Use slopes to show that A (1,1) , B (11, 3) C (10, 8) and D (0,6) are vertices of a rectangle.
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It's not enough to show that quadrilateral has 2 sets of parallel sides, since all we can tell from this is that quadrilateral is a parallelogram. To be a rectangles, adjacent sides must be perpendicular, i.e. they must have slopes that are negative reciprocals of each other.
So we need slope AB = slope CD = m
and slope BC = slope DA = −1/m
Slope AB = (3−1)/(11−1) = 2/10 = 1/5
Slope CD = (8−6)/(10−0) = 2/10 = 1/5
Slope BC = (3−8)/(11−10) = −5/1 = −5
Slope DA = (1−6)/(1−0) = −5/1 = −5
Opposite sides AB and CD are parallel
Opposite sides BC and DA are parallel
Pairs of adjacent sides are perpendicular (since their slopes are 1/5 and −5)
Therefore, ABCD is a rectangle.
So we need slope AB = slope CD = m
and slope BC = slope DA = −1/m
Slope AB = (3−1)/(11−1) = 2/10 = 1/5
Slope CD = (8−6)/(10−0) = 2/10 = 1/5
Slope BC = (3−8)/(11−10) = −5/1 = −5
Slope DA = (1−6)/(1−0) = −5/1 = −5
Opposite sides AB and CD are parallel
Opposite sides BC and DA are parallel
Pairs of adjacent sides are perpendicular (since their slopes are 1/5 and −5)
Therefore, ABCD is a rectangle.
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You first need to plot these 4 points on the X-Y coordinate to visualize the problem.
You need to prove the following conditions in order to conclude that it is a rectangle.
slope of AB = slope of CD (i.e. AB is parallel to CD)
slope of BC = slope of AD (i.e. BC is parallel to AD)
Finally show slope of AB = -1/slope of AD (i.e. AB is perpendicular to AD)
slope of the line = m = (y2-y1)/(x2-x1)
Work it out. If the above conditions are met, it is a rectangle.
You need to prove the following conditions in order to conclude that it is a rectangle.
slope of AB = slope of CD (i.e. AB is parallel to CD)
slope of BC = slope of AD (i.e. BC is parallel to AD)
Finally show slope of AB = -1/slope of AD (i.e. AB is perpendicular to AD)
slope of the line = m = (y2-y1)/(x2-x1)
Work it out. If the above conditions are met, it is a rectangle.
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A (1,1) , B (11, 3) C (10, 8) and D (0,6) are vertices of a quadrilateral ABCD whence
slope of AB is (3 -- 1)/(11 -- 1) = 1/5
slope of DC is (8 -- 6)/(10 -- 0) = 1/5 hence AB II DC
slope of AD is (6 -- 1)/(0 -- 1) = -- 5
slope of BC is (8 -- 3)/(10 -- 11) = -- 5 hence AD II BC
as 1/5*(--5) = --1, AB is perpendicular to AD and DC is perpendicular to BC
hence ABCD is a rectangle.
slope of AB is (3 -- 1)/(11 -- 1) = 1/5
slope of DC is (8 -- 6)/(10 -- 0) = 1/5 hence AB II DC
slope of AD is (6 -- 1)/(0 -- 1) = -- 5
slope of BC is (8 -- 3)/(10 -- 11) = -- 5 hence AD II BC
as 1/5*(--5) = --1, AB is perpendicular to AD and DC is perpendicular to BC
hence ABCD is a rectangle.
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Well let's see. an easy way to determine would be that A and B have slope 1/5....and C and D have slope 1/5 from the two points, this automatically means they're parallel. Rectangles have two pairs of parallel sides.
Also A to D is -5 and B to C is -5 too. This rectangle must be slanted, because there is a slope for it's length.
@Iceman: if you're going to give people who had the right answer before you thumbs down they I will also give you a thumbs down :)
Also A to D is -5 and B to C is -5 too. This rectangle must be slanted, because there is a slope for it's length.
@Iceman: if you're going to give people who had the right answer before you thumbs down they I will also give you a thumbs down :)
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slopes:
AB = (3 - 1)/(11 - 1) = 2/10 = 1/5
BC = (8 - 3)/(10 - 11) = -5
AB & BC are perpendicular since the slopes are negative reciprocal
AD = (6 - 1)/(0 - 1) = -5
AD and BC are parallel since they have the same slope
ETC
AB = (3 - 1)/(11 - 1) = 2/10 = 1/5
BC = (8 - 3)/(10 - 11) = -5
AB & BC are perpendicular since the slopes are negative reciprocal
AD = (6 - 1)/(0 - 1) = -5
AD and BC are parallel since they have the same slope
ETC
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AB^2 = 104, BC^2 = 26, CD^2 = 104, DA^2 = 26
AC^2 = 130, BD^2 = 130
so AC^2 = AB^2 + BC^2, and BD^2 = BC^2 + CD^2
END Provement of rectangle.
AC^2 = 130, BD^2 = 130
so AC^2 = AB^2 + BC^2, and BD^2 = BC^2 + CD^2
END Provement of rectangle.