Q5. Show, using a determinant, that the points (2, 3), (4, 7) and (−1, −3) are collinear.
The determinant with rows 2, 3, 1; 4, 7, 1; −1, −3, 1 is 2(7 + 3) − 3(4 + 1) + 1(−12 + 7) = 20 − 15 − 5 = 0. The area is 0, so the points lie on one line (it is y = 2x − 1).
Q6. For which values of k are the points (k, 2 − 2k), (1 − k, 2k) and (−4 − k, 6 − 2k) collinear?
Subtract row 1 from rows 2 and 3 (a property of determinants, beyond the NCERT text): the rows become k, 2 − 2k, 1; 1 − 2k, 4k − 2, 0; −4 − 2k, 4, 0. Expanding along column 3 gives (1 − 2k)(4) − (4k − 2)(−4 − 2k) = 8k2 + 4k − 4 = 4(2k − 1)(k + 1). This is 0 when k = 1/2 or k = −1.
Q7. Show that the triangle with vertices (0, 0), (4, 0) and (x, 3) has area 6 whatever x is. Why?
The determinant with rows 0, 0, 1; 4, 0, 1; x, 3, 1 is, along row 1, 1 × (4 × 3 − 0 × x) = 12, so the area is 12/2 = 6. Geometrically, the base from (0, 0) to (4, 0) has length 4 and the third vertex is always at height 3 above it, so the area is ½ × 4 × 3.
Q8. List the vertices of NCERT Example 6 in the order (5, 1), (−4, 2), (3, 8). What is the determinant now, and does the area change?
This swaps the first and third rows, which changes the sign of the determinant: it is −61 instead of 61. The area is half the absolute value, 61/2, as before. This is why the area is defined with the absolute value.