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

Question 19

Step 1: Three points are collinear if the area of the triangle formed by them is 0.

Step 2: Use the coordinate area formula:

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

Step 3: Substitute the coordinates:

• \(A(x_1,y_1) = (k+1,\,2k)\)
• \(B(x_2,y_2) = (3k,\,2k+3)\)
• \(C(x_3,y_3) = (5k-1,\,5k)\)

Step 4: Compute the required differences:

• \(y_2 - y_3 = (2k+3) - 5k = 3 - 3k\)
• \(y_3 - y_1 = 5k - 2k = 3k\)
• \(y_1 - y_2 = 2k - (2k+3) = -3\)

Step 5: Put into the formula (inside the absolute value):

\((k+1)(3-3k) + (3k)(3k) + (5k-1)(-3)\)

Step 6: Simplify step by step:

• \((k+1)(3-3k) = 3(k+1) - 3k(k+1) = (3k+3) - (3k^2+3k) = 3 - 3k^2\)
• \((3k)(3k) = 9k^2\)
• \((5k-1)(-3) = -15k + 3\)

Add them:

\((3 - 3k^2) + 9k^2 + (-15k + 3) = 6 + 6k^2 - 15k\)

Step 7: For collinearity, area must be zero:

\(\dfrac{1}{2}|6 + 6k^2 - 15k| = 0\ \Rightarrow\ 6 + 6k^2 - 15k = 0\)

Divide by 3:

\(2k^2 - 5k + 2 = 0\)

Step 8: Factorise:

\(2k^2 - 5k + 2 = (2k-1)(k-2) = 0\)

Step 9: Solve:

\(2k-1=0 \Rightarrow k=\tfrac{1}{2}\), or \(k-2=0 \Rightarrow k=2\).

Final Step (choose the valid value): \(k=\tfrac{1}{2}\) makes the points coincide (all three become the same point), so it does not form a proper set of distinct collinear points.

Therefore, \(k=2\).