The equation of one of the sides of an isosce | vedclass

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Question

(A) Let the vertex opposite to the hypotenuse be $B(2, 2)$. Since the triangle is an isosceles right-angled triangle,the angles at the base are $45^\c

Vedclass · Accepted Answer

$x - 7y + 12 = 0$

Vedclass · Answer

(A) Let the vertex opposite to the hypotenuse be $B(2, 2)$. Since the triangle is an isosceles right-angled triangle,the angles at the base are $45^\circ$.Let the slope of one of the sides passing through $B(2, 2)$ be $m$.The equation of this side is $y - 2 = m(x - 2)$,which simplifies to $mx - y + (2 - 2m) = 0$.The angle between this line and the hypotenuse $3x + 4y - 4 = 0$ is $45^\circ$.Using the formula $	an 	heta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight|$,where $m_1 = m$ and $m_2 = -\frac{3}{4}$:$	an 45^\circ = \left| \frac{m - (-3/4)}{1 + m(-3/4)} ight| = 1$$\left| \frac{4m + 3}{4 - 3m} ight| = 1$This gives two cases:Case $1$: $\frac{4m + 3}{4 - 3m} = 1$ $\Rightarrow 4m + 3 = 4 - 3m$ $\Rightarrow 7m = 1$ $\Rightarrow m = \frac{1}{7}$.Substituting $m = \frac{1}{7}$ into $y - 2 = m(x - 2)$:$y - 2 = \frac{1}{7}(x - 2)$ $\Rightarrow 7y - 14 = x - 2$ $\Rightarrow x - 7y + 12 = 0$.Case $2$: $\frac{4m + 3}{4 - 3m} = -1$ $\Rightarrow 4m + 3 = -4 + 3m$ $\Rightarrow m = -7$.Substituting $m = -7$ into $y - 2 = m(x - 2)$:$y - 2 = -7(x - 2)$ $\Rightarrow y - 2 = -7x + 14$ $\Rightarrow 7x + y - 16 = 0$.Comparing with the given options,$x - 7y + 12 = 0$ is the correct equation.

The equation of one of the sides of an isosceles right-angled triangle,whose hypotenuse is $3x + 4y = 4$ and the vertex opposite to the hypotenuse is $(2, 2)$,is:

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