Aspire Faculty ID #18687 · Topic: NIMCET 2026 · Just now
NIMCET 2026

A triangle has a vertex at (1, 2) and the mid points of the two sides through it are (-1, 1) and (2, 3). Then the area of this triangle is

Solution

Let vertex be $A(1,2)$. 
Midpoints of sides through $A$ are: 
$M_1(-1,1)$ and $M_2(2,3)$ 

Let other two vertices be $B(x_1,y_1)$ and $C(x_2,y_2)$. 

Using midpoint formula: $\left(\frac{1+x_1}{2},\frac{2+y_1}{2}\right)=(-1,1)$ 
$x_1=-3,\ y_1=0$ 
So, $B(-3,0)$ 

Similarly, $\left(\frac{1+x_2}{2},\frac{2+y_2}{2}\right)=(2,3)$ $x_2=3,\ y_2=4$ 
So, $C(3,4)$ 

Area of triangle: 
$\frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|$ 
$=\frac{1}{2}|1(0-4)+(-3)(4-2)+3(2-0)|$ 
$=\frac{1}{2}|-4-6+6|$ 
$=\frac{1}{2}\times 4$ 
$=2$
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