Answer:
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Step-by-step explanation:
Be: p(-3,-1) q(-1,-7) r(3,3)
we must find the value of the middle point a and b
[tex]a=\frac{pq}{2}\\b=\frac{rq}{2}[/tex]
remember that distance between two points is:
[tex]d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2} }[/tex]
pq=distance between p and q
qr =distance between q and r[tex]pq=\sqrt{(-1+3)^{2}+(-7+1)^{2}}=\sqrt{(2)^{2}+(-6)^{2}}=\sqrt{40}\\ pq=2\sqrt{10}\\ a=\frac{pq}{2}=\frac{2\sqrt{10}}{2}=\sqrt{10}\\ coordinates\\a=\frac{pq}{2}=\frac{(-3-1,-7-1)}{2}=(-2,-4)\\\\qr=\sqrt{(3+1)^{2}+(3+7)^{2}}=\sqrt{(4)^{2}+(10)^{2}}=\sqrt{116}\\qr=2\sqrt{29}\\ b=\frac{qr}{2}=\frac{2\sqrt{29}}{2}=\sqrt{29}\\ coordinates\\b=\frac{pq}{2}=\frac{(-1+3,-7+3)}{2}=(1,-2)\\[/tex]
ab is parallel to pr if slope ab is equal to slope pr
[tex]m_{ab}=\frac{-2+4}{1+2}=\frac{2}{3}\\m_{pr}=\frac{3+1}{3+3}=\frac{2}{3}\\ m_{ab}=m_{pr}[/tex]
finally distance ab is equal to distance pr/2
[tex]d_{ab=}\sqrt{(-2+4)^{2}+(1+2)^{2}}=\sqrt{13}\\ d_{pr=}\sqrt{(3+1)^{2}+(3+3)^{2}}=\sqrt{52}\\ d_{ab=}\frac{d_{pr}}{2}[/tex]
