Step-by-step explanation:
Let's get an important definition out of the way first. A parallelogram is a quadrilateral with two pairs of parallel sides. We can translate this idea to coordinate geometry by saying that each pair of opposite sides must have the same slope.
Slope, as you might remember, is a measure of change. Specifically, it's the change in a line's y value (Δy) divided by the corresponding change in its x value (Δx). This means that if we can prove that:
- The slope of [tex]\overleftrightarrow{KA}[/tex] = the slope of [tex]\overleftrightarrow{IT}[/tex]
- The slope of [tex]\overleftrightarrow{KT}[/tex] = the slope of [tex]\overleftrightarrow{AI}[/tex]
We'll have proved our figure is a quadrilateral. Let's prove those cases, then. For KA and IT:
[tex]slope_{KA}\stackrel{?}{=}slope_{IT}[/tex]
To find the slope of any line segment, we can find the difference between the corresponding coordinates of its endpoints and divide the difference in y-coordinates by the difference in x-coordinates. Since the endpoints of KA are K(0, 0) and A(a, 0), the slope of these points is
[tex]slope_{KA}=\frac{0-0}{a-0} =\frac{0}{a} =0[/tex]
We can do this same for IT using its endpoints I(a + b, c) and T(b, c):
[tex]slope_{IT}=\frac{c-c}{b-(a+b)} =\frac{0}{-a}=0[/tex]
Since 0 = 0, we can say that [tex]slope_{KA}=slope_{IT}[/tex], and we've proven our first condition. Now we just need to prove that [tex]slope_{KT}=slope_{AI}[/tex]:
[tex]slope_{KT}\stackrel{?}{=}slope_{AI}[/tex]
Using K(0, 0) and T(b, c), [tex]slope_{KT}=\frac{c-0}{b-0}=\frac{c}{b}[/tex], and using A(a, 0) and I(a + b, c), [tex]slope_{AI}=\frac{c-0}{(a+b)-a}=\frac{c}{b}[/tex], proving that [tex]slope_{KT}=slope_{AI}[/tex].
Since we've proven that [tex]slope_{KA}=slope_{IT}[/tex] and [tex]slope_{KT}=slope_{AI}[/tex], we've proven that the quadrilateral KAIT is, by definition, a parallelogram.
