Question

Determine whether the points are collinear.
P (-2, 3), B (1, 2), C (4, 1)

Answer 1

Hint: Here first we will calculate the lengths of PB, BC and PC using the following formula:-
The length of the line having A \[\left( {{x_1},{y_1}} \right)\]and Q \[\left( {{x_2},{y_2}} \right)\] as end points is given by:-
\[AQ = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} \]
Then we will check whether \[PB + BC = PC\], if it turns out to be true then the given points are collinear otherwise they are not collinear.

Complete step-by-step answer:
The given points are:-
P (-2, 3), B (1, 2), C (4, 1)
Now we know that:-
The length of the line having A\[\left( {{x_1},{y_1}} \right)\] and Q\[\left( {{x_2},{y_2}} \right)\] as end points is given by:-
\[AQ = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} \]
Applying this formula for points P and B we get:-
\[PB = \sqrt {{{\left( {1 - ( - 2} \right)}^2} + {{\left( {2 - 3} \right)}^2}} \]
Simplifying it we get:-
\[PB = \sqrt {{{\left( 3 \right)}^2} + {{\left( { - 1} \right)}^2}} \]
\[ \Rightarrow PB = \sqrt {9 + 1} \]
Solving it further we get:-
\[PB = \sqrt {10} \]……………………… (1)
Now we will calculate the length of BC:-
Now we know that:-
The length of the line having A\[\left( {{x_1},{y_1}} \right)\] and Q\[\left( {{x_2},{y_2}} \right)\] as end points is given by:-
\[AQ = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} \]
Applying this formula for points B and C we get:-
\[BC = \sqrt {{{\left( {4 - 1} \right)}^2} + {{\left( {1 - 2} \right)}^2}} \]
Simplifying it we get:-
\[BC = \sqrt {{{\left( 3 \right)}^2} + {{\left( { - 1} \right)}^2}} \]
\[ \Rightarrow BC = \sqrt {9 + 1} \]
Solving it further we get:-
\[BC = \sqrt {10} \]……………………… (2)
Now we will calculate the length of PC:-
Now we know that:-
The length of the line having A\[\left( {{x_1},{y_1}} \right)\] and Q\[\left( {{x_2},{y_2}} \right)\] as end points is given by:-
\[AQ = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} \]
Applying this formula for points P and C we get:-
\[PC = \sqrt {{{\left( {4 - \left( { - 2} \right)} \right)}^2} + {{\left( {1 - 3} \right)}^2}} \]
Simplifying it we get:-
\[PC = \sqrt {{{\left( 6 \right)}^2} + {{\left( { - 2} \right)}^2}} \]
\[ \Rightarrow PC = \sqrt {36 + 4} \]
\[ \Rightarrow PC = \sqrt {40} \]
Solving it further we get:-
\[PC = 2\sqrt {10} \]……………………… (3)
Now we know that for the points to be collinear \[PB + BC = PC\] should be true.
Let us first consider LHS:-
\[LHS = PB + BC\]
Putting values from equation 1 and 2 we get:-
\[LHS = \sqrt {10} + \sqrt {10} \]
\[ \Rightarrow LHS = 2\sqrt {10} \]
Now let us consider RHS:-
\[RHS = PC\]
Putting value from equation 3 we get:-
\[RHS = 2\sqrt {10} \]
Hence, \[LHS = RHS\]
Therefore, \[PB + BC = PC\]

Therefore, the given points are collinear.

Note: Students should note that if two lines AB and BC are collinear then they can be expressed as:-
\[AB = \lambda BC\] and the points A, B, C would be collinear.
Also, if the points A, B and C are collinear, then the area of the triangle formed by these points would be equal to zero.