Find the perimeter if the third vertex lies on a line
The area of a triangle is 5 square units. Two of its vertices are ( 2 , 1 ) and ( 3 , − 2 ) . The third vertex lies on the line y = x + 3 . What is the shortest perimeter of the triangle?
The answer is 12.399.
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3 solutions
The vertices of the triangle are A ( 2 , 1 ) , B ( 3 , − 2 ) , C ( x , x + 3 ) . By the shoelace formula , the area is
2 1 ∣ ( 2 × ( − 2 ) + 3 × ( x + 3 ) + x × 1 ) − ( 1 × 3 + ( − 2 ) × x + ( x + 3 ) × 2 ) ∣ = ∣ 2 x − 2 ∣
so we have ∣ 2 x − 2 ∣ = 5 . So either 2 x − 2 = 5 or 2 − 2 x = 5 , giving the two solutions x = 2 7 or x = − 2 3 respectively. These give two different perimeters, 1 7 . 3 7 7 … or (the smaller answer) 1 2 . 3 9 8 … .
You're absolutely right Chris. I was just going to point this out. Saw your post and controlled myself. :)
Area = 1/2 * (x1(y2 − y3) + x2(y3 − y1) + x3(y1 − y2))
Let the third point be (x, y)
5 = 0.5 * (x *(-2-1) + 3 * ( 1 - y) + 2 * (y +2)
10 = (-3x + 3 -3y + 2y + 4)
3 = -3x - y
y = -3x - 3
But, (x, y) also lies on the line y = x +3
-3x - 3 = x + 3
-4x = 6
Therefore, x = -1.5 and y = -1.5 + 3 = 1.5
The third vertex is (-1.5, 1.5)
Side 1 = SQRT((2-3)^2+(1+2)^2) = SQRT(10)
Side 2 = SQRT((2+1.5)^2+(1-1.5)^2) = SQRT(12.5)
Side 3 = SQRT((3+1.5)^2 + (-2-1.5)^2) = SQRT(32.5)
Follows that perimeter =12.399
There are two valid answers - I've posted a solution showing this but you might want to modify the problem wording slightly.
Good point , just drew a graph and verified it. Thanks for pointing it out.
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No worries, thanks for updating the problem! Incidentally, the different signs of the area in the shoelace formula correspond to whether the order of the points A B C is clockwise or anticlockwise.