Question

Triangle PQR has vertices P(-3, -1), Q(-1, 7), and R(3, 3), and point A and B are midpoints of PQ and RQ, respectively. Use coordinate geometry to prove
(a)
AB is parallel to PR
(b)
AB is half the length of PR


Please include graph photos

Answer 1

Answer:

Part a) The figure a shows that AB is parallel to PR. Please see attached figure a.

Part b) The length of AB is half the length of PR

Step-by-step explanation:

Part A)

Let M be the midpoint formula between points A(x₁, y₁) and B(x₂, y₂)

$M=({\displaystyle {\frac {x_{1}+x_{2}}{2}}, {\frac {y_{1}+y_{2}}{2})$

Let A be the midpoint of P(-3, -1) and Q(-1, 7)

The coordinates of the midpoint of P(-3, -1) and Q(-1, 7):

$Mpq{} =({\displaystyle {\frac {-3+(-1)}{2}}, {\frac {-1+7}{2})$

$Mpq{} =(-2, 3)$

Let B be the midpoint of Q(-1, 7) and R(3, 3):

$Mqr{} =({\displaystyle {\frac {-1+(3)}{2}}, {\frac {7+3}{2})$

$Mqr{} =(1, 5)$

The figure a shows that AB is parallel to PR. Please see attached figure a.

Part B)

The point A has the coordinates: A(-2, 3)

The point B has the coordinates: B(1, 5)

The distance formula is: $d={\sqrt {(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}$

So, the length of AB $={\sqrt {(1-(-2))^{2}+(5-3)^{2}$

                                         = $\sqrt{9+4}$

                                         = $\sqrt{13}$

                                          = $3.6056$

And, the length of PR $={\sqrt {(3-(-3))^{2}+(3-(-1)^{2}$

                                           = $\sqrt{36+16}$

                                           = $\sqrt{52}$

                                           = $2\sqrt{13}$

                                           = = $7.2111$

As the length of AB = 3.6056 and the length of PR = 7.2111

So, it is clear that the length of AB is half the length of PR.

Keywords: distance formula, midpoints, triangle, coordinate plane

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