Finding vertices on a graph with variable points.

[richierichard]

Here is the problem: If "a" and "b" are not both 0, show that the points (2b,a),(b,b), and (a,0) are the vertices of a right triangle.

Do I need to solve for the variables given or can I just plug in any number?

[stapel_eliz]

Here is the problem: If "a" and "b" are not both 0, show that the points (2b,a),(b,b), and (a,0) are the vertices of a right triangle.

Do I need to solve for the variables given or can I just plug in any number?

Infinitely-many points would "solve" the triangle, so there is little sense in attempting to "solve" anything. Also, I don't know what you mean by "just plugging in any number".

Instead, try doing as the exercise directs: You are given the three points, in generic form, forming the vertices of some triangle. You are asked to show that the three points in this particular format form a right triangle. You know the Distance Formula and the Pythagorean Theorem. So a good start might be to create expressions for the lengths of the three sides of the triangle and then, by plugging those distances into the Pythagorean Theorem, confirming that the lengths show the triangle to be right.

[little_dragon]

U get
d^2=(2b-b)^2+(a-b)^2=b^2+a^2-2ab+b^2=2b^2-2ab+a^2
d^2=(2b-a)^2+(a-0)^2=4b^2-4ab+a^2+a^2=4b^2-4ab+2a^2
d^2=(a-b)^2+(0-b)^2=a^2-2ab+b^2+b^2=2b^2-2ab+a^2
U can see wat U get when U add!

[Greenboyb1]

U get
d^2=(2b-b)^2+(a-b)^2=b^2+a^2-2ab+b^2=2b^2-2ab+a^2
d^2=(2b-a)^2+(a-0)^2=4b^2-4ab+a^2+a^2=4b^2-4ab+2a^2
d^2=(a-b)^2+(0-b)^2=a^2-2ab+b^2+b^2=2b^2-2ab+a^2
U can see wat U get when U add!

the above translates into




hope it helps