Question

A triangle has a vertex at \((1,2)\) and the midpoints of two sides through it are \((-1,1)\) and \((2,3)\). Find its area.

• A. \(1\)
• B. \(2\)
• C. \(3\)
• D. \(4\)

Hint:

If a midpoint and one endpoint are known, the other endpoint is obtained using \[ (x,y)=(2x_m-x_1,\;2y_m-y_1). \]

Answer 1

The Correct Option is B

Step 1: Use the midpoint-doubling idea.
If $A=(1,2)$ and $M$ is the midpoint of side $AB$, then $B$ is found by $B=2M-A$, that is double the midpoint and subtract $A$.
Step 2: Recover vertex $B$.
With midpoint $(-1,1)$: $B=(2(-1)-1,\;2(1)-2)=(-3,0)$.
Step 3: Recover vertex $C$.
With midpoint $(2,3)$: $C=(2(2)-1,\;2(3)-2)=(3,4)$.
Step 4: Set up the area formula.
Use Area $=\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$ with the three vertices.
Step 5: Substitute the coordinates.
With $A(1,2),B(-3,0),C(3,4)$: Area $=\frac{1}{2}|1(0-4)+(-3)(4-2)+3(2-0)|=\frac{1}{2}|-4-6+6|$.
Step 6: Finish.
This is $\frac{1}{2}\times 4=2$, matching option 2.
\[ \boxed{2} \]