question_answer

Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are\[(0,-\,\,1),\](2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle.

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\[=\frac{\text{Area}\,\,\text{of}\,\,\Delta \text{NPM}}{\text{Area}\,\,\text{of}\,\,\Delta \text{ABC}}=\frac{1}{4}\]