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Prove that point (1, 1), (-2, 7) and (3,-3) are collinear.

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Last updated date: 13th Jun 2024
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Answer
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Hint: Here in this question we should know the definition of collinear points and how we can prove the points are collinear.
Definition of collinear points: - Three or more points are said to be collinear if they lie on a single straight line.
We will use distance formula between the two points ${x_1},{y_1}$ and ${x_2},{y_2}$ that is mentioned below: -
$d = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} $ d= distance between two points.

Complete step-by-step solution:
Let the three given points be named as A (1, 1), B (-2, 7) and C (3,-3). To prove these three points are collinear we have to prove that they lie on a single straight line. For this we will use distance formula.
$d = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} $
And to prove they are collinear we have to prove one of the three conditions mentioned below: -
AB+BC=AC
BC+AC=AB
AB+AC=BC
Finding length of line segments AB, BC and AC
Points for AB are A (1, 1) and B (-2, 7)
$ \Rightarrow AB = \sqrt {{{[ - 2 - 1]}^2} + {{[7 - 1]}^2}} $ (Putting values in distance formula $d = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} $ )
$ \Rightarrow AB = \sqrt {{{[ - 3]}^2} + {{[6]}^2}} $
$ \Rightarrow AB = \sqrt {9 + 36} $
 \[ \Rightarrow AB = \sqrt {45} \]
$\therefore AB = 6.70$
 Points for BC are B (-2, 7) and C (3, -3)
$ \Rightarrow BC = \sqrt {{{[3 + 2]}^2} + {{[ - 3 - 7]}^2}} $ (Putting values in distance formula $d = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} $ )
$ \Rightarrow BC = \sqrt {{{[5]}^2} + {{[10]}^2}} $
$ \Rightarrow BC = \sqrt {25 + 100} $
$ \Rightarrow BC = \sqrt {125} $ (Finding square root)
$\therefore BC = 11.18$
 Points for AC are A (1, 1) and C (3, -3)
$ \Rightarrow AC = \sqrt {{{[3 - 1]}^2} + {{[ - 3 - 1]}^2}} $ (Putting values in distance formula $d = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} $ )
$ \Rightarrow AC = \sqrt {{{[2]}^2} + {{[ - 4]}^2}} $
$ \Rightarrow AC = \sqrt {4 + 16} $
$ \Rightarrow AC = \sqrt {20} $ (Finding Square root)
$\therefore AC = 4.47$
Checking conditions: -
AB+BC=AC
$ \Rightarrow 6.70 + 11.18 \ne 4.47$ First condition is not satisfied.
BC+AC=AB
$ \Rightarrow 11.18 + 4.47 \ne 6.70$ Second condition is not satisfied.
AB+AC=BC
$ \Rightarrow 6.70 + 4.47 = 11.18$ Third condition is satisfied.

Therefore one condition for collinearity is satisfied so the given points (1, 1), (-2, 7) and (3,-3) are collinear.

Note: Students must apply distance formula carefully as the common mistake which is done by most of the students is that they get confused between subtraction sign instead of addition sign in distance formula.