Question

(i) If the points A (a, −11), B (5, b), C(2, 15) and D (1, 1) are the vertices of a parallelogram ABCD, find the values of a and b.

(ii) Point A(3, 1), B(5, 1), C(a, b) and D(4, 3) are vertices of a parallelogram ABCD. Find the values of a and b.

Answer 1

(i)

Let ABCD be a parallelogram in which the co-ordinates of the vertices are A (a,−11); B (5, b); C (2, 15) and D (1, 1).

Since ABCD is a parallelogram, the diagonals bisect each other. Therefore the mid-point of the diagonals of the parallelogram will coincide.

In general to find the mid-point of two pointsand we use section formula as,

The mid-point of the diagonals of the parallelogram will coincide.

So,

Therefore,

Now equate the individual terms to get the unknown value. So,

Similarly,

Therefore,

(ii) Given: Point A(3, 1), B(5, 1), C(a, b) and D(4, 3) are vertices of a parallelogram ABCD.

Diagonals of a parallelogram bisect each other.

∴ Mid point of AC = Mid point of BD

$$\begin{gathered} \mathrm{Mid}\mathrm{point}\mathrm{of}\left(x_1,y_1\right)\mathrm{and}\left(x_2,y_2\right)\mathrm{is}\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right). \\ \mathrm{Mid}\mathrm{point}\mathrm{of}AC=\left(\frac{3+a}{2},\frac{1+b}{2}\right) \\ \mathrm{Mid}\mathrm{point}\mathrm{of}BD=\left(\frac{5+4}{2},\frac{1+3}{2}\right) \\ =\left(\frac{9}{2},\frac{4}{2}\right)=\left(\frac{9}{2},2\right) \\ \therefore \left(\frac{3+a}{2},\frac{1+b}{2}\right)=\left(\frac{9}{2},2\right) \\ \Rightarrow \frac{3+a}{2}=\frac{9}{2}\mathrm{and}\frac{1+b}{2}=2 \\ \Rightarrow 3+a=9\mathrm{and}1+b=4 \\ \Rightarrow a=6\mathrm{and}b=3 \end{gathered}$$

Hence, the values of a and b is 6 and 3, respectively.