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

Given:

The points $A(3,1), B(5,1), C(a,b)$ and $D(4,3)$ are the vertices of a parallelogram $ABCD$.

To do:

We have to find the values of $a$ and $b$.

Solution:

Let the diagonals $AC$ and $BD$ bisect each other at $O$.

Using the mid-point formula, we get,

\( \mathrm{O} \) is the mid-point of \( \mathrm{AC} \)

The coordinates of \( \mathrm{O} \) are \( \left(\frac{3+a}{2}, \frac{1+b}{2}\right) \)

Similarly,

\( \mathrm{O} \) is the mid-point of \( \mathrm{BD} \).

The coordinates of \( \mathrm{O} \) are \( \left(\frac{5+4}{2}, \frac{1+3}{2}\right) \)

\( =\left(\frac{9}{2}, \frac{4}{2}\right) \) 

\( =\left(\frac{9}{2}, 2\right) \)

On comparing, we get,

\( \frac{9}{2}=\frac{3+a}{2} \)

\( \Rightarrow 3+a=9 \)

\( \Rightarrow a=9-3=6 \)

\( \frac{1+b}{2}=2 \)

\( \Rightarrow b+1=4 \)

\( \Rightarrow b=4-1=3 \)

The values of $a$ and $b$ are $6$ and $3$ respectively.