Suppose the hypotenuse and its opposite verte | vedclass

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(C) The equation of the hypotenuse $BC$ is $3x + 4y - 4 = 0$. Let $A = (2, 2)$ be the vertex opposite to the hypotenuse. The altitude $AD$ from $A$ to

Vedclass · Accepted Answer

$7x + y - 16 = 0$

Vedclass · Answer

(C) The equation of the hypotenuse $BC$ is $3x + 4y - 4 = 0$. Let $A = (2, 2)$ be the vertex opposite to the hypotenuse. The altitude $AD$ from $A$ to $BC$ is perpendicular to $BC$. The slope of $BC$ is $m_{BC} = -\frac{3}{4}$. Since $AD \perp BC$,the slope of $AD$ is $m_{AD} = \frac{4}{3}$. In an isosceles right-angled triangle,the altitude from the vertex to the hypotenuse bisects the angle at the vertex. Thus,the sides $AB$ and $AC$ make an angle of $45^{\circ}$ with the altitude $AD$. Let $m$ be the slope of side $AB$ or $AC$. Using the formula $	an 	heta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight|$: $	an 45^{\circ} = \left| \frac{m - \frac{4}{3}}{1 + m(\frac{4}{3})} ight|$ $\Rightarrow 1 = \left| \frac{3m - 4}{3 + 4m} ight|$. This gives two cases: $1 = \frac{3m - 4}{3 + 4m}$ $\Rightarrow 3 + 4m = 3m - 4$ $\Rightarrow m = -7$. $-1 = \frac{3m - 4}{3 + 4m}$ $\Rightarrow -3 - 4m = 3m - 4$ $\Rightarrow 7m = 1$ $\Rightarrow m = \frac{1}{7}$. Using the point-slope form $y - y_1 = m(x - x_1)$ with point $(2, 2)$: For $m = -7$: $y - 2 = -7(x - 2)$ $\Rightarrow y - 2 = -7x + 14$ $\Rightarrow 7x + y - 16 = 0$. For $m = \frac{1}{7}$: $y - 2 = \frac{1}{7}(x - 2)$ $\Rightarrow 7y - 14 = x - 2$ $\Rightarrow x - 7y + 12 = 0$. Comparing with the options,$7x + y - 16 = 0$ is present.

Suppose the hypotenuse and its opposite vertex of an isosceles right-angled triangle are $3x + 4y - 4 = 0$ and $(2, 2)$ respectively. Then,which of the following is another side of the triangle?

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