Geometry

Triangles

Hard

Find the value of k so that the area of the triangle with vertices A(k + 1, 1), B(4, -3) and C(7, -k) is 6 square units.

✅ Correct Option: 1

Area $= \dfrac{1}{2}|x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|$

$= \dfrac{1}{2}|(k+1)(k-3) + 4(-k-1) + 7(4)|$

$= \dfrac{1}{2}|k^2 - 6k + 21|$

Setting this equal to 6 gives $k^2 - 6k + 21 = 12$ , so $k^2 - 6k + 9 = 0$ and $k = 3$ . The other case gives no real root.

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Find the value of k so that the area of the triangle with vertices A(k + 1, 1), B(4, -3) and C(7, -k) is 6 square units.

Solution

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