isoceles right angled triangle the angles should be $ 45^{\circ}, 45^{\circ}$ and $ 90^{\circ}$
$ \tan 45^{\circ} = \bigg| \large\frac{m- \bigg( -\large\frac{3}{4} \bigg)}{1+m\bigg(-\large\frac{3}{4} \bigg) }\bigg|$
But $\tan 45^{\circ}=1$
$1 = \bigg| \large\frac{m+\large\frac{3}{4}}{1-\large\frac{3m}{4}} \bigg|$
$1-\large\frac{3m}{4} = \pm \bigg( m + \large\frac{3}{4} \bigg)$
Step 2 :
If $ 1 - \large\frac{3m}{4}=m+\large\frac{3}{4}$
$ \Rightarrow m+ \large\frac{3m}{4}$$=1-\large\frac{3}{4}$
$ \large\frac{7m}{4}=\large\frac{1}{4}$
$ \therefore m = \large\frac{1}{7}$
The coordinates of B is (2,2)
Hence the equation of the line AB is
$ 7y-14=x-2$
$ \therefore x-7y=-12$
Hence $x-7y+12$ is the required equation.
answered Jul 1, 2014 by thanvigandhi_1
edited Jul 1, 2014 by thanvigandhi_1

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