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(a) Prove that $P(3, -3), Q(5, -2), R(6, 0)$ and $S(4, -1)$ are the vertices of a rhombus $PQRS$. Also, find if it is a square or not.
OR
(b) Find the ratio in which the point $(-5, p)$ divides the line segment joining $(-7, -2)$ and $(-4, 5)$. Hence, find the value of $p$.
OR
(b) Find the ratio in which the point $(-5, p)$ divides the line segment joining $(-7, -2)$ and $(-4, 5)$. Hence, find the value of $p$.
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(a) $RS = \sqrt{(6 - 4)^2 + (0 - (-1))^2} = \sqrt{5}$
$SP = \sqrt{(4 - 3)^2 + (-1 + 3)^2} = \sqrt{5}$
$PQ = \sqrt{(5 - 3)^2 + (-2 + 3)^2} = \sqrt{5}$
$QR = \sqrt{(6 - 5)^2 + (0 + 2)^2} = \sqrt{5}$
$\implies PQRS$ is a rhombus
$PR = \sqrt{(6 - 3)^2 + (0 + 3)^2} = \sqrt{18} = 3\sqrt{2}$
$QS = \sqrt{(5 - 4)^2 + (-2 + 1)^2} = \sqrt{2}$
$\implies PR \neq QS$ So, $PQRS$ is not a square
OR
(b)
$-5 = \frac{-4k-7}{k+1} \implies k = 2$
$\therefore$ Ratio is $2 : 1$
$p = \frac{2(5) + 1(-2)}{3} = \frac{8}{3}$
$SP = \sqrt{(4 - 3)^2 + (-1 + 3)^2} = \sqrt{5}$
$PQ = \sqrt{(5 - 3)^2 + (-2 + 3)^2} = \sqrt{5}$
$QR = \sqrt{(6 - 5)^2 + (0 + 2)^2} = \sqrt{5}$
$\implies PQRS$ is a rhombus
$PR = \sqrt{(6 - 3)^2 + (0 + 3)^2} = \sqrt{18} = 3\sqrt{2}$
$QS = \sqrt{(5 - 4)^2 + (-2 + 1)^2} = \sqrt{2}$
$\implies PR \neq QS$ So, $PQRS$ is not a square
OR
(b)
$-5 = \frac{-4k-7}{k+1} \implies k = 2$
$\therefore$ Ratio is $2 : 1$
$p = \frac{2(5) + 1(-2)}{3} = \frac{8}{3}$