Question

Three vertices of a parallelogram are $(a+b,a-b),(2a+b,2a-b),(a-b,a+b)$. Find the fourth vertex.

• A. $(-b,b)$
• B. $(b,b)$
• C. $(-b,-b)$
• D. None of these

Answer 1

The correct option is A $(-b,b)$

Let $ABCD$ be a parallelogram.

Let $A(a+b,a-b),B(2a+b,2a-b)$ and $C(a-b,a+b)$. We have to find co-ordinates of the fourth vertex.

Let fourth vertex be $D(x,y)$

Since, $ABCD$ is a parallelogram, the diagonals bisect each other.

$\therefore$ The mid-point of the diagonals of the parallelogram will coincide.

Mid - point formula,

$P(x,y)=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

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

Si, co-ordinate of mid-point of $AC=$ Co-ordinate of mid-point of $BD$

$\therefore$ $(\frac{a+b+a-b}{2},\frac{a-b+a+b}{2})=(\frac{2a+b+x}{2},\frac{2a-b+y}{2})$

$\Rightarrow$ $(a,a)=(\frac{2a+b+x}{2},\frac{2a-b+y}{2})$

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

$\Rightarrow$ $x=-b$

$\Rightarrow$ $y=b$

$\therefore$ The fourth vertex is $D(-b,b)$