## Finding vertices on a graph with variable points.

Simplificatation, evaluation, linear equations, linear graphs, linear inequalities, basic word problems, etc.
richierichard
Posts: 1
Joined: Sun Feb 12, 2012 12:19 am
Contact:

### Finding vertices on a graph with variable points.

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
Posts: 1628
Joined: Mon Dec 08, 2008 4:22 pm
Contact:
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
Posts: 226
Joined: Mon Dec 08, 2008 5:18 pm
Contact:

### Re: Finding vertices on a graph with variable points.

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
Posts: 7
Joined: Tue Jan 10, 2012 4:38 pm
Location: U.S.A.
Contact:

### Re: Finding vertices on a graph with variable points.

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
$d^2=(2b-b)^2+(a-b)^2=b^2+a^2-2ab+b^2=2b^2-2ab+a^2$
$d^2=(2b-a)^2+(a-0)^2=4b^2-4ab+a^2+a^2=4b^2-4ab+2a^2$
$d^2=(a-b)^2+(0-b)^2=a^2-2ab+b^2+b^2=2b^2-2ab+a^2$

hope it helps