The vertices of a triangle are $P (1, 2)$ , $Q (5, 8)$ and $R (k, k^{2})$ . If the area of the triangle is $12$ square units, find all possible values of $k$ .

hard 5 marks Class 10 · Mathematics · Coordinate Geometry

Answer

k = \frac{3 + 113}{4} or k = \frac{3 - 113}{4}

The correct answer is $k = \frac{3 + 113}{4} or k = \frac{3 - 113}{4}$ — Therefore the real values of $k$ are $k = \frac{3 + 113}{4}$ or $k = \frac{3 - 113}{4}$ .

Step-by-step solution

Use the determinant formula for area: $Area = \frac{1}{2} x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})$ .
Substitute $P (1, 2), Q (5, 8), R (k, k^{2})$ :
$E = 1 (8 - k^{2}) + 5 (k^{2} - 2) + k (2 - 8)$ .
Simplify: $E = (8 - k^{2}) + 5 k^{2} - 10 + k (- 6) = 4 k^{2} - 6 k - 2$ .
Given area $12$ , set $\frac{1}{2} ∣ E ∣ = 12 \Rightarrow ∣ E ∣ = 24$ .
This yields two equations: $4 k^{2} - 6 k - 2 = 24$ or $4 k^{2} - 6 k - 2 = - 24$ .
First case: $4 k^{2} - 6 k - 26 = 0 \Rightarrow 2 k^{2} - 3 k - 13 = 0$ .
Solve: $k = \frac{3 \pm 9 + 104}{4} = \frac{3 \pm 113}{4}$ .
Second case: $4 k^{2} - 6 k + 22 = 0 \Rightarrow 2 k^{2} - 3 k + 11 = 0$ , whose discriminant $9 - 88 = - 79$ is negative, so no real $k$ .
Therefore the real values of $k$ are $k = \frac{3 + 113}{4}$ or $k = \frac{3 - 113}{4}$ .

Common mistake: Ignoring the absolute‑value condition, which can lead to missing the negative case or extraneous solutions.

Concept tested: Area of triangle using determinant

Published 23 August 2026 · How this solution was prepared and checked

The vertices of a triangle are P(1,2), Q(5,8) and R(k,k2). If the area of the triangle is 12 square units, find all possible values of k.

Step-by-step solution

Related questions

The vertices of a triangle are $P (1, 2)$ , $Q (5, 8)$ and $R (k, k^{2})$ . If the area of the triangle is $12$ square units, find all possible values of $k$ .