Question

If A(6, 1), B(8, 2), C(9, 4) and D(7, 3) are the vertices of ${\displaystyle \square }$ABCD, show that ${\displaystyle \square }$ABCD is a parallelogram.

Solution:

Slope of line = ${\displaystyle \frac{{\text{y}}_2-{\text{y}}_1}{{\text{x}}_2-{\text{x}}_1}}$

∴ Slope of line AB = ${\displaystyle \frac{2-1}{8-6}=\square }$ .......(i)

∴ Slope of line BC = ${\displaystyle \frac{4-2}{9-8}=\square }$ .....(ii)

∴ Slope of line CD = ${\displaystyle \frac{3-4}{7-9}=\square }$ .....(iii)

∴ Slope of line DA = ${\displaystyle \frac{3-1}{7-6}=\square }$ .....(iv)

∴ Slope of line AB = ${\displaystyle \square }$ ......[From (i) and (iii)]

∴ line AB || line CD

∴ Slope of line BC = ${\displaystyle \square }$ ......[From (ii) and (iv)]

∴ line BC || line DA

Both the pairs of opposite sides of the quadrilateral are parallel.

∴ ${\displaystyle \square }$ABCD is a parallelogram.

Answer 1

Slope of line = ${\displaystyle \frac{{\text{y}}_2-{\text{y}}_1}{{\text{x}}_2-{\text{x}}_1}}$

∴ Slope of line AB = ${\displaystyle \frac{2-1}{8-6}=\frac{1}{2}}$ .......(i)

∴ Slope of line BC = ${\displaystyle \frac{4-2}{9-8}=2}$ .....(ii)

∴ Slope of line CD = ${\displaystyle \frac{3-4}{7-9}=\frac{1}{2}}$ .....(iii)

∴ Slope of line DA = ${\displaystyle \frac{3-1}{7-6}=2}$ .....(iv)

∴ Slope of line AB = Slope of line CD ......[From (i) and (iii)]

∴ line AB || line CD

∴ Slope of line BC = Slope of line DA ......[From (ii) and (iv)]

∴ line BC || line DA

Both the pairs of opposite sides of the quadrilateral are parallel.

∴ ${\displaystyle \square }$ABCD is a parallelogram.