Question

Triangle pqr has vertices p(-3 -1) q(-1,-7) and r(3,3) and points a and b are midpoints of segment pq and segment rq respectively.Use coordinate geometry to prove that segment ab is parallel to segment pr and is half the length of segment pr

Answer 1

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

$a=\frac{pq}{2}\\b=\frac{rq}{2}$

remember that distance between two points is:

$d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2} }$

pq=distance between p and q

qr =distance between q and r$pq=\sqrt{(-1+3)^{2}+(-7+1)^{2}}=\sqrt{(2)^{2}+(-6)^{2}}=\sqrt{40}\\ pq=2\sqrt{10}\\ a=\frac{pq}{2}=\frac{2\sqrt{10}}{2}=\sqrt{10}\\ coordinates\\a=\frac{pq}{2}=\frac{(-3-1,-7-1)}{2}=(-2,-4)\\\\qr=\sqrt{(3+1)^{2}+(3+7)^{2}}=\sqrt{(4)^{2}+(10)^{2}}=\sqrt{116}\\qr=2\sqrt{29}\\ b=\frac{qr}{2}=\frac{2\sqrt{29}}{2}=\sqrt{29}\\ coordinates\\b=\frac{pq}{2}=\frac{(-1+3,-7+3)}{2}=(1,-2)$

ab is parallel to pr if slope ab is equal to slope pr

$m_{ab}=\frac{-2+4}{1+2}=\frac{2}{3}\\m_{pr}=\frac{3+1}{3+3}=\frac{2}{3}\\ m_{ab}=m_{pr}$

finally distance ab is equal to distance pr/2

$d_{ab=}\sqrt{(-2+4)^{2}+(1+2)^{2}}=\sqrt{13}\\ d_{pr=}\sqrt{(3+1)^{2}+(3+3)^{2}}=\sqrt{52}\\ d_{ab=}\frac{d_{pr}}{2}$