Question

Find the value of ‘p’ for which the points A (– 5, 1), B(1, p) and C(4,– 2) are collinear.

Answer 1

Hint: Collinear points are points that lie on the same line. Find the vector AB and the vector AC, they should be parallel, hence, one vector is a constant multiple of the other. Hence, equate them and find the value of ‘p’.

Complete step-by-step answer:
Collinear points are points that all lie in the same line, whether they are close together, far apart, or form a ray, line segment, or line. In Latin ‘col’ means together and ‘linear’ means line.
We are given three points A (– 5, 1), B(1, p), and C(4,– 2). We need to find the value of ‘p’ such that these three points are collinear.
We find the vectors AB and AC.
\[\overrightarrow {AB} = (1 - ( - 5))\hat i + (p - 1)\hat j\]
Simplifying, we have:
\[\overrightarrow {AB} = 6\hat i + (p - 1)\hat j............(1)\]
\[\overrightarrow {AC} = (4 - ( - 5))\hat i + ( - 2 - 1)\hat j\]
Simplifying, we have:
\[\overrightarrow {AC} = 9\hat i - 3\hat j.........(2)\]
If the three lines are collinear, then the vectors AB and AC are parallel to each other and differ only by a constant multiple, say a. Then, we have:
\[\overrightarrow {AB} = a\overrightarrow {AC} \]
Substituting equation (1) and equation (2) in the above equation, we have:
$\Rightarrow$ \[6\hat i + (p - 1)\hat j = a(9\hat i - 3\hat j)\]
Simplifying, we have:
$\Rightarrow$ \[6\hat i + (p - 1)\hat j = 9a\hat i - 3a\hat j\]
Equality implies that the components of the vectors are equal. Hence, we have:
$\Rightarrow$ \[6 = 9a.............(3)\]
$\Rightarrow$ \[p - 1 = - 3a...........(4)\]
From equation (3), we get the value of a as follows:
$\Rightarrow$ \[a = \dfrac{6}{9}\]
$\Rightarrow$ \[a = \dfrac{2}{3}................(5)\]
Substituting equation (5) in equation (4), we have:
$\Rightarrow$ \[p - 1 = - 3 \times \dfrac{2}{3}\]
Simplifying, we have:
$\Rightarrow$ \[p - 1 = - 2\]
$\Rightarrow$ \[p = - 2 + 1\]
$\Rightarrow$ \[p = - 1\]
Hence, the value of ‘p’ is – 1.

Note: You can also solve this using geometry and matrix methods. In the geometric method, find the straight line equation joining the points A and C and substitute point B in it. In the matrix method, use the fact that the area of collinear points is zero to solve for p.