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
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
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