Question

ABCD is a quadrilateral.

Coordinates are:
A (3,2)
B (8,1)
C (7,6)
D (2,7)

Find the midpoint of:
AC
BD

Find the slope of:
AC
BD

Find the distance of:
AC
BD

What most specific type of quadrilateral is ABCD based on the info...

Thanks.!!

Answer 1

Answer on Question#$40149 – Math – Geometry

ABCD is a quadrilateral.

Coordinates are:

A (3,2)

B (8,1)

C (7,6)

D (2,7)

Find the midpoint of:

AC

BD

Find the slope of:

AC

BD

Find the distance of:

AC

BD

What most specific type of quadrilateral is ABCD based on the info.

Solution:

We get parallelogram - most specific type of quadrilateral.

Let us find the coordinates of midpoint P.

AC: $\frac{x - 3}{7 - 3} = \frac{y - 2}{6 - 2}$, $\frac{x - 3}{4} = \frac{y - 2}{4}$,

$$x - 3 = y - 2, \quad y = x - 1.$$

BD: $\frac{x - 8}{2 - 8} = \frac{y - 1}{7 - 1}$, $\frac{x - 8}{6} = \frac{y - 1}{6}$, $y = -x + 9$;

Whence coordinates of midpoint P:

$$\left\{ \begin{array}{l} y = x - 1 \\ y = -x + 9. \end{array} \right\} \iff -x + 9 = x - 1, -2x = -10,$$ $$\iff \quad x = 5,$$ $$y = 4.$$

The slope of AC (with respect to the axis Ox):

As $\mathrm{AC}: y = x - 1$ , then slope is 45 degrees or $\frac{\pi}{4}$

The slope of BD (with respect to the axis Ox)

As $\mathrm{BD}:, y = -x + 9$ , then slope is (-45) degrees or $-\frac{\pi}{4}$

Let find the distance of AC and BD:

$$| A C | = \sqrt {(7 - 3) ^ {2} + (6 - 2) ^ {2}} = \sqrt {1 6 + 1 6} = 4 \sqrt {2}.$$ $$| B D | = \sqrt {(2 - 8) ^ {2} + (7 - 1) ^ {2}} = \sqrt {3 6 + 3 6} = 6 \sqrt {2}.$$

Answer: (5;4)

$$\frac {\pi}{4} \text { and } - \frac {\pi}{4}$$ $$4 \sqrt {2} \text { and } 6 \sqrt {2}$$ $$p a r a l l e l o g r a m$$