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