The equations of two equal sides of an isosce | vedclass

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(C) Let the slope of the third side be $m$. Since it passes through $(1, -10)$,its equation is $y + 10 = m(x - 1)$.Since the third side makes equal an

Vedclass · Accepted Answer

$3x + y + 7 = 0$ or $x - 3y - 31 = 0$.

Vedclass · Answer

(C) Let the slope of the third side be $m$. Since it passes through $(1, -10)$,its equation is $y + 10 = m(x - 1)$.Since the third side makes equal angles with the two given sides,we use the formula for the angle between two lines:$\frac{m - 7}{1 + 7m} = \frac{m - (-1)}{1 + m(-1)}$$\frac{m - 7}{1 + 7m} = \frac{m + 1}{1 - m}$$(m - 7)(1 - m) = (m + 1)(1 + 7m)$$m - m^2 - 7 + 7m = m + 7m^2 + 1 + 7m$$8m^2 + m + 8 = 0$ is incorrect; let's re-evaluate:$m - m^2 - 7 + 7m = m + 7m^2 + 1 + 7m$$-m^2 + 8m - 7 = 7m^2 + 8m + 1$$8m^2 = -8 \implies m^2 = -1$ (This implies the angle bisector approach is needed).Alternatively,the slopes of the given lines are $m_1 = 7$ and $m_2 = -1$. The angle bisectors of these lines have slopes $m = -3$ and $m = 1/3$.For $m = -3$,the line is $y + 10 = -3(x - 1) \implies 3x + y + 7 = 0$.For $m = 1/3$,the line is $y + 10 = \frac{1}{3}(x - 1) \implies x - 3y - 31 = 0$.Thus,the equations are $3x + y + 7 = 0$ or $x - 3y - 31 = 0$.

The equations of two equal sides of an isosceles triangle are $7x - y + 3 = 0$ and $x + y - 3 = 0$. If the third side passes through the point $(1, -10)$,find the equation of the third side.

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