Use the distance formula to prove that triangle ABC is an isosceles triangle.?

Question

http://i40.photobucket.com/albums/e215/xxs3xp1stolxx/9.jpg

Answer 1

Line Segment AB=square root of ((-a-0)^2+(0-b)^2)
AB=square root of(a^2+b^2)
Line Segment CB=square root of ((a-0)^2+(0-b)^2)
CB=square root of(a^2+b^2)
Thus AB=CB
Thus the triangle is isosceles

Answer 2

√( (x1 – x2)^2 + (y1 – y2)^2 )
A (-a, 0)
B (0, b)
C (a, 0)
NOTE: √( ), everything in the () is under the radical.

Distance between A & B = AB = √( (x1 – x2)^2 + (y1 – y2)^2 ) = √( (-a – 0)^2 + (0 – b)^2 ) = √( (-a)^2 + (-b)^2) = √( (a)^2 + (-b)^2)
Distance between C & B = BC= √( (x1 – x2)^2 + (y1 – y2)^2 ) = √( (a – 0) + (0 – b) ) = √( (a)^2 + (-b)^2)
AB = BC

EDIT: Be careful to not make the mistake that many of the answerers have which is to drop the negative from the first a...it doesn't drop until you square it. Proofs are about showing EVERY step.

Answer 3

AB = sqrt(a^2 + b^2) CB = sqrt(a^2+b^2) subsequently AB = CB subsequently triangle ABC is isosceles anticipate AB || CD Then attitude a million = attitude 2 yet attitude a million is given to be not = attitude 2 subsequently AB not || CD

Answer 4

basically you want to prove that the two sides AB and BC are equal.

so let's find the distance of both sides.

let's do the distance of AB first.
A (-a,0) B(0,b)
so the distance is Sqrt( (0--a)^2 +(b-0)^2)
which is sqrt(a^2+b^2)

now let's do the distance of BC.
B(0,b) C(a,0)
so the distance is Sqrt( (a-0)^2 + (0-b)^2)
which is sqrt(a^2+b^2)

distance of AB and BC are equal meaning that this triangle is isosceles.

note my notation sqrt( ) means the square root of

Answer 5

AB = sqrt((0-a)^2+ (b-0)^2)= sqrt(a^2+b^2)
BC = sqrt((0-a)^2 + (b-0)^2) = sqrt(a^2+b^2
Therefore AB = BC because they both = sqrt(a^2+b^2
Thus triangle ABC is isosceles by definition

Answer 6

I think:

d(P,Q)=sq. of (x^2+y^2)

-a(-a) and a(a) (Looks less awkward than aa, huh?) give us the same result. Both points are far from B by the same number of y because both of them are the x-intercept of the lines they're on. Now, what's different?

P.S.
This is Pythagoras Theorem. I'm not sure if it's really the "distance formula."

Answer 7

AB: x1=-a y1=0 x2=0 y2=b
distance=sqrt[(x1-x2)^2+(y1-y2)^2]
=sqrt(a^2+b^2)
AC: x1=-a y1=0 x2=0 y2=b
=sqrt(a^2+b^2)
AB=AC ,so the triangle is isoceles.