Question

PLEASE HELP ME. QUAD has coordinates Q(-4, 9), U(2, 3), A(-3, -2), and D(-9, 4). Prove that quadrilateral QUAD is a

rectangle.

Answer 1

To prove that quadrilateral QUAD is a rectangle, we need to show that it has four right angles. We can do this by showing that the slopes between the consecutive pairs of points are negative reciprocals of each other.

Step-by-step explanation:

To prove that quadrilateral QUAD is a rectangle, we need to show that it has four right angles. We can do this by showing that the slopes between the consecutive pairs of points are negative reciprocals of each other.

Answer 2

Check the picture below.

`~\hfill \stackrel{\textit{\large distance between 2 points}}{d = √(( x_2- x_1)^2 + ( y_2- y_1)^2)}~\hfill~ \\\\[-0.35em] ~\dotfill\\\\ Q(\stackrel{x_1}{-4}~,~\stackrel{y_1}{9})\qquad U(\stackrel{x_2}{2}~,~\stackrel{y_2}{3}) ~\hfill QU=√([ 2- (-4)]^2 + [ 3- 9]^2) \\\\\\ ~\hfill \boxed{QU=√(72)} \\\\\\ U(\stackrel{x_1}{2}~,~\stackrel{y_1}{3})\qquad A(\stackrel{x_2}{-3}~,~\stackrel{y_2}{-2}) ~\hfill UA=√([ -3- 2]^2 + [ -2- 3]^2) \\\\\\ ~\hfill \boxed{UA=√(50)}`

`A(\stackrel{x_1}{-3}~,~\stackrel{y_1}{-2})\qquad D(\stackrel{x_2}{-9}~,~\stackrel{y_2}{4}) ~\hfill AD=√([ -9- (-3)]^2 + [ 4- (-2)]^2) \\\\\\ ~\hfill \boxed{AD=√(72)} \\\\\\ D(\stackrel{x_1}{-9}~,~\stackrel{y_1}{4})\qquad Q(\stackrel{x_2}{-4}~,~\stackrel{y_2}{9}) ~\hfill DQ=√([ -4- (-9)]^2 + [ 9- 4]^2) \\\\\\ ~\hfill \boxed{DQ=√(50)}`

now, let's take a peek at that above, DQ = UA and QU = AD, so opposite sides of the polygon are equal.

Now, let's check the slopes of DQ and QU

`D(\stackrel{x_1}{-9}~,~\stackrel{y_1}{4})\qquad Q(\stackrel{x_2}{-4}~,~\stackrel{y_2}{9}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{9}-\stackrel{y1}{4}}}{\underset{run} {\underset{x_2}{-4}-\underset{x_1}{(-9)}}} \implies \cfrac{9 -4}{-4 +9}\implies \cfrac{5}{5}\implies \cfrac{1}{1}\implies 1 \\\\[-0.35em] ~\dotfill`

`Q(\stackrel{x_1}{-4}~,~\stackrel{y_1}{9})\qquad U(\stackrel{x_2}{2}~,~\stackrel{y_2}{3}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{3}-\stackrel{y1}{9}}}{\underset{run} {\underset{x_2}{2}-\underset{x_1}{(-4)}}} \implies \cfrac{3 -9}{2 +4}\implies \cfrac{-6}{6}\implies \cfrac{-1}{1}\implies -1`

keeping in mind that perpendicular lines have negative reciprocal slopes, let's notice that the reciprocal of 1/1 is just 1/1 and the negative of that is just -1/1 or -1, so QU has a slope that is really just the negative reciprocal of DQ, those two lines are perpendicular, thus making a 90° angle, and their congruent opposite sides will also make a 90°, that makes QUAD hmmm yeap, you guessed it.