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Question
The area of a triangle is 5 sq. units. Two of its vertices are (2, 1) and (3, −2). The third vertex is (x, y) where y = x + 3. Find the coordinates of the third vertex.
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Solution
Let the vertices A(2, 1), B(3, –2) and C(x, y)
Area of a triangle = 5 sq. units
`1/2[x_1y_2 + x_2y_3 + x_3y_1 - (x_2y_1 + x_3y_2 + x_1y_3)]` = 5
`1/2[- 4 + 3y + x - (3 - 2x + 2y)]` = 5
– 4 + 3y + x – 3 + 2x – 2y = 10
3x + y – 7 = 10
3x + y = 17 ...(1)
Given y = x + 3
Substitute the value ofy = x + 3 in (1)
3x + x + 3 = 17
4x = 17 – 3
4x = 14
x = `14/4 = 7/2`
Substitute the value of x in y = x + 3
y = `7/2 + 3`
⇒ y = `(7 + 6)/2`
= `13/2`
∴ The coordinates of the third vertex is `(7/2, 13/2)`
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