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Worked Examples · Example 5

Q.Are the points A(3,6,9)A(3, 6, 9), B(10,20,30)B(10, 20, 30) and C(25,−41,5)C(25, -41, 5), the vertices of a right angled triangle?

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We check whether the three points form a right triangle by computing the squared distances between each pair and testing if the Pythagorean relation holds. The distances are AB=714AB = 7\sqrt{14}, BC=6686BC = \sqrt{6686}, and CA=3101CA = \sqrt{3101}; none of the squared sums match, so the triangle is not right-angled.

The question is: do three given points in 3D space form a right-angled triangle? In coordinate geometry, the cleanest way to test this is to compute the side lengths (using the distance formula) and then check if the square of the longest side equals the sum of the squares of the other two. That is the Pythagorean theorem in reverse — if it holds, the triangle is right-angled; if not, it isn’t.

We don’t need to find angles or slopes. Just distances. And because we’re in three dimensions, the distance between two points (x1,y1,z1)(x_1, y_1, z_1) and (x2,y2,z2)(x_2, y_2, z_2) is:

d=(x2−x1)2+(y2−y1)2+(z2−z1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}

We’ll compute ABAB, BCBC, and CACA, then compare squares.


  1. Distance ABAB between A(3,6,9)A(3,6,9) and B(10,20,30)B(10,20,30):

AB2=(10−3)2+(20−6)2+(30−9)2=72+142+212AB^2 = (10-3)^2 + (20-6)^2 + (30-9)^2 = 7^2 + 14^2 + 21^2

=49+196+441=686= 49 + 196 + 441 = 686

So AB=686=49×14=714AB = \sqrt{686} = \sqrt{49 \times 14} = 7\sqrt{14}.

  1. Distance BCBC between B(10,20,30)B(10,20,30) and C(25,−41,5)C(25,-41,5):

BC2=(25−10)2+(−41−20)2+(5−30)2=152+(−61)2+(−25)2BC^2 = (25-10)^2 + (-41-20)^2 + (5-30)^2 = 15^2 + (-61)^2 + (-25)^2

=225+3721+625=4571= 225 + 3721 + 625 = 4571

So BC=4571BC = \sqrt{4571}.

  1. Distance CACA between C(25,−41,5)C(25,-41,5) and A(3,6,9)A(3,6,9):

CA2=(3−25)2+(6−(−41))2+(9−5)2=(−22)2+472+42CA^2 = (3-25)^2 + (6-(-41))^2 + (9-5)^2 = (-22)^2 + 47^2 + 4^2

=484+2209+16=2709= 484 + 2209 + 16 = 2709

So CA=2709CA = \sqrt{2709}. …

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