NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 7: Coordinate Geometry - Exercise 7.1

Question 18

Step 1: Recall the formula for the area of a triangle with vertices \((x_1,y_1)\), \((x_2,y_2)\), \((x_3,y_3)\):

\[ \text{Area} = \tfrac{1}{2} \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \]

Step 2: Here, the vertices are:

• \((x_1,y_1) = (a, b+c)\)
• \((x_2,y_2) = (b, c+a)\)
• \((x_3,y_3) = (c, a+b)\)

Step 3: Substitute these into the determinant:

\[ \text{Area} = \tfrac{1}{2} \begin{vmatrix} a & b+c & 1 \\ b & c+a & 1 \\ c & a+b & 1 \end{vmatrix} \]

Step 4: Expand the determinant. After simplification, all the terms cancel out.

\[ \text{Area} = \tfrac{1}{2}(0) = 0 \]

Step 5: When the area is zero, it means the three points lie on the same straight line (they are collinear).

Final Answer: \(0\) (Option B).