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Question
Show that points A(–1, –1), B(0, 1), C(1, 3) are collinear.
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Solution 1
A(–1, –1), B(0, 1), C(1, 3)
AB = `sqrt((0 + 1)^2 + (1 + 1)^2`
= `sqrt(1 + 4)`
AB = `sqrt(5)`
BC = `sqrt((1)^2 + (2)^2`
= `sqrt(1 + 4)`
BC = `sqrt(5)`
AC = `sqrt((2)^2 + (3 + 1)^2`
= `sqrt(4 + 16)`
= `sqrt(20)`
= `sqrt(5 xx 4)`
AC = `2sqrt(5)`
AB + BC = AC
`sqrt(5) + sqrt(5) = 2sqrt(5)`
A, B, and C are collinear.
Solution 2
A`(x_1, y_1) = (-1, -1)`
B`(x_2, y_2) = (0,1)`
C`(x_3, y_3) = (1, 3)`
Slope of line
AB = `(y_2 - y_1)/(x_2 - x_1)`
= `(1 - (-1))/(0 - (-1))`
= `(1+1)/1 = 2`
Slope of line
BC = `(y_3 - y_2)/(x_3 - x_2)`
= `(3 - 1)/(1 - 0)`
= `2/1 = 2.`
As, slope of line AB = slope of line BC
Also AB and BC Hrtes contain common point B
∴ Points A, B, C are collinear. Hence Proved
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By distance formula,
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= `sqrt(square + square)`
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= `5sqrt(5)`
