2010-I-12 – Solving Master

Ans: (a)

3 x - 4 y + 78 = 0

(b)

(24, 0)

(c)

150

(d)

\frac{3}{2}

The slope of $A B$
$\begin{array}{cl} = & \frac{24 - 18}{6 - (- 2)} \\ = & \frac{3}{4} \end{array}$

Therefore, the equation of the straight line passing through $A$ and $B$ is

$\begin{array}{rcl} y - 24 & = & \frac{3}{4} (x - 6) \\ 4 y - 96 & = & 3 x - 18 \\ 3 x - 4 y + 78 & = & 0 \end{array}$
Let $(x, 0)$ be the coordinates of $C$ . Since $A B ⊥ A C$ , then
$\begin{array}{rcl} m_{A B} \times m_{A C} & = & - 1 \\ \frac{3}{4} \times \frac{24 - 0}{6 - x} & = & - 1 \\ 72 & = & - 4 (6 - x) \\ 4 x & = & 96 \\ x & = & 24 \end{array}$

Therefore, the coordinates of $C$ are $(24, 0)$ .
The area of $Δ A B C$
$\begin{array}{cl} = & \frac{1}{2} | \begin{array}{cc} 6 & 24 \\ - 2 & 18 \\ 24 & 0 \\ 6 & 24 \end{array} | \\ = & \frac{1}{2} (6 \times 18 + (- 2) \times 0 + 24 \times 24 - 24 \times (- 2) - 18 \times 24 - 0 \times 6) \\ = & 150 square units \end{array}$
From the graph above, we can see that the heights of $Δ A B D$ and $Δ A C D$ are the same. Hence, we have

$\begin{array}{cl} B D : D C \\ = & Area of Δ A B D : Area of Δ A D C \\ = & 90 : 150 - 90 \\ = & 3 : 2 \\ = & \frac{3}{2} : 1 \end{array}$

Therefore, the value of $r$ is $\frac{3}{2}$ .