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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?

 May 8, 2019
 #1
avatar+19908 
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?

 May 8, 2019
 #2
avatar+63 
+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
avatar+106533 
+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

 

 

 

cool cool cool

 May 8, 2019
 #4
avatar+23861 
+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:

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

 

laugh

 May 9, 2019

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