Let A(x_{1}, y_{1}) and B(x_{2}, y_{2}) be two endpoints of the line segment. To find the length of the line segment or distance between two points, we use the formula given below.
d = √(x_{2} – x_{1})^{2} + (y_{2 }– y_{1})^{2}
Find a given that :
Problem 1 :
P(2, 3) and Q(a, -1) are 4 units apart
Solution :
d = √(x_{2} – x_{1})^{2} + (y_{2 }– y_{1})^{2}
Let x_{1} = 2, y_{1} = 3, x_{2} = a, y_{2} = -1
Here d = 4
4 = √(a – 2)^{2} + (-1_{ }– 3)^{2}
4 = √(a – 2)^{2} + (-4)^{2}
4 = √(a – 2)^{2} + 16
Taking square on both sides.
4^{2} = (a – 2)^{2} + 16
(a – 2)^{2} = 0
a - 2 = 0
a = 2
So, the missing coordinates are a = 2.
Problem 2 :
P(-1, 1) and Q(a, -2) are 5 units apart
Solution :
d = √((x_{2} – x_{1})^{2} + (y_{2 }– y_{1})^{2})
Let x_{1} = -1, y_{1} = 1, x_{2} = a, y_{2} = -2
Here d = 5
5 = √(a + 1)^{2} + (-2_{ }– 1)^{2}
5 = √(a + 1)^{2} + (-3)^{2}
5 = √(a + 1)^{2} + 9
Taking squares on both sides.
25 = (a + 1)^{2} + 9
Subtracting 9 on both sides,
16 = (a + 1)^{2}
Again take squares on both sides.
(a + 1) = √16
a + 1 = ± 4
a + 1 = 4 a = 4 - 1 a = 3 |
a + 1 = -4 a = -4 - 1 a = -5 |
So, the missing coordinates are a = 3 or
a = -5.
Problem 3 :
X (a, a) is √8 units from the origin
Solution :
X(a, a) and O(0, 0)
d = √((x_{2} – x_{1})^{2} + (y_{2 }– y_{1})^{2})
Let x_{1} = a, y_{1} = a, x_{2} = 0, y_{2} = 0
Here d = √8
√8 = √(0 - a)^{2} + (0_{ }– a)^{2}
√8 = √a^{2} + a^{2}
√8 = √2a^{2}
Taking squares on both sides.
8 = 2a^{2}
8/2 = a^{2}
4 = a^{2}
a = ±2
Problem 4 :
A (0, a) is equidistant from P(3, -3) and Q(-2, 2)
Solution :
AP = AQ
d = √((x_{2} – x_{1})^{2} + (y_{2 }– y_{1})^{2})
√((3 – 0)^{2} + (-3_{ }– a)^{2})= √((-2 – 0)^{2} + (2_{
}– a)^{2})
√(9 + (-3)^{2} + a^{2} – 2(-3)(a))) = √(4 + 2^{2} + a^{2} – 2(2)(a))
√(9 + 9 + a^{2} + 6a) = √(4 + 4 + a^{2} – 4a)
√18 + a^{2} + 6a = √(8 + a^{2} – 4a)
Taking square on both sides.
18 + a^{2} + 6a = 8 + a^{2} – 4a
18 + a^{2} + 6a - 8 - a^{2} + 4a = 0
10 + 10a = 0
10a = -10
a = -1
Problem 5 :
Find b given that A(3, -2) and B(b, 1) are √13 units apart.
Solution :
d = √((x_{2} – x_{1})^{2} + (y_{2 }– y_{1})^{2})
Let x_{1} = 3, y_{1} = -2, x_{2} = b, y_{2} = 1
Here d = √13
√13 = √((b – 3)^{2} + (1_{ }+ 2)^{2})
√13 = √(b – 3)^{2} + (3)^{2}
√13 = √(b – 3)^{2} + 9
Taking square on both sides.
13 = (b – 3)^{2} + 9
Subtracting 9 on both sides.
13 - 9 = (b – 3)^{2}
4 = (b – 3)^{2}
b - 3 = √4
b - 3 = √4
b - 3 = ±2
b - 3 = 2 b = 2 + 3 b = 5 |
b - 3 = -2 b = -2 + 3 b = 1 |
So, the missing coordinates are b = 1 or b = 5.
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM