NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 7: Coordinate Geometry - Exercise 7.3

Question 9

Step 1: Let the vertices be \(A(-8,4)\), \(B(-6,6)\), \(C(-3,9)\).

Step 2: Use the coordinate geometry formula for the area of a triangle:

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

Step 3: Substitute \((x_1,y_1)=(-8,4)\), \((x_2,y_2)=(-6,6)\), \((x_3,y_3)=(-3,9)\):

\(\text{Area} = \dfrac{1}{2}\left|(-8)(6-9) + (-6)(9-4) + (-3)(4-6)\right|\)

Step 4: Simplify each term:

• \((-8)(6-9) = (-8)(-3) = 24\)
• \((-6)(9-4) = (-6)(5) = -30\)
• \((-3)(4-6) = (-3)(-2) = 6\)

Step 5: Add them:

\(24 + (-30) + 6 = 0\)

Step 6: Compute the area:

\(\text{Area} = \dfrac{1}{2}|0| = 0\)

Final Answer: The area of the triangle is 0 square units (so the three points are collinear).