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.

1. Answer :

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.

2. Answer :

Distance between A and B :

AB = √[(x2 - x1)2 + (y2 - y1)2]

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

= √[(4 - 5)2 + (-1 + 2)2]

= √[(-1)2 + 12]

= √[1 + 1]

= √2

Distance between B and C :

BC = √[(1 - 4)2 + (2 + 1)2]

= √[(-3)2  + 32]

= √[9 + 9]

= √18

= 3√2

Distance between A and C :

AC = √[(5 - 1)2 + (-2 - 2)2]

= √[42 + (-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

(y - y1)/(y2 - y1) = (x - x1)/(x2 - x1)

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

(y + 2)/(-1 + 2) = (x - 5)/(4 - 5)

(y + 2)/1 = (x - 5)/(-1)

y + 2 = -x + 5

y = -x + 3

Substitute 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 :

Slope of AB = (y2 - y1)/(x2 - x1)

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

= (-1 + 2)/(4 - 5)

= 1/(-1)

= -1

Slope of BC = (2 + 1)/(1 - 4)

= 3/(-3)

= -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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