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".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?
the above translates intoU 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!