Question

# Two vertices of a triangle are (−2, −1) and (3, 2) and third vertex lies on the line x + y = 5. If the area of the triangle is 4 square units, then the third vertex is (a) (0, 5) or, (4, 1) (b) (5, 0) or, (1, 4) (c) (5, 0) or, (4, 1) (d) (0, 5) or, (1, 4)

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Solution

## Let (h, k) be the third vertex of the triangle. It is given that the area of the triangle with vertices (h, k), (−2, −1) and (3, 2) is 4 square units. $\frac{1}{2}\left|h\left(-1-2\right)-3\left(-1-k\right)-2\left(2-k\right)\right|=4$ $⇒3h-5k+1=±8$ Taking positive sign, we get, $3h-5k+1=8$ $3h-5k-7=0$ ... (1) Taking negative sign, we get, $3h-5k+9=0$ ... (2) The vertex (h, k) lies on the line x + y = 5. $h+k-5=0$ ... (3) On solving (1) and (3), we find (4, 1) to be the coordinates of the third vertex. Similarly, on solving (2) and (3), we find (2, 3) to be the coordinates of the third vertex. Disclaimer: The correct option is not given in the question of the book.

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