The equations of two equal sides of an isosce | vedclass

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(C) Let the slope of the third side be $m$. The equation of the line passing through $(1, -10)$ is $y + 10 = m(x - 1)$,which simplifies to $mx - y - (

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let the slope of the third side be $m$. The equation of the line passing through $(1, -10)$ is $y + 10 = m(x - 1)$,which simplifies to $mx - y - (m + 10) = 0$. Since the triangle is isosceles,the third side makes equal angles with the two given sides. The slopes of the given lines are $m_1 = 7$ and $m_2 = -1$. Using the formula for the angle between two lines,$	an 	heta = |\frac{m - m_1}{1 + m \cdot m_1}| = |\frac{m - m_2}{1 + m \cdot m_2}|$,we get: $|\frac{m - 7}{1 + 7m}| = |\frac{m - (-1)}{1 + m(-1)}| = |\frac{m + 1}{1 - m}|$. Solving $|\frac{m - 7}{1 + 7m}| = |\frac{m + 1}{1 - m}|$: Case $1$: $\frac{m - 7}{1 + 7m} = \frac{m + 1}{1 - m}$ $\Rightarrow (m - 7)(1 - m) = (m + 1)(1 + 7m)$ $\Rightarrow m - m^2 - 7 + 7m = m + 7m^2 + 1 + 7m$ $\Rightarrow 8m^2 - m + 8 = 0$. This gives no real roots. Case $2$: $\frac{m - 7}{1 + 7m} = -(\frac{m + 1}{1 - m})$ $\Rightarrow (m - 7)(1 - m) = -(m + 1)(1 + 7m)$ $\Rightarrow -m^2 + 8m - 7 = -(7m^2 + 8m + 1)$ $\Rightarrow 6m^2 + 16m - 6 = 0$ $\Rightarrow 3m^2 + 8m - 3 = 0$. Factoring gives $(3m - 1)(m + 3) = 0$,so $m = \frac{1}{3}$ or $m = -3$. For $m = \frac{1}{3}$,the equation is $y + 10 = \frac{1}{3}(x - 1)$ $\Rightarrow 3y + 30 = x - 1$ $\Rightarrow x - 3y - 31 = 0$. For $m = -3$,the equation is $y + 10 = -3(x - 1)$ $\Rightarrow y + 10 = -3x + 3$ $\Rightarrow 3x + y + 7 = 0$.

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 $(1, -10)$,then the equation of the third side is:

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