Triangle $M N O$ has vertices $M (2, 3)$ , $N (8, - 1)$ and $O (5, 7)$ . Find the length of the altitude from vertex $O$ to side $M N$ . Give your answer correct to two decimal places (in units).

hard 3 marks Class 10 · Mathematics · Triangles

Answer

4.99 units

The correct answer is 4.99 units — Round to two decimal places: $h \approx 4.99$ units.

Step-by-step solution

First find the length of side $M N$ (the base).
Vector $M N = (8 - 2, - 1 - 3) = (6, - 4)$ .
Length $M N = 6^{2} + (- 4)^{2} = 36 + 16 = 52 = 213 \approx 7.21 units .$
Compute the area of $△ M N O$ using the determinant formula:
$Area = \frac{1}{2} ∣ x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2}) ∣$ .
Substitute $(x_{1}, y_{1}) = M (2, 3), (x_{2}, y_{2}) = N (8, - 1), (x_{3}, y_{3}) = O (5, 7)$ :
$Area = \frac{1}{2} ∣2 ((- 1) - 7) + 8 (7 - 3) + 5 (3 - (- 1)) ∣$
Calculate each term: $2 ((- 8)) = - 16$ , $8 (4) = 32$ , $5 (4) = 20$ . Sum $= - 16 + 32 + 20 = 36$ .
Absolute value $∣36∣ = 36$ ; therefore $Area = \frac{1}{2} \times 36 = 18 square units$ .
Altitude $h$ from $O$ to $M N$ satisfies $Area = \frac{1}{2} \times M N \times h$ .
Solve for $h$ : $h = \frac{2 \times Area}{M N} = \frac{2 \times 18}{7.211} = \frac{36}{7.211} \approx 4.99$ .
Round to two decimal places: $h \approx 4.99$ units.

Common mistake: Forgetting the absolute value when evaluating the determinant, leading to a negative area.

Concept tested: Altitude length using area and coordinate geometry