Exercise 4.2 answers: Determinants

Open NCERT p. 83Matches NCERT’s answer

(i) (1, 0), (6, 0), (4, 3)

1. Use the area formula: Area = ½ |x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|.
2. With (1,0), (6,0), (4,3): 1×(0−3) + 6×(3−0) + 4×(0−0) = −3 + 18 + 0 = 15.
3. Area = ½ × |15| = 15/2 sq units.

Answer15/2 sq units

(ii) (2, 7), (1, 1), (10, 8)

1. With (2,7), (1,1), (10,8): 2×(1−8) + 1×(8−7) + 10×(7−1) = −14 + 1 + 60 = 47.
2. Area = ½ × |47| = 47/2 sq units.

Answer47/2 sq units

(iii) (−2, −3), (3, 2), (−1, −8)

1. With (−2,−3), (3,2), (−1,−8): (−2)×(2−(−8)) + 3×(−8−(−3)) + (−1)×(−3−2) = −20 − 15 + 5 = −30.
2. Area = ½ × |−30| = 15 sq units.

Answer15 sq units

Watch this explained “One triangle, all the way through”, 12:34 into Area of a triangle from its vertices, and why three collinear points give zero

Open NCERT p. 83One way to think about it

1. Points are collinear if the area of the triangle they form is zero
2. Area = |1/2 · det([a b+c 1; b c+a 1; c a+b 1])|
3. Expanding the determinant:
4. det = a[(c+a) − (a+b)] − (b+c)[b − c] + 1[b(a+b) − c(c+a)]
5. = a(c − b) − (b+c)(b − c) + (ab + b² − c² − ca)
6. = a(c − b) − (b² − c²) + (ab + b² − c² − ca)
7. = ac − ab − b² + c² + ab + b² − c² − ca = 0
8. Since the area is 0, the three points are collinear.

In shortShown

Watch this explained “Letters instead of numbers”, 16:29 into Area of a triangle from its vertices, and why three collinear points give zero

Open NCERT p. 83Matches NCERT’s answer

(i) (k, 0), (4, 0), (0, 2)

1. Area expression: k×(0−2) + 4×(2−0) + 0×(0−0) = −2k + 8.
2. ½|−2k + 8| = 4, so |−2k + 8| = 8.
3. Either −2k + 8 = 8 (giving k = 0) or −2k + 8 = −8 (giving k = 8).
4. So k = 0 or k = 8.

Answerk = 0 or 8

(ii) (−2, 0), (0, 4), (0, k)

1. Area expression: (−2)×(4−k) + 0×(k−0) + 0×(0−4) = −8 + 2k.
2. ½|2k − 8| = 4, so |2k − 8| = 8.
3. Either 2k − 8 = 8 (giving k = 8) or 2k − 8 = −8 (giving k = 0).
4. So k = 0 or k = 8.

Answerk = 0 or 8

Watch this explained “Running the problem backwards”, 10:08 into Area of a triangle from its vertices, and why three collinear points give zero

Open NCERT p. 83Matches NCERT’s answer

(i) Find equation of line joining (1, 2) and (3, 6)

1. Let P(x, y) be any point on the line through (1, 2) and (3, 6). The three points lie on one line, so the triangle they form has area 0.
2. So ½ |x y 1; 1 2 1; 3 6 1| = 0.
3. Expand along the first row: x(2 × 1 − 1 × 6) − y(1 × 1 − 1 × 3) + 1(1 × 6 − 2 × 3) = −4x + 2y + 0.
4. So ½(−4x + 2y) = 0, which gives 2y = 4x, that is y = 2x.

Answery = 2x

(ii) Find equation of line joining (3, 1) and (9, 3)

1. Let P(x, y) be any point on the line through (3, 1) and (9, 3). The triangle formed by the three points has area 0.
2. So ½ |x y 1; 3 1 1; 9 3 1| = 0.
3. Expand along the first row: x(1 × 1 − 1 × 3) − y(3 × 1 − 1 × 9) + 1(3 × 3 − 1 × 9) = −2x + 6y + 0.
4. So ½(−2x + 6y) = 0, which gives x = 3y, that is x − 3y = 0.

Answerx − 3y = 0

Watch this explained “Declaring the area to be nothing”, 13:58 into Area of a triangle from its vertices, and why three collinear points give zero

Open NCERT p. 83Matches NCERT’s answer

1. Area expression: 2×(4−4) + 5×(4−(−6)) + k×(−6−4) = 0 + 50 − 10k = 50 − 10k.
2. ½|50 − 10k| = 35, so |50 − 10k| = 70.
3. Either 50 − 10k = 70 (giving k = −2) or 50 − 10k = −70 (giving k = 12).
4. So k = 12 or k = −2, which matches option (D).

Answer(D) 12, −2

Watch this explained “Four shapes of question”, 17:47 into Area of a triangle from its vertices, and why three collinear points give zero