Question

WILL GIVE BRAINLIEST! For quadrilateral ABCD with vertices A(–3, –4), B(3, 0), C(5, 6) and D(–7, 2), the points M, N, P, and Q are the midpoints of AB, BC, CD, and DA respectively. Show that MN = PQ and PN = QM. Provide your complete solutions and proofs in your paper homework and enter the numeric answers online.

Answer 1

Answer:

The distances between the given points are;

MN = PQ = √41

PN = QM = √26

Step-by-step explanation:

The given coordinates of the vertices of the quadrilateral are;

A(-3, -4), B(3, 0), C(5, 6), D(-7, 2)

The midpoint of AB = The point M

The midpoint of BC = The point N

The midpoint of CD = The point P

The midpoint of DA = The point Q

Therefore;

The coordinates of the point M = ((-3 + 3)/2, (-4 + 0)/2) = (0, -2)

The coordinates of the point N = ((3 + 5)/2, (0 + 6)/2) = (4, 3)

The coordinates of the point P = ((5 + (-7))/2, (6 + 2)/2) = (-1, 4)

The coordinates of the point Q = ((-7 + (-3))/2, (2 + (-4))/2) = (-5, -1)

The formula for the distance, d, between two coordinates is given as follows;

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

The distance of MN using an online distance between coordinates calculation tool is therefore;

$d_{MN} = \sqrt{\left (3-(-2) \right )^{2}+\left (4 - 0 \right )^{2}} = \sqrt{41}$

The distance of PQ is given as follows;

$d_{PQ} = \sqrt{\left ((-1)-4 \right )^{2}+\left (-5 - (-1) \right )^{2}} = \sqrt{41}$

Therefore;

MN = PQ = √41

The distance of PN is given as follows;

$d_{PN} = \sqrt{\left (3-4 \right )^{2}+\left (4 - (-1) \right )^{2}} = \sqrt{26}$

The distance of QM is given as follows;

$d_{QM} = \sqrt{\left (-2-(-1) \right )^{2}+\left (0 - (-5) \right )^{2}} = \sqrt{26}$

PN = QM = √26