Determine whether the points are collinear.P(-2, 3), Q(1, 2), R(4, 1)
Determine whether the points are collinear.P(-2, 3), Q(1, 2), R(4, 1)
P(–2, 3), Q(1, 2), R(4, 1)
\[PQ = \sqrt{\left( - 2 - 1 \right)^2 + \left( 3 - 2 \right)^2}\]
\[ = \sqrt{9 + 1} = \sqrt{10}\]
\[QR = \sqrt{\left( 1 - 4 \right)^2 + \left( 2 - 1 \right)^2}\]
\[ = \sqrt{9 + 1}\]
\[ = \sqrt{10}\]
\[PR = \sqrt{\left( - 2 - 4 \right)^2 + \left( 3 - 1 \right)^2}\]
\[ = \sqrt{36 + 4}\]
\[ = \sqrt{40}\]
\[ = 2\sqrt{10}\]
PQ + QR = PR
So, the given points lie on the same line. Hence, the given points are collinear.
Explanation:-
We are given three points: $P(-2, 3)$, $Q(1, 2)$, and $R(4, 1)$, and we need to determine if they are collinear (i.e., if they all lie on the same line).
To do this, we can use the distance formula to find the distances between each pair of points, and then check if the sum of the two shorter distances is equal to the longest distance. If it is, then the three points are collinear; if not, then they are not collinear.
Using the distance formula, we find:
$PQ = \sqrt{(1 – (-2))^2 + (2 – 3)^2} = \sqrt{10}$
$QR = \sqrt{(4 – 1)^2 + (1 – 2)^2} = \sqrt{10}$
$PR = \sqrt{(-2 – 4)^2 + (3 – 1)^2} = \sqrt{40} = 2\sqrt{10}$
Since $PQ + QR = 2\sqrt{10} + \sqrt{10} = 3\sqrt{10} = PR$, we can see that the sum of the two shorter distances is equal to the longest distance. Therefore, the given points are collinear.
Chapter 5. Co-ordinate Geometry – Practice Set 5.1 (Page 107)