SOLUTION: Find a and b that a>0,b>0 and the point (a,b),(1,b) and (a,6) are vertices of a triangle of a base 3 units and area equal to 6 square units.

Hi, there--

The Problem:
Find a and b that a>0,b>0 and the point (a,b),(1,b) and (a,6) are vertices of a triangle of a base 3 
units and area equal to 6 square units.

A Solution:
The formula for the area of a triangle is A = (1/2)*B*H, where B is the base and H is the height. In 
your problem, the base is 3 units and the area is 6 square units. We substitute these known values 
into the formula to find the height of the triangle.

A = (1/2)*B*H
6 = (1/2)*(3)*H
6 = (3/2)*H
H = 6*(2/3)
H = 4

The height of the triangle is 4 units.

No we need to find values of a and b in the three ordered pairs that give us a base of 3 units and 
a height of 4 units.

Let's assume that we have a right triangle. It will simplify our calculations. Choose the point (1,b) 
to be the vertex at the right angle of the triangle. We'll use the distance formula to write equations 
using information from the problem.



Let's have the points (1,b) and (a,b) form the leg that is the base of the triangle. (Since they both 
have the same y-coordinate, they must form a horizontal line segment.) Then





Now we know that the points are (4,b),(1,b) and (4,6). Let's have the points (4,b) and (4,6) form the 
leg that is the height of the triangle. (Since they both have the same x-coordinate, they must form 
a vertical line segment.)







Now we know that the three points are (4,10), (1,10) and (4,6).

I hope this helps. Feel free to email me if you have questions about this solution.

Ms.Figgy
math.in.the.vortex@gmail.com