Question

The area of triangle ABC is 5 sq units. Two of its vertices are A (2, 1) and B (3,-2)
Third vertex C lies on the line given by y - x + 3 = 0. Find the coordinates of C.​

Answer 1

The coordinates of C is (5, 2).

Step-by-step explanation:

Given,

The area of triangle ABC is 5 sq units.

Two of its vertices are A (2, 1) and B (3,-2).

Let C(x,y) be the third vertices.

∵ y - x + 3 = 0  ......(1)

∴ The area of triangle ABC = $\frac{1}{2}$ ×{2× (- 2 -y) + 3(y -1)+ x(1+2)}

⇒ 10 = - 4 - 2y + 3y - 3 +3x

⇒ 3x + y = 17  ......(2)

Solving equations (1) and (2), we get

∴ x = 5 and y = 2

Hence, the coordinates of C is (5, 2).

Answer 2

Answer:

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