Question

Vertices of triangle are (k, -2k), (2, k) and (k, 2) has area 10 sq. unit, then the number of possible integral values of k is

• A. 1
• B. 2
• C. 4
• D. 5

Answer 1

Answer: (B)

2

Answer 2

Let the vertices be $A(k, -2k)$, $B(2, k)$, and $C(k, 2)$.

The area of a triangle with coordinates $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by:

$$\text{Area} = \frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|$$

Substitute the coordinates:

$$\text{Area} = \frac{1}{2} \left| k(k-2) + 2(2 - (-2k)) + k((-2k) - k) \right|$$ $$= \frac{1}{2}\left| k^2 - 2k + 2(2+2k) - 3k^2 \right|$$ $$= \frac{1}{2}\left| k^2 - 2k + 4 + 4k - 3k^2 \right|$$ $$= \frac{1}{2}\left| -2k^2 + 2k + 4 \right|$$

Given that the area is $10$ sq. units, we have:

$$\frac{1}{2}\left| -2k^2 + 2k + 4 \right| = 10 \implies \left| -2k^2 + 2k + 4 \right| = 20$$

This leads to two cases:

Case 1:

$$-2k^2 + 2k + 4 = 20$$ $$-2k^2 + 2k -16 = 0 \implies 2k^2 - 2k + 16 = 0$$ $$k^2 - k + 8 = 0$$

The discriminant is:

$$\Delta = (-1)^2 - 4(1)(8) = 1 - 32 = -31 \quad \text{(No real solutions)}$$

Case 2:

$$-2k^2 + 2k + 4 = -20$$ $$-2k^2 + 2k + 24 = 0 \implies 2k^2 - 2k - 24 = 0$$ $$k^2 - k - 12 = 0$$

Solve using the quadratic formula:

$$k = \frac{1 \pm \sqrt{1 + 48}}{2} = \frac{1 \pm 7}{2}$$

Thus,

$$k = \frac{8}{2} = 4 \quad \text{or} \quad k = \frac{-6}{2} = -3$$

Both are integral values.

Therefore, the number of possible integral values of $k$ is 2.