SHOW THAT THE POINTS ARE THE VERTICES OF A RIGHT TRIANGLE

Example 1 :

Show that the following are the vertices of a right angled triangle. 

A(-3, -4), B(2, 6), C(-6, 10) 

Solution :

Distance between A and B :

Formula to find the distance between two points :

d = √[(x₂ - x₁)² + (y₂ - y₁)²]

Substitute (x₁, y₁) = A(-3, -4) and (x₂, y₂) = B(2, 6).

= √[(2 + 3)² + (6 + 4)²]

= √[5² + 10²]

= √[25 + 100]

AB = √125

AB² = 125

Distance between D and C :

= √[(x₂ - x₁)² + (y₂ - y₁)²]

Substitute (x₁, y₁) = B(2, 6) and (x₂, y₂) = C(-6, 10).

= √[(-6 - 2)² + (10 - 6)²]

= √[(-8)² + 4²]

= √[64 + 16]

DC = √80

DC² = 80

Distance between A and C :

= √[(x₂ - x₁)² + (y₂ - y₁)²]

Substitute (x₁, y₁) = A(-3, -4) and (x₂, y₂) = C(-6, 10).

= √[(-6 + 3)² + (10 + 4)²]

= √[(-3)² + 14²]

= √[9 + 196]

AC = √205

AC² = 205

From the above workings, 

125 + 80 = 205

AB² + DC² = AC²

The side lengths AB, BC and AC satisfy Pythagorean Theorem.

So, the given points form a right triangle. 

Example 2 :

Prove that (2, -2) (-2, 1) and (5, 2) are the vertices of right angled triangle. Find the area of the triangle and length of hypotenuse.

Solution :

Let the given coordinates as A(2, -2) B(-2, 1) and C(5, 2).

Distance between A and B :

d = √[(x₂ - x₁)² + (y₂ - y₁)²]

AB = √[(-2 - 2)² + (1 - (-2))²]

= √[(-4)² + (1 + 2)²]

= √[(-4)² + 3²]

= √16 + 9

= √25

Distance between B and C :

B(-2, 1) and C(5, 2).

BC = √[(5 - (-2))² + (2 - 1)²]

= √[(5 + 2)² + 1²]

= √[7² + 1²]

= √49 + 1

= √50

Distance between C and A :

A(2, -2) and C(5, 2).

CA = √[(5 - 2)² + (2 - (-2))²]

= √[3² + (2 + 2)²]

= √[3² + 4²]

= √9 + 16

= √25

BC² = AB² + CA²

√50² = √25² + √25²

50 = 25 + 25

50 = 50

Since it satisfies the Pythagorean theorem, it given coordinates are the points of the right triangle.

BC is the longest side and it must be the hypotenuse.

Example 3 :

The points 𝐴(4, 7), 𝐵(𝑝, 3) and 𝐶(7, 3) are the vertices of a right triangle, right–angled at 𝐵. Find the value of 𝑝.

Solution :

Let the given coordinates as 𝐴(4, 7), 𝐵(𝑝, 3) and 𝐶(7, 3)

Distance between A and B :

d = √[(x₂ - x₁)² + (y₂ - y₁)²]

AB = √[(p - 4)² + (3 - 7)²]

= √[(p - 4)² + (-4)²]

= √[(p - 4)² + 16]

Distance between B and C :

𝐵(𝑝, 3) and 𝐶(7, 3)

BC = √[(7 - p)² + (3 - 3)²]

= √[(7 - p)² + 0²]

= 7 - p

Distance between C and A :

𝐴(4, 7) and 𝐶(7, 3)

CA = √[(7 - 4)² + (3 - 7)²]

= √[3² + (-4)²]

= √[3² + 4²]

= √9 + 16

= √25

Since right angled at B, AB is perpendicular to BC.

AC² = AB² + BC²

√25² = √[(p - 4)² + 16]² + (7 - p)²

25 = p² - 8p + 16 + 16 + 7² - 14p + p²

25 = 2p² - 22p + 32 + 49

2p² - 22p + 81 - 25 = 0

2p² - 22p + 56 = 0

p² - 11p + 28 = 0

(p - 4) (p - 7) = 0

p = 4 and p = 7

So, the possible values of p are 4 and 7.

Example 4 :

Show that the points (−2, 5);(3, −4) and (7, 10) are the vertices of a right angled isosceles triangle.

Solution :

Let the given points be A (−2, 5) B (3, −4) and C (7, 10)

Distance between AB :

AB = √[(3 + 2)² + (-4 - 5)²]

= √[5² + (-9)²]

= √(25 + 81)

= √106

Distance between BC :

BC = √[(7 - 3)² + (10 + 4)²]

= √[4² + 14²]

= √(16 + 196)

= √212

Distance between CA :

CA = √[(7 + 2)² + (10 - 5)²]

= √[9² + 5²]

= √(81 + 25)

= √106

BC² = AB² + AC²

√212² = √106² + √106²

212 = 106 + 106

212 = 212

So, these are the points of right triangle. Since two sides are equal, it must be isosceles triangle.

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