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Let  \mathrm{A(1,1), B(-4,3), C(-2,-5)} be vertices of a triangle \mathrm{A B C, P} be a point on side \mathrm{ B C,}  and \mathrm{ \Delta_{1}} and  \mathrm{ \Delta_{2}}  be the areas of triangles \mathrm{ A P B} and \mathrm{ A B C,} respectively.
If \mathrm{ \Delta_{1}: \Delta_{2}=4: 7, } then the area enclosed by the lines \mathrm{A P, A C } and the \mathrm{ x-axis } is

Option: 1

\frac{1}{4}


Option: 2

\frac{3}{4}


Option: 3

\frac{1}{2}


Option: 4

1


Answers (1)

best_answer

given 

\Delta _{1}=\frac{1}{2} \begin{vmatrix} \mathrm{x} & \mathrm{y} & 1 \\ 1 & 1 & 1 \\ -4 & 3 & 1 \end{vmatrix} \\

\Delta _{2}=\frac{1}{2} \begin{vmatrix} 1 & 1 & 1 \\ -4 & 3 & 1 \\ -2 & -5 & 1 \end{vmatrix}

\text{Given}\; \frac{\Delta _{1}}{\Delta_{2} }=\frac{4}{7}\Rightarrow \frac{-2\mathrm{x}-5\mathrm{y}+7}{36}=\frac{4}{7}

\begin{aligned} \Rightarrow 14 \mathrm{x}+35\mathrm{y}=-95............................(1)\\ \text{equation of BC is } 4\mathrm{x}+y=-13.........(2) \end{aligned}

Solve eq (1) & (2)

\text{Point}\;\; \mathrm{P}\left ( \frac{-20}{7},\frac{-11}{7} \right )\\

\text{ Here Point}\;\; \mathrm{Q}\left ( \frac{-1}{2},0 \right ) \& \;\; \mathrm{R}\left ( \frac{1}{2},0 \right )\\\text{ So Area of}\;\; \Delta \mathrm{AQR}=\frac{1}{2}\times 1 \times 1 =\frac{1}{2}

Posted by

Sanket Gandhi

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