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Points $\text{P}(6, 0)$, $\text{Q}(2, 8)$ and $\text{R}(-2, 4)$ are vertices of $\triangle \text{PQR}$. It is given that $\text{MN \| QR}$ such that $\frac{\text{PM}}{\text{MQ}} = \frac{1}{3}$. Using distance formula and ratio formula, show that $\frac{\text{MN}}{\text{QR}} = \frac{1}{4}$.
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Solution:
$\text{MN \| QR} \Rightarrow \frac{\text{PM}}{\text{MQ}} = \frac{\text{PN}}{\text{NR}} = \frac{1}{3}$ (
frac{1}{2} Mark)
Coordinates of $\text{M}$ are $(\frac{3 \times 6 + 1 \times 2}{3+1}, \frac{3 \times 0 + 1 \times 8}{3+1}) = (\frac{18+2}{4}, \frac{0+8}{4}) = (5, 2)$ (
frac{1}{2} Mark)
Coordinates of $\text{N}$ are $(\frac{3 \times 6 + 1 \times (-2)}{3+1}, \frac{3 \times 0 + 1 \times 4}{3+1}) = (\frac{18-2}{4}, \frac{0+4}{4}) = (4, 1)$ (
frac{1}{2} Mark)
$\text{MN} = \sqrt{(5-4)^2 + (2-1)^2} = \sqrt{1^2 + 1^2} = \sqrt{2}$ (
frac{1}{2} Mark)
$\text{QR} = \sqrt{(2-(-2))^2 + (8-4)^2} = \sqrt{4^2 + 4^2} = \sqrt{16+16} = \sqrt{32} = 4\sqrt{2}$ (
frac{1}{2} Mark)
Therefore $\frac{\text{MN}}{\text{QR}} = \frac{\sqrt{2}}{4\sqrt{2}} = \frac{1}{4}$ (
frac{1}{2} Mark)
$\text{MN \| QR} \Rightarrow \frac{\text{PM}}{\text{MQ}} = \frac{\text{PN}}{\text{NR}} = \frac{1}{3}$ (
frac{1}{2} Mark)
Coordinates of $\text{M}$ are $(\frac{3 \times 6 + 1 \times 2}{3+1}, \frac{3 \times 0 + 1 \times 8}{3+1}) = (\frac{18+2}{4}, \frac{0+8}{4}) = (5, 2)$ (
frac{1}{2} Mark)
Coordinates of $\text{N}$ are $(\frac{3 \times 6 + 1 \times (-2)}{3+1}, \frac{3 \times 0 + 1 \times 4}{3+1}) = (\frac{18-2}{4}, \frac{0+4}{4}) = (4, 1)$ (
frac{1}{2} Mark)
$\text{MN} = \sqrt{(5-4)^2 + (2-1)^2} = \sqrt{1^2 + 1^2} = \sqrt{2}$ (
frac{1}{2} Mark)
$\text{QR} = \sqrt{(2-(-2))^2 + (8-4)^2} = \sqrt{4^2 + 4^2} = \sqrt{16+16} = \sqrt{32} = 4\sqrt{2}$ (
frac{1}{2} Mark)
Therefore $\frac{\text{MN}}{\text{QR}} = \frac{\sqrt{2}}{4\sqrt{2}} = \frac{1}{4}$ (
frac{1}{2} Mark)