Math question Pre calc?

Question

Use the distance formula to prove that the points (5,7), (4,0) and (2,1) are vertices of a right triangle

Answer 1

if it is a right triangle, the a^2 + b^2 = c^2, for sides a, b, hypotenuse c.

Just by looking at the points, you really do not know which is which. DRAW!

If A=(5,7), B=(4,0) and C=(2,1),

looks like AC is the hypotenuse (but then, I have only a rough sketch).

Take distances

(AB)^2 = (5-4)^2 + (7-0)^2, where AB is the distance between A and B.

(AB)^2 = 1 + 49 = 50

(BC)^2 = (4-2)^2 + (0-1)^2
= 4 + 1 = 5

(CA)^2 = (2-5)^2 + (1-7)^2
= 9 + 36 = 45

You see now that (CA)^2 + (BC)^2 = (AB)^2, hence is a right triangle.

Answer 2

Hello,

You need to find the length of the three sides and then see if the satisfy the theorem of Pythagros.


c^2 = a^2 + b^2 Where c is the hypotenuse and a is the length of one side and b the length of the other side.

I'll help you start.

distance between (5,7) and (4,0) sqrt((5-4)^2 + (7-0)^2)

sqrt(1 + 49) = sqrt50 or 5sqrt(2)

Now (4,0) and (2,1) sqrt((4-2)^2 + (0-1)^2)

sqrt (4 +1) = sqrt(5)

Now I'll let you finish.

Hope This Helps!!

Answer 3

let A = (5,7)

B = (4,0)

C = (2,1)

using distance formula d = sqrt[(x1 - x2)^2 + (y1 - y2)^2]

AB = sqrt[(5 - 4)^2 + (7-0)^2] = sqrt(50)

BC = sqrt[(4-2)^2 + (0 -1)^2] = sqrt(5)

AC = sqrt[(5 - 2)^2 + (7 -1)^2] = sqrt(45)

AC^2 + BC^2 = 45 + 5 = 50

AB^2 = 50

so AC^2 + BC^2 = AB^2

so ABC is a right angled triangle and

Answer 4

c = √[(4-5)² + (0-7)² ] = √50
a =√[(4-2)² + (0-1)² ] = √5
b =√[(5-2)² + (7-1)² ] = √45
a² + b² = c² . . . . this will prove if the sides form the right triangle
(√5)² + (√45)² = (√50)²
5 + 45 = 50
50 = 50
it is a right triangle