First slide

Cartesian plane

Question

If $Δ_{1}$ is the area of the triangle formed by the centroid and two vertices of a triangle, $Δ_{2}$ is the area of the triangle formed by the mid-points of the sides of the same triangle, then $Δ_{1} : Δ_{2} =$

Moderate

A

$3 : 4$
B

$4 : 1$
C

$4 : 3$
D

$2 : 1$

Solution

Let $A (x_{1}, y_{1}), B (x_{2}, y_{2})$ and $C (x_{3}, y_{3})$ be the vertices of a $∆ A B C$ , and let $G$ be its centroid. Then,
$Δ_{1} = Area of Δ G B C$
$\begin{aligned} \Rightarrow Δ_{1} = \frac{1}{2} |\begin{array}{ccc} \frac{x_{1} + x_{2} + x_{3}}{3} & \frac{y_{1} + y_{2} + y_{3}}{3} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array}| \\ \Rightarrow Δ_{1} = \frac{1}{6} |\begin{array}{ccc} x_{1} + x_{2} + x_{3} & y_{1} + y_{2} + y_{3} & 3 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array}| Applying R_{1} \to R_{1} (3) \end{aligned}$
$\Rightarrow Δ_{1} = \frac{1}{6} |\begin{array}{lll} x_{1} y_{1} 1 \\ x_{2} y_{2} 1 \\ x_{3} y_{3} 1 \end{array}|$
$Applying R_{1} \to R_{1} - R_{2} - R_{3}$
$\Rightarrow Δ_{1} = \frac{Δ}{3}$ where $∆$ is the area of $∆ A B C$
$Δ_{2} =$ Area of triangle formed by the mid-points of the sides
$\begin{aligned} \Rightarrow Δ_{2} = |\begin{array}{lll} \frac{x_{1} + x_{2}}{2} & \frac{y_{1} + y_{2}}{2} & 1 \\ \frac{x_{2} + x_{3}}{2} & \frac{y_{2} + y_{3}}{2} & 1 \\ \frac{x_{3} + x_{1}}{2} & \frac{y_{3} + y_{1}}{2} & 1 \end{array}| 1 \\ \Rightarrow Δ_{2} = \frac{1}{8} |\begin{array}{lll} x_{1} + x_{2} & y_{1} + y_{2} & 1 \\ x_{2} + x_{3} & y_{2} + y_{3} & 1 \\ x_{3} + x_{1} & y_{3} + y_{1} & 1 \end{array}| \end{aligned}$ $\begin{aligned} Applying C_{1} \to C_{1} (2), \\ C_{2} \to C_{2} (2) \end{aligned}$
$\Rightarrow Δ_{2} = \frac{1}{8} |\begin{array}{ccc} 2 (x_{1} + x_{2} + x_{3}) & 2 (y_{1} + y_{2} + y_{3}) & 3 \\ x_{2} + x_{3} & y_{2} + y_{3} & 1 \\ x_{3} + x_{1} & y_{3} + y_{1} & 1 \end{array}|$
$Applying R_{1} \to R_{1} + R_{2} + R_{3}$
$\Rightarrow Δ_{2} = \frac{2}{8} |\begin{array}{ccc} x_{1} + x_{2} + x_{3} & y_{1} + y_{2} + y_{3} & \frac{3}{2} \\ x_{2} + x_{3} & y_{2} + y_{3} & 1 \\ x_{3} + x_{1} & y_{3} + y_{1} & 1 \end{array}|$ Applying $R_{1} \to R_{1} (1 / 2)$
$\Rightarrow Δ_{2} = \frac{1}{4} |\begin{array}{ccc} x_{1} + x_{2} + x_{3} & y_{1} + y_{2} + y_{3} & 3 / 2 \\ - x_{1} & - 1 / 1 & - 1 / 2 \\ - x_{2} & - 1 / 2 & - 1 / 2 \end{array}|$ $\begin{aligned} R_{2} \to R_{2} - R_{1} \\ R_{3} \to R_{3} - R_{1} \end{aligned}$
$\Rightarrow Δ_{2} = \frac{1}{8} |\begin{array}{ccr} x_{1} + x_{2} + x_{3} & y_{1} + y_{2} + y_{3} & 3 \\ - x_{1} & - y_{1} & - 1 \\ - x_{2} & - y_{2} & - 1 \end{array}|$ Applying $C_{3} \to C_{3}$
$\Rightarrow Δ_{2} = \frac{1}{8} |\begin{array}{rrr} x_{3} y_{3} 1 \\ - x_{1} - y_{1} - 1 \\ - x_{2} - y_{2} - 1 \end{array}|$ $Applying R_{1} \to R_{1} + R_{2} + R_{3}$
$\begin{aligned} \Rightarrow Δ_{2} = \frac{1}{4} Δ \\ ∴ Δ_{1} : Δ_{2} = 4 : 3 \end{aligned}$

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