A) 5 : 6
B) 3 : 4
C) 1 : 4
D) 4 : 1
Correct Answer: C
Solution :
Mid-point of M\[=\left( \frac{0+0}{2},\frac{3-1}{2} \right)\] |
\[\left[ \because \text{Mid-point=}\left( \frac{{{x}_{1}}+{{x}_{2}}}{2},\frac{{{y}_{1}}+{{y}_{2}}}{2} \right) \right]\] |
= (0, 1) |
Mid-point of N \[=\left( \frac{0+2}{2},\frac{-\,\,1+1}{2} \right)=(1,0)\] |
And Mid-point of \[P=\left( \frac{0+2}{2},\frac{3+1}{2} \right)=(1,2)\] |
Let N \[(1,0)=({{x}_{1}},{{y}_{1}})\] and \[P\,\,(1,2)=({{x}_{2}},{{y}_{2}})\] and \[M\,\,(0,1)=({{x}_{3}},{{y}_{3}})\] |
\[\therefore \]Area of \[\Delta NPM\] |
\[=\frac{1}{2}[{{x}_{1}}({{y}_{2}}-{{y}_{3}})+{{x}_{2}}({{y}_{3}}-{{y}_{1}})+{{x}_{3}}({{y}_{1}}-{{y}_{2}})]\] |
\[=\frac{1}{2}[1\,\,(2-1)+1\,\,(1-0)+0(0-2)]\] |
\[=\frac{1}{2}[1\,\,(1)+1+0]=\frac{2}{2}=1=1\] |
Let \[A=({{x}_{1}},{{y}_{1}})=(0,-1),\]\[B=({{x}_{2}},{{y}_{2}})=(2,1)\]and\[C=({{x}_{3}},{{y}_{3}})=(0,3)\] |
\[\therefore \]Area of \[\Delta ABC\] |
\[=\frac{1}{2}[{{x}_{1}}({{y}_{2}}-{{y}_{3}})+{{x}_{2}}({{y}_{3}}-{{y}_{1}})+{{x}_{3}}({{y}_{1}}-{{y}_{2}})]\] \[=\frac{1}{2}[0\,\,(1-3)+2(3+1)+0(-1-1)]\] |
\[=\frac{1}{2}(0+8+0)=4\]sq units |
\[\therefore \]Required ratio\[=\frac{\text{Area}\,\,\text{of}\,\,\Delta \text{NPM}}{\text{Area}\,\,\text{of}\,\,\Delta \text{ABC}}=\frac{1}{4}\] |
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