The points $A (1, 2)$ , $B (5, 6)$ and $C (k, 14)$ are vertices of a triangle.

medium 4 marks Class 10 · Mathematics · Coordinate Geometry

Answer

(i)

k = 13

, (ii)

22

square units

The correct answer is (i) $k = 13$ , (ii) $22$ square units — Area $= \frac{1}{2} ∣44∣ = 22$ .

Step-by-step solution

Calculate the slope of $A B$ : $m_{A B} = \frac{6 - 2}{5 - 1} = \frac{4}{4} = 1$ .
For collinearity, the slope of $A C$ must also be $1$ : $m_{A C} = \frac{14 - 2}{k - 1} = \frac{12}{k - 1} = 1$ .
Solve $\frac{12}{k - 1} = 1 \Rightarrow k - 1 = 12 \Rightarrow k = 13$ .
Thus the points are collinear when $k = 13$ .
Now take $k = 2$ and 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 $(x_{1}, y_{1}) = (1, 2)$ , $(x_{2}, y_{2}) = (5, 6)$ , $(x_{3}, y_{3}) = (2, 14)$ .
Compute: $1 (6 - 14) + 5 (14 - 2) + 2 (2 - 6) = 1 (- 8) + 5 (12) + 2 (- 4) = - 8 + 60 - 8 = 44$ .
Area $= \frac{1}{2} ∣44∣ = 22$ .

Common mistake: Ignoring the absolute value when computing area or mixing up the order of coordinates in the determinant.

Concept tested: Collinearity and area of triangle using determinant