This question here is giving me a hard time can any help me?

1) which angle in the quadrilateral with vertices A(2,-4),, B(-5,-2),C(-4,2) and D(4,0) is a right angle?

Masterx4020 May 8, 2019

#1**0 **

Slope of the line between (-5,-2) and (-4,2) is rise/run = 4/1 = 4

slope from (-4.2) to (4,0) is rise/run = -2/8 = -1/4

does that give you a clue?

ElectricPavlov May 8, 2019

#2**+1 **

sort of I can see that they are perpendicular, but I am confused. Geometry is just realy hard for me to comprehend.

Masterx4020
May 8, 2019

#3**+1 **

A(2,-4),, B(-5,-2),C(-4,2) and D(4,0)

Look at the graph

It seems clear that Angles D and A are not right angles

The slope between C and D = [ 0 - 2] / [ 4 - - 4 ] = -2 / 8 = -1/4

The slope between A and B = [ -2 - - 4 ] / [ -5 - 2] = 2/-7 = - 2/7

The solpe between C and B = [ 2 - - 2 ] / [ -4 - - 5 ] = 4 / 1 = 4

Since CD and CB have negative reciprocal slopes....then angle C is the right angle

CPhill May 8, 2019

#4**+1 **

**This question here is giving me a hard time can any help me?**

**1)**

**which angle in the quadrilateral with vertices A(2,-4),, B(-5,-2),C(-4,2) and D(4,0) is a right angle?**

\(\begin{array}{|rcll|} \hline \overline{CD}^2 = 8^2+2^2 = 68 \\ \overline{CB}^2 = 4^2+1^2 = 17 \\ \overline{BD}^2 = 9^2+2^2 = 85 \\ \hline \end{array}\)

**Pythagoras:**

**\(\begin{array}{|rcll|} \hline \overline{CD}^2+\overline{CB}^2 &\overset{?}{=}& \overline{BD}^2 \quad | \quad \text{If true, then C is a right angle!} \\ 68+17 &\overset{?}{=}& 85 \\ 85 &=& 85 \quad | \quad \text{So C is a right angle!} \\ \hline \end{array}\)**

heureka May 9, 2019