## HOW TO CHECK WHETHER THE GIVEN POINTS FORM A RIGHT TRIANGLE

Here we are going see how to check whether the given points form a right triangle.

## How to check whether the given points form a right triangle ?

Step 1 :

After naming the given points, we have to find the length of each side of the given triangle using the formula distance between two points.

Step 2 :

We can pythagorean theorem,

That is,

Square of length of larger side  =  sum of the squares of other two sides

Step 3 :

If the given points satisfies the above condition, we can decide that the given points form a right triangle.

## How to prove a right triangle with coordinates - Examples

Example 1 :

Examine whether the given points  A (-3,-4) and B (2,6) and C(-6,10) forms a right triangle.

Solution :

Distance Between Two Points (x ₁, y₁) and (x₂ , y₂)

√(x₂ - x₁) ² + (y₂ - y₁) ²

The three points are  A (-3,-4) and B (2,6) and C(-6,10)

Distance between the points A and B

√(x₂ - x₁)² + (y₂ - y₁)²

Here x₁ = -3, y₁ = -4, x₂ = 2  and  y₂ = 6

√(2-(-3))² + (6-(-4))²

=   √(2+3)² + (6+4)²

=  √5² + 10²

=   √25 + 100

=   √125 units

Distance between the points B and C

√(x₂ - x₁)² + (y₂ - y₁)²

Here x₁ = 2, y₁ = 6, x₂ = -6  and  y₂ = 10

=    √(-6-2)² + (10-6)²

=    √(-8)² + (4)²

=    √64 + 16

=    √80 units

Distance between the points C and A

√(x₂ - x₁)² + (y₂ - y₁)²

Here x₁ = -6, y₁ = 10, x₂ = -3  and  y₂ = -4

=    √(-3-(-6))² + (-4-10)²

=    √(-3+6)² + (-14)²

=    √3² + (-14)²

=    √9 + 196

=    √205 units

AB = √125 units

BC = √80 units

CA = √205 units

(CA)² = (AB)² + (BC)²

(√205)²  = (√125)² + (√80)²

205 = 125 + 80

205 = 205

Therefore A,B and C forms a right triangle.

Let us see the next example problem on "How to check whether the given points form a right triangle"

Example 2 :

Examine whether the given points  P (7,1) and Q (-4,-1) and R (4,5) forms a right triangle.

Solution :

Distance Between Two Points (x ₁, y₁) and (x₂ , y₂)

√(x₂ - x₁)² + (y₂ - y₁)²

The three points are  P (7,1) and Q (-4,-1) and R (4,5)

Distance between the points P and Q

√(x₂ - x₁)² + (y₂ - y₁)²

Here x₁ = 7, y₁ = 1, x₂ = -4  and  y₂ = -1

=  √(-4-7)² + (-1-1)²

=   √(-11)² + (-2)²

=   √121 + 4

=   √125 units

Distance between the points Q and R

√(x₂ - x₁)² + (y₂ - y₁)²

Here x₁ = -4, y₁ = -1, x₂ = 4  and  y₂ = 5

=   √(4-(-4))² + (5-(-1))²

=    √(4+4)² + (5+1)²

=    √8² + 6²

=    √64 + 36

=    √100 units

Distance between the points R and P

√(x₂ - x₁)² + (y₂ - y₁)²

Here x₁ = 4, y₁ = 5, x₂ = 7  and  y₂ = 1

=  √(7-4)² + (1-5)²

=    √(3)² + (-4)²

=    √9 + 16

=    √25 units

PQ = √125 units

QR = √100 units

RP = √25 units

(PQ)² = (QR)² + (RP)²

(√125)²  = (√100)² + (√25)²

125 = 100 + 25

125 = 125

Hence, the given points  P,Q and R forms a right triangle.

Let us see the next example problem on "How to check whether the given points form a right triangle"

Example 3 :

Examine whether the given points  P (4,4) and Q (3,5) and R (-1,-1) forms a right triangle.

Solution :

Distance Between Two Points (x ₁, y₁) and (x₂ , y₂)

√(x₂ - x₁)² + (y₂ - y₁)²

The three points are  P (4,4) and Q (3,5) and R (-1,-1)

Distance between the points P and Q

√(x₂ - x₁)² + (y₂ - y₁)²

Here x₁ = 4, y₁ = 4, x₂ = 3  and  y₂ = 5

√(3-4)² + (5-4)²

=  √(-1)² + (1)²

=  √1 + 1

=  √2 units

Distance between the points Q and R

√(x₂ - x₁)² + (y₂ - y₁)²

Here x₁ = 3, y₁ = 5, x₂ = -1  and  y₂ = -1

√(-1-3)² + (-1-5)²

=   √(-4)² + (-6)²

√16 + 36

=  √52 units

Distance between the points R and P

√(x₂ - x₁)² + (y₂ - y₁)²

Here x₁ = -1, y₁ = -1, x₂ = 4  and  y₂ = 4

=  √(4-(-1))² + (4-(-1))²

=   √(4+1)² + (4+1)²

=   √5² + (5)²

=  √25 + 25

=  √50 units

PQ = √2 units

QR = √52 units

RP = √50 units

(QR)² = (PQ)² + (RP)²

(√52)²  = (√2)² + (√50)²

52 = 2 + 50

52 = 52

Hence, the given points P,Q and R forms a right triangle.

Let us see the next example problem on "How to check whether the given points form a right triangle" "How to check whether the given points form a right triangle"

Example 4 :

Examine whether the given points  A (2,0) and B (-2,3) and C (-2,-5) forms a right triangle.

Solution :

Distance Between Two Points (x ₁, y₁) and (x₂ , y₂)

√(x₂ - x₁)² + (y₂ - y₁)²

The three points are  A (2,0) and B (-2,3) and C (-2,-5)

Distance between the points A and B

√(x₂ - x₁)² + (y₂ - y₁)²

Here x₁ = 2, y₁ = 0, x₂ = -2  and  y₂ = 3

=  √(-2-2))² + (3-0)²

=    √(-4)² + (3)²

=   √16 + 9

=   √25  units

Distance between the points B and C

√(x₂ - x₁)² + (y₂ - y₁)²

Here x₁ = -2, y₁ = 3, x₂ = -2  and  y₂ = -5

=  √(-2-(-2))² + (-5-3)²

=    √(-2+2)² + (-8)²

=    √0² + 64

=    √64 units

Distance between the points C and A

√(x₂ - x₁)² + (y₂ - y₁)²

Here x₁ = -2, y₁ = -5, x₂ = 2  and  y₂ = 0

=  √(2-(-2))² + (0-(-5))²

=   √(2+2)² + (0+5)²

=    √4² + (5)²

=    √16 + 25

=    √41 units

AB = √25 units

BC = √64 units

CA = √41 units

(BC)² = (AB)² + (CA)²

(√64)²  = (√25)² + (√41)²

64 = 25 + 41

64 ≠ 66

Hence, the given points A,B and C will not form a right triangle.   "How to check whether the given points form a right triangle"

Let us see the next example problem on "How to check whether the given points form a right triangle"

Example 5 :

Examine whether the given points  A (0,0) and B (5,0) and C (0,6) forms a right triangle.

Solution :

To show that the given points forms a right triangle we need to find the distance between three points. The sum of squares of two sides is equal to the square of remaining side.

Distance Between Two Points (x ₁, y₁) and (x₂ , y₂)

√(x₂ - x₁)² + (y₂ - y₁)²

The three points are  A (0,0) and B (5,0) and C (0,6)

Distance between the points A and B

√(x₂ - x₁)² + (y₂ - y₁)²

Here x₁ = 0, y₁ = 0, x₂ = 5  and  y₂ = 0

=    √(5-0)² + (0-0)²

=    √(5)² + (0)²

=    √5² + 0²

=    √25 + 0

=    √25 units

Distance between the points B and C

√(x₂ - x₁)² + (y₂ - y₁)²

Here x₁ = 5, y₁ = 0, x₂ = 0  and  y₂ = 6

=    √(0-5)² + (6-0)²

=    √(-5)² + (6)²

=    √25 + 36

=    √61 units

Distance between the points C and A

√(x₂ - x₁)² + (y₂ - y₁)²

Here x₁ = 0, y₁ = 6, x₂ = 0  and  y₂ = 0

=    √(0-0)² + (0-6)²

=    √(0)² + (-6)²

=    √0 + 36

=    √36 units

AB = √25 units

BC = √61 units

CA = √36 units

(BC)² = (AB)² + (CA)²

(√61)²  = (√25)² + (√36)²

61 = 25 + 36

61 = 61

Hence, the given points A,B and C forms a right triangle.

Let us see the next example problem on "How to check whether the given points form a right triangle"

Example 6 :

Examine whether the given points  P (4,4) and Q (3,5) and R (-1,-1) forms a right triangle.  "How to check whether the given points form a right triangle"

Solution :

To show that the given points forms a right triangle we need to find the distance between three points. The sum of squares of two sides is equal to the square of remaining side.

Distance Between Two Points (x ₁, y₁) and (x₂ , y₂)

√(x₂ - x₁)² + (y₂ - y₁)²

The three points are  P (4,4) and Q (3,5) and R (-1,-1)

Distance between the points P and Q

√(x₂ - x₁)² + (y₂ - y₁)²

Here x₁ = 4, y₁ = 4, x₂ = 3  and  y₂ = 5

=    √(3-4)² + (5-4)²

=    √(-1)² + (1)²

=    √1 + 1

=    √2 units

Distance between the points Q and R

√(x₂ - x₁)² + (y₂ - y₁)²

Here x₁ = 3, y₁ = 5, x₂ = -1  and  y₂ = -1

=    √(-1-3)² + (-1-5)²

=    √(-4)² + (-6)²

=    √16 + 36

=    √52 units

Distance between the points R and P

√(x₂ - x₁)² + (y₂ - y₁)²

Here x₁ = -1, y₁ = -1, x₂ = 4  and  y₂ = 4

=    √(4-(-1))² + (4-(-1))²

=    √(4+1)² + (4+1)²

=    √5² + (5)²

=    √25 + 25

=    √50 units

PQ = √2 units

QR = √52 units

RP = √50 units

(QR)² = (PQ)² + (RP)²

(√52)²  = (√2)² + (√50)²

52 = 2 + 50

52 = 52

Hence, the given points P,Q and R forms a right triangle.

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