# WORKSHEET ON COLLINEAR POINTS

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.

Using the concept of area of triangle, if the three points (x1, y1), (x2, y2) and (x3, y3) are collinear, then

x1y2 + x2y3 + x3y1x2y1 + x3y2 + x1y3

(x1, y1) = A(5, -2)

(x2, y2) = B(4, -1)

(x3, y3) = C(1, 2)

x1y2 + x2y3 + x3y= 5(-1) + 4(2) + 1(-2)

= -5 + 8 - 2

= 1 ----(1)

x2y1 + x3y2 + x1y3 =  4(-2) + 1(-1) + 5(2)

= -8 - 1 + 10

= 1 ----(2)

From (1) and (2), we get

x1y2 + x2y3 + x3y1 = x2y1 + x3y2 + x1y3

Hence, the three points A, B and C are collinear.

Formula to find the distance between two points (x1, y1) and (x2, y2) :

Substitute (x1, y1) = A(5, -2) and (x2, y2) = 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 + 12]

= √[1 + 1]

= √2

Distance between B(4, -1) and C(1, 2) :

BC = √[(1 - 4)2 + (2 + 1)2]

= √[(-3)2  + 32]

= √[9 + 9]

= √18

= 3√2

Distance between A(5, -2) and C(1, 2) :

AC = √[(1 - 4)2 + (2 + 2)2]

= √[(-4)+ 42]

= √[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.

Equation of the straight line in two-point form is

Substitute (x1, y1) = A(5, -2) and (x2, y2) = 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⁾⁄₍₋

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.

Formula to find slope of a line joining two points (x1, y1) and (x2, y2) :

Substitute (x1, y1) = A(5, -2) and (x2, y2) = 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 (x1, y1) = B(4, -1) and (x2, y2) = 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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