Let A 1 , 1 , B - 4 , 3 , C - 2 , - 5 be vertices of a triangle A B C , P be a point on side B C , and Δ 1 and Δ 2 be the areas of triangle A P B and A B C . Respectively. If Δ 1 : Δ 2 = 4 : 7 , then

Question

Let A 1 , 1 , B - 4 , 3 , C - 2 , - 5 be vertices of a triangle A B C , P be a point on side B C , and Δ 1 and Δ 2 be the areas of triangle A P B and A B C . Respectively. If Δ 1 : Δ 2 = 4 : 7 , then the area enclosed by the lines A P , A C and the x -axis is

TestHub · Accepted Answer

Plotting the given value in diagram we have,
Now let $P (x, y)$ be a point on $B C$ , so area formed by $△ A B P$ will be,
$Δ_{1} = \frac{1}{2} |\begin{matrix} x & y & 1 \\ 1 & 1 & 1 \\ - 4 & 3 & 1 \end{matrix}| = \frac{1}{2} (- 2 x - 5 y + 7)$
And area of triangle formed by $△ A B C$ will be, $Δ_{2} = \frac{1}{2} |\begin{matrix} 1 & 1 & 1 \\ - 4 & 3 & 1 \\ - 2 & - 5 & 1 \end{matrix}| = \frac{36}{2}$
Given $\frac{Δ_{1}}{Δ_{2}} = \frac{4}{7} \Rightarrow \frac{- 2 x - 5 y + 7}{36} = \frac{4}{7}$
$\Rightarrow 14 x + 35 y = - 95 \dots (1)$
Equation of $B C$ by two point form will be $4 x + y = - 13 \dots (2)$
Solving equation $(1)$ and $(2)$ we get,
Point $P (\frac{- 20}{7}, \frac{- 11}{7})$
Now equation of $A P$ will be $y - 1 = \frac{1 + \frac{11}{7}}{1 + \frac{20}{7}} (x - 1)$
$\Rightarrow 3 y - 2 x = 1$ ,
So $x -$ intercept will be point $Q (\frac{- 1}{2}, 0)$ and similarly from $A C$ we get $R (\frac{1}{2}, 0)$
So Area of triangle $A Q R = \frac{1}{2} \times 1 \times 1 = \frac{1}{2}$

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