1. Using the concept of area of triangle, show that the points A(5, -2), B(4, -1) and C(1, 2) are collinear.
2. Using the concept of distance between two points, show that the points A(5, -2), B(4, -1) and C(1, 2) are collinear.
3. Using the concept of equation of line, show that the points A(5, -2), B(4, -1) and C(1, 2) are collinear.
4. Using the concept of slope, show that the points A(5, -2), B(4, -1) and C(1, 2) are collinear.
1. Answer :
Using the concept of area of triangle, if the three points (x_{1}, y_{1}), (x_{2}, y_{2}) and (x_{3}, y_{3}) are collinear, then
x_{1}y_{2} + x_{2}y_{3} + x_{3}y_{1} = x_{2}y_{1} + x_{3}y_{2} + x_{1}y_{3}
(x_{1}, y_{1}) = A(5, -2)
(x_{2}, y_{2}) = B(4, -1)
(x_{3}, y_{3}) = C(1, 2)
x_{1}y_{2} + x_{2}y_{3} + x_{3}y_{1 }= 5(-1) + 4(2) + 1(-2)
= -5 + 8 - 2
= 1 ----(1)
x_{2}y_{1} + x_{3}y_{2} + x_{1}y_{3} = 4(-2) + 1(-1) + 5(2)
= -8 - 1 + 10
= 1 ----(2)
From (1) and (2), we get
x_{1}y_{2} + x_{2}y_{3} + x_{3}y_{1} = x_{2}y_{1} + x_{3}y_{2} + x_{1}y_{3}
Hence, the three points A, B and C are collinear.
2. Answer :
Formula to find the distance between two points (x_{1}, y_{1}) and (x_{2}, y_{2}) :
Substitute (x_{1}, y_{1}) = A(5, -2) and (x_{2}, y_{2}) = B(4, -1) into the above formula to find the distance between the two points A and B.
= √[(4 - 5)^{2} + (-1 + 2)^{2}]
= √[(-1)^{2} + 1^{2}]
= √[1 + 1]
= √2
Distance between B(4, -1) and C(1, 2) :
BC = √[(1 - 4)^{2} + (2 + 1)^{2}]
= √[(-3)^{2} + 3^{2}]
= √[9 + 9]
= √18
= 3√2
Distance between A(5, -2) and C(1, 2) :
AC = √[(1 - 4)^{2} + (2 + 2)^{2}]
= √[(-4)^{2 }+ 4^{2}]
= √[16 + 16]
= √32
= 4√2
AB + BC = √2 + 3√2 = 4√2 = AC
AB + BC = AC
Hence, the three points A, B and C are collinear.
3. Answer :
Equation of the straight line in two-point form is
Substitute (x_{1}, y_{1}) = A(5, -2) and (x_{2}, y_{2}) = B(4, -1) into the above formula to get equation of a line through the points A and B.
⁽ʸ ⁺ ²⁾⁄₍₋₁ ₊ ₂₎ = ⁽ˣ ⁻ ⁵⁾⁄₍₄ ₋ ₅₎
⁽ʸ ⁺ ²⁾⁄₁ = ⁽ˣ ⁻ ⁵⁾⁄₍₋₁₎
y + 2 = -x + 5
y = -x + 3
Substitute the third point C(1, 2).
2 = -1 + 3
2 = 2
Equation of the line passing through A(5, -2) and(B(4, -1) is satisfied by C(1, 2).
So, all the three points A, B and C are on the same line.
Hence, the three points A, B and C are collinear.
4. Answer :
Formula to find slope of a line joining two points (x_{1}, y_{1}) and (x_{2}, y_{2}) :
Substitute (x_{1}, y_{1}) = A(5, -2) and (x_{2}, y_{2}) = B(4, -1) into the above formula to get the slope of the line joining the two points A and B.
= ⁽⁻¹ ⁺ ²⁾⁄₍₄ ₋ ₅₎
= -1
Substitute (x_{1}, y_{1}) = B(4, -1) and (x_{2}, y_{2}) = C(1, 2) into the above formula to get the slope of the line joining the two points A and B.
= ⁽² ⁺ ¹⁾⁄₍₁ ₋ ₄₎
= -1
Slope of AB = slope of BC and B is the common point.
Hence, the three points A, B and C are collinear.
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