Triangle Area Problem - Detailed Solution and Interactive Quiz

Triangle Area Problem [IIT 1963]

Question: The area of the triangle formed by the points \( (a,\, b+c) \), \( (b,\, c+a) \), \( (c,\, a+b) \) is:

(a) \( abc \)

(b) \( a^2 + b^2 + c^2 \)

(c) \( ab + bc + ca \)

(d) \( 0 \)

Detailed Step-by-Step Explanation

Step 1: Area Formula
The area of a triangle with vertices \( (x_1, y_1) \), \( (x_2, y_2) \), \( (x_3, y_3) \) is given by:

\( \text{Area} = \frac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right| \)

Step 2: Identify the Coordinates
For the points:
\( A = (a,\, b+c) \), \( B = (b,\, c+a) \), \( C = (c,\, a+b) \)
We have:
\( x_1 = a,\; y_1 = b+c \)
\( x_2 = b,\; y_2 = c+a \)
\( x_3 = c,\; y_3 = a+b \)

Step 3: Substitute into the Formula
Substitute the coordinates into the area formula:

\( \text{Area} = \frac{1}{2} \left| a[(c+a)-(a+b)] + b[(a+b)-(b+c)] + c[(b+c)-(c+a)] \right| \)

Step 4: Simplify Each Term
For the first term:
\( a[(c+a)-(a+b)] = a(c+a-a-b) = a(c-b) \)

For the second term:
\( b[(a+b)-(b+c)] = b(a+b-b-c) = b(a-c) \)

For the third term:
\( c[(b+c)-(c+a)] = c(b+c-c-a) = c(b-a) \)

Step 5: Sum the Terms
Adding the three terms, we get:

\( a(c-b) + b(a-c) + c(b-a) \)

Expand each term:
\( = ac - ab + ab - bc + bc - ac \)
\( = 0 \)

Step 6: Compute the Area
Therefore, the area is:

\( \text{Area} = \frac{1}{2} \times 0 = 0 \)

Hence, the area of the triangle is \( 0 \), which means the points are collinear. This corresponds to option (d).