PS is the median of the triangle with vertice | vedclass

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(C) $S$ is the midpoint of $QR = \left(\frac{6+7}{2}, \frac{-1+3}{2}\right) = \left(\frac{13}{2}, 1\right)$.Slope of $PS = \frac{2-1}{2-\frac{13}{2}}

Vedclass · Accepted Answer

$\frac{-7}{2}, \frac{-7}{9}$

Vedclass · Answer

(C) $S$ is the midpoint of $QR = \left(\frac{6+7}{2}, \frac{-1+3}{2}\right) = \left(\frac{13}{2}, 1\right)$.Slope of $PS = \frac{2-1}{2-\frac{13}{2}} = \frac{1}{-\frac{9}{2}} = -\frac{2}{9}$.The line is parallel to $PS$,so its slope is $m = -\frac{2}{9}$.The equation of the line passing through $(1, -1)$ with slope $m = -\frac{2}{9}$ is $y - (-1) = -\frac{2}{9}(x - 1)$.$9(y + 1) = -2(x - 1)$ $\Rightarrow 9y + 9 = -2x + 2$ $\Rightarrow 2x + 9y + 7 = 0$.To find the $X$-intercept,set $y = 0$: $2x + 7 = 0 \Rightarrow x = -\frac{7}{2}$.To find the $Y$-intercept,set $x = 0$: $9y + 7 = 0 \Rightarrow y = -\frac{7}{9}$.Thus,the intercepts are $-\frac{7}{2}$ and $-\frac{7}{9}$.

$PS$ is the median of the triangle with vertices at $P(2, 2)$,$Q(6, -1)$,and $R(7, 3)$. The intercepts on the coordinate axes of the line passing through point $(1, -1)$ and parallel to $PS$ are respectively:

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