Problem 1 :
Examine whether the given points A (-3,-4) and B (2,6) and C(-6,10) forms a right triangle.
Problem 2 :
Examine whether the given points P (7,1) and Q (-4,-1) and R (4,5) forms a right triangle.
Problem 3 :
Examine whether the given points P (4,4) and Q (3,5) and R (-1,-1) forms a right triangle.
Problem 4 :
Examine whether the given points A (2,0) and B (-2,3) and C (-2,-5) forms a right triangle.
Problem 5 :
Examine whether the given points A (0,0) and B (5,0) and C (0,6) forms a right triangle.
Problem 6 :
Examine whether the given points P (4,4) and Q (3,5) and R (-1,-1) forms a right triangle.
Problem 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.
Problem 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.
Problem 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.
Problem 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.
Problem 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.
Problem 6 :
Examine whether the given points P (4,4) and Q (3,5) and R (-1,-1) 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 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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