In an isosceles right-angled triangle,if the | vedclass

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(A) The equation of the hypotenuse is $3x + 4y = 4$,which can be written as $y = -\frac{3}{4}x + 1$. Thus,the slope of the hypotenuse is $m_1 = -\frac

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$\frac{1}{7}, -7$

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(A) The equation of the hypotenuse is $3x + 4y = 4$,which can be written as $y = -\frac{3}{4}x + 1$. Thus,the slope of the hypotenuse is $m_1 = -\frac{3}{4}$. Let the slopes of the other two sides be $m$. Since it is an isosceles right-angled triangle,the angle between the hypotenuse and each of the other two sides is $45^{\circ}$. Using the formula $	an 	heta = \left| \frac{m - m_1}{1 + m \cdot m_1} ight|$,we have: $	an 45^{\circ} = \left| \frac{m - (-3/4)}{1 + m(-3/4)} ight| = 1$. $1 = \left| \frac{4m + 3}{4 - 3m} ight|$. This gives two cases: Case $1$: $\frac{4m + 3}{4 - 3m} = 1$ $\Rightarrow 4m + 3 = 4 - 3m$ $\Rightarrow 7m = 1$ $\Rightarrow m = \frac{1}{7}$. Case $2$: $\frac{4m + 3}{4 - 3m} = -1$ $\Rightarrow 4m + 3 = -4 + 3m$ $\Rightarrow m = -7$. Therefore,the slopes of the remaining two sides are $\frac{1}{7}$ and $-7$.

In an isosceles right-angled triangle,if the equation of the hypotenuse is $3x + 4y = 4$ and its opposite vertex is $(2, 2)$,then the slopes of the remaining two sides are:

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