Use the distance formula to show that the following set of points describes an isosceles triangle.?

Question

Please help me understand the following problem...

Use the distance formula to show that the following set of points describes an isosceles triangle.
(-3,0), (2,3) and (1,-1)

Answer 1

Find the lengths of the three sides. The distance formula between two points is:

sqrt((x2 - x1)^2 + (y2-y1)^2)

I'll just give the lengths for now, you can calculate it:

(-3,0) to (2,3) : sqrt(34)
(-3,0) to (1,-1): sqrt(17)
(2,3) to (1,-1) : sqrt(17)

Two of the sides have the same length. This is the definition of an isoceles triangle.

Answer 2

To do this, make a right triangle, drawing a line between 2 point and calling that the hypotenuse


Distance from (-3,0) to (2,3)
sqrt(5^2+3^2)=sqrt34

Distance from (-3,0) to (1,-1)
sqrt(4^2+1^2)=sqrt17

Distance from (2,3) to (1,-1)
sqrt(1^2+4^2)=sqrt17

Since 2 of these are swrt17, it is isosceles

Answer 3

The distance formula says that the straight line between two points (x1,y1) and (x2,y2) in 2-D space is sqrt( (x1-x2)^2 + (y1-y2)^2 ). You can extend it to 3-D as well.

Call the three points A,B and C. Compute the distances AB, BC and CA using the distance formula. You should find that two of these lines are of equal length - by definition, ABC is an isosceles triangle.

AB = sqrt( (-3-2)^2 + (0-3)^2) = sqrt( 25+9)
BC = sqrt( (2-1)^2 + (3+1)^2) = sqrt(1+16)
CA = sqrt( (1+3)^2 + (-1)^2) = sqrt(16+1)

BC and CA are of equal length, so ABC is an isosceles triangle.

Answer 4

Let the points A(-2,4) B(2,7) AND C(5,3) be the vertices of an isosceles triangle then Using distance formula find the length of one side of triangle i.e AB, which will be AB=5 Again using this formula to find length of second side BC,which is BC=5 so, AB=BC Triangle is isosceles. If you find length of third side it will be AC=5 square root of 2.

Answer 5

A(-3 , 0)
B(2 , -3)
C(1 , -1)
AB² = (2 + 3)² + (-3 - 0)²
AB² = 25 + 9 = 34
AC² = (1 + 3)² + (-1)²
AC² = 16 + 1 = 17
BC² = (1 - 2)² + (-1 - 3)²
BC² = 1 + 16 = 17
Thus AB² = AC² + BC² so triangle is in fact a right angled isosceles triangle
The equal sides are AC and BC

Answer 6

Find the distance between any two points. If you discover the distance is the same between two sets of points, then an isisceles triangle is formed.

Plot the points first to see the triangle and then apply the distance formula to see if two of the sides are the same length.

Answer 7

Get delta x and delta y by subtracting points.

(-3,0) - (1, -1) = |4 + 1|

(1, -1) - (2, 3) = |1+ 4|

Having same deltas means same distances.
Hence triangle is an isosceles.

Answer 8

Use the distance formula, d = √((x2-x1)²+(y2-y1)²), for each pairing of points (1st and 2nd, 2nd and 3rd, 3rd and 1st)

If two of the three distances are the same, then the points describe an isosceles triangle.

Answer 9

let A(-3,0,B(2,3); C(1,-1) be the three points.
AB=sqrt(34), BC=sqrt(17); CA=sqrt(17)
AB=CA=sqrt(17)
so the points form an isosceles triangle.