Let A(1,1), B(-4,3), C(-2,-5) be vertices of a triangle ABC, P be a point on side BC, and …

Question

Let $A(1,1),\ B(-4,3),\ C(-2,-5)$ be vertices of a triangle $ABC$, $P$ be a point on side $BC$, and $\Delta_1$ and $\Delta_2$ be the areas of triangles $APB$ and $ABC$, respectively. If $\Delta_1:\Delta_2=4:7$, then the area enclosed by the lines $AP, AC$ and the $x$-axis is:

Accepted Answer

$\dfrac{\Delta_1}{\Delta_2} = \dfrac{BP}{BC}=\dfrac47$, so $P$ divides $BC$ such that $BP:PC=4:3$.

$$P = \frac{3B+4C}7 = \frac{3(-4,3)+4(-2,-5)}7 = \left(\frac{-12-8}7,\frac{9-20}7\right) = \left(-\frac{20}7,-\frac{11}7\right)$$

Line $AC$: through $(1,1),(-2,-5)$, slope $=2$: $y-1=2(x-1) \implies y=2x-1$. Meets $x$-axis at $x=\dfrac12$: point $\left(\dfrac12,0\right)$.

Line $AP$: through $(1,1)$ and $\left(-\dfrac{20}7,-\dfrac{11}7\right)$, slope $=\dfrac{1+\frac{11}7}{1+\frac{20}7}=\dfrac{18/7}{27/7}=\dfrac23$: $y-1=\dfrac23(x-1)$. Meets $x$-axis at $x=-\dfrac12$: point $\left(-\dfrac12,0\right)$.

Vertices of the enclosed triangle: $(1,1),\left(\dfrac12,0\right),\left(-\dfrac12,0\right)$.

$$\text{Area} = \frac12\times1\times1 = \frac12\text{ sq. unit}$$

Solutions · 1