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  4. Two sides of a parallelogram are along the lines, x + y = 3 and x - y + 3 = 0. If its diagonals intersect at (2, 4) and one of its vertex is (a, b), then a + b is

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Q. Two sides of a parallelogram are along the lines, $x + y = 3$ and $x - y + 3 = 0$. If its diagonals intersect at $(2, 4)$ and one of its vertex is $(a, b)$, then $a + b$ is

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Solution:

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Since, $x - y + 3 = 0$ and $x + y = 3$ are perpendicular lines
and intersection point of $x - y + 3 = 0$ and $x + y = 3$ is $P (0, 3)$.
$\Rightarrow M$ is mid-point of $ PR \Rightarrow R(4, 5)$
Let $S(x_1, x_2 + 3)$ and $Q(x_2, 3 - x_2)$
$M$ is mid-point of $SQ$
$\Rightarrow x_1 + x_2 = 4, x_1 + 3 + 3 - x_2 = 8$
$\Rightarrow x_1 = 3, x_2 = 1$
Then, the vertex $D$ is $(3, 6)$.
$\Rightarrow (a, b) = (3, 6)$
$\Rightarrow a = 3, b = 6$
$\therefore a + b = 3 + 6 = 9$