+0  
 
0
165
5
avatar+20 

Find the slope of (a) AB, (b) any line parallel to AB, and (c) any line perpendicular to AB.

  1. A(-2, 0) and B(4, 4)

 

Use slopes to show that RST is a right triangle.

2. R(-3, -4), S(2, 2), T(14, -8)

 

Prove that the points of ABCD below form a parallelogram by
showing that opposite sides are parallel and opposite sides are congruent.

3. A(-6, -4), B(4, 2), C(6, 8), D(-4, 2)

 

Determine what special type of quadrilateral HIJK is without graphing.

4. H(0, 0), I(5, 0), J(7, 9), K(1, 9)
 

5. H(0, 1), I(2, -3), J(-2, -1), K(-4, 3)

 Apr 25, 2021
 #1
avatar+121048 
+2

1)  a) Slope  of  AB    =  (4 - 0)  / ( 4 - - 2)  =  4/ ( 4 + 2)  =  4 / 6  =  2 / 3 

 

     b)  Any parallel line to AB  will have  the same slope

 

     c)  A perpendicular line has a  negative reciprocal slope =  -3 /  2

 

 

Use slopes to show that △RST is a right triangle.

2. R(-3, -4), S(2, 2), T(14, -8)

 

Slope of RS  = ( 2 - - 4)  / ( 2 - - 3)  =  6 / 5

 

Slope of  ST    = ( -8-2) /( 14 - 2)  =  -10 / 12  =  - 5 / 6

 

RS  and  ST  have  negative reciprocal slopes......thus......they meet at right  angles

 

 

cool cool cool

 Apr 25, 2021
 #2
avatar+121048 
+2

Prove that the points of ABCD below form a parallelogram by
showing that opposite sides are parallel and opposite sides are congruent.

3. A(-6, -4), B(4, 2), C(6, 8), D(-4, 2)

 

Slope  of  AB  =   (2 - - 4)  / (4 - - 6)  =  6 / 12  =  1/2

Slope of  CD  = ( 2 - 8)  /( -4 - 6)  =  -6 / -12  = 1/2

AB  =   sqrt   [ ( -6 - 4)^2  +  ( -4 - 2)^2  ]   =  sqrt  [ (-10)^2  +(- 6)^2 ]  = sqrt 136

CD  =sqrt  [ ( -4 - 6)^2  + ( 2 - 8)^2 ]    =sqrt [ 10^2  + (-6)^2  ] = sqrt 136

 

Now...using the same procedure....

 

Find  the slope of AD    and  BC

Then....using the distance formula   find   AD  and BC

 

Let me know  if  you get  stuck

 

 

 

cool cool cool

 Apr 25, 2021
 #4
avatar+20 
+2

Thank you👍

jasminerramirez10  Apr 25, 2021
 #5
avatar+121048 
0

No prob.....

 

 

cool cool cool

CPhill  Apr 25, 2021
 #3
avatar+121048 
+2

Determine what special type of quadrilateral HIJK is without graphing.

4. H(0, 0), I(5, 0), J(7, 9), K(1, 9)

This  is  a trapezoid

Reasons  :   

HI  is  parallel to  JK

HK  and IJ  are not parallel


 

5. H(0, 1), I(2, -3), J(-2, -1), K(-4, 3)

Slope of  HI   =  -2

HI  =  sqrt (20)

Slope of  JK =  -2

JK   = sqrt (20)

Parallelogram....opposite sidex are  equal and parallel

 

cool cool cool

 Apr 25, 2021

57 Online Users