Answer:
Step-by-step explanation:
Given: In ΔPQR, the coordinates of the vertices are P(0, 0), Q(2a, 0), and R(2b, 2c).
To prove: The line containing the midpoints of two sides of a triangle is parallel to the third side.
Proof: In ΔPQR, the coordinates of the vertices are P(0, 0), Q(2a, 0), and R(2b, 2c).
Let, A, B and C be the mid-points of PQ, PR and QR respectively. Thus, the coordinates of S are:
[tex]A=(\frac{0+2a}{2},\frac{0+0}{2})=(a,0)[/tex]
The coordinates of B are:
[tex]B=(\frac{0+2b}{2},\frac{0+2c}{2})=(b,c)[/tex]
And the coordinates of C are:
[tex]C=(\frac{2a+2b}{2},\frac{0+2c}{2})=(a+b,c)[/tex]
Now, slope of AB is given as:
[tex]s={\frac{c-0}{b-a}}={\frac{c}{b-a}}[/tex]
And slope of QR is given as:
[tex]s={\frac{2c-0}{2b-2a}}={\frac{c}{b-a}}[/tex]
Since the slopes of AB and QR are equal, hence they must be parallel.
Hence proved.
Also, Since A is the midpoint of PQ, therefore teh coordinates are:
A=[tex](\frac{0+2a}{2},\frac{0+0}{2})=(a,0)[/tex]
