2 vertices of a triangle are 1,-4 and 6,-4. List 2 possible coordinates of the third vertex so that the triangle has an area of 20 units.
Let the line joining (1,-4) and (6,-4) be the base of your triangle.
This is a horizontal line, and it should be easy to see that its length is 5
let the height be h
(1/2)(5)h = 20
5h = 40
h = 4
so any point on the x-axis (or 4 units above your base) would be such a third point
possible cases:
(5,0) and (8,0)
you could also have points along a line 4 units below your base, or a y value of -8
two such points are:
(3,-8) and (7, -8)
how do you do this on a coordnate plane.
Just to clarify I think the first one is wrong 5x 4= 20 and it has to be 20 but triangles you divide by 1/2 and it'd be ten
To find the possible coordinates of the third vertex, we need to consider the formula for calculating the area of a triangle. The area of a triangle can be determined using the coordinates of its vertices.
The formula for the area of a triangle, given its vertices (x1, y1), (x2, y2), and (x3, y3), is:
Area = (1/2) * |x1*(y2 - y3) + x2*(y3 - y1) + x3*(y1 - y2)|
In this case, we already know the coordinates of two vertices as (1, -4) and (6, -4), and we want the triangle to have an area of 20 units. Therefore, we can substitute these known values into the formula and solve for the third coordinate.
Area = (1/2) * |1*(y2 - y3) + 6*(y3 - (-4)) + x3*(-4 - (-4))| = 20
Simplifying the equation, we have:
|y2 - y3 + 6*(y3 + 4)| = 40
Now, we can list the two possible coordinates of the third vertex by solving this equation:
1. Let's assign various values to y3 and solve for y2.
- Let y3 = 0
Substituting y3 = 0 into the equation, we have:
|y2 - 0 + 6*(0 + 4)| = 40
|y2 + 24| = 40
y2 + 24 = 40 or y2 + 24 = -40
y2 = 16 or y2 = -64
Therefore, one possible coordinate of the third vertex is (1, -4) and the other possible coordinate is (6, 16).
2. Let's assign different values to y3 and solve for y2 again.
- Let y3 = 8
Substituting y3 = 8 into the equation, we have:
|y2 - 8 + 6*(8 + 4)| = 40
|y2 + 80| = 40
y2 + 80 = 40 or y2 + 80 = -40
y2 = -40 or y2 = -120
Therefore, one possible coordinate of the third vertex is (1, -4) and the other possible coordinate is (6, -40).
In summary, two possible coordinates for the third vertex of the triangle, where the triangle has an area of 20 units, are (1, -4) and (6, 16), as well as (1, -4) and (6, -40).