Question

If the points (p, 0), (0, q) and (1, 1) are collinear than $\dfrac{1}{p} + \dfrac{1}{q}$ is equal to
$
  (a){\text{ - 1}} \\
  (b){\text{ 1}} \\
  (c){\text{ 2}} \\
  (d){\text{ 0}} \\
$

Answer 1

Hint – In this question use the concept that if 3 points are collinear then the slope of any 2 pairs of points amongst them must be equal. Use direct formula to find slope if two coordinates are given as $\left( {{x_1},{y_1}} \right),\left( {{x_2},{y_2}} \right)$ , that is $\left( {\dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}} \right)$. This will help to find the solution.

Complete step-by-step answer:

${\text{A}}\left( {p,0} \right),{\text{ B}}\left( {0,q} \right),{\text{ and C}}\left( {1,1} \right)$
Now as we know that if three points are collinear then their slope of any two points must be equal to the other two points.
Therefore slope of AB $ = $ Slope of BC = slope of CA
Collinearity of points: - Collinear points always lie on the same line.
Now we know
Slope between two points${\text{ = }}\left( {\dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}} \right)$
Consider $A\left( {p,0} \right) \equiv \left( {{x_1},{y_1}} \right),{\text{ }}B\left( {0,q} \right) \equiv \left( {{x_2},{y_2}} \right),{\text{ }}C\left( {1,1} \right) \equiv \left( {{x_3},{y_3}} \right)$
Therefore slope of AB${\text{ = }}\left( {\dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}} \right){\text{ = }}\dfrac{{q - 0}}{{0 - p}} = \dfrac{{ - q}}{p}$
Therefore slope of BC${\text{ = }}\left( {\dfrac{{{y_3} - {y_2}}}{{{x_3} - {x_2}}}} \right){\text{ = }}\dfrac{{1 - q}}{{1 - 0}} = 1 - q$
Therefore slope of CA${\text{ = }}\left( {\dfrac{{{y_3} - {y_1}}}{{{x_3} - {x_1}}}} \right){\text{ = }}\dfrac{{1 - 0}}{{1 - p}} = \dfrac{1}{{1 - p}}$
Points are collinear
Therefore slope of AB $ = $ Slope of BC
$ \Rightarrow \dfrac{{ - q}}{p} = 1 - q$
Divide by q on both sides we have,
$ \Rightarrow \dfrac{{ - q}}{{pq}} = \dfrac{1}{q} - \dfrac{q}{q}$
Now simplify we have,
$ \Rightarrow \dfrac{{ - 1}}{p} = \dfrac{1}{q} - 1$
$ \Rightarrow \dfrac{1}{p} + \dfrac{1}{q} = 1$
So this is the required value of$\dfrac{1}{p} + \dfrac{1}{q}$.
Hence option (B) is correct.

Note – Collinear means that the points must be lying on the same line. Now there can too be other conditions to prove that points are collinear, like if three points are collinear that is $\left( {{x_1},{y_1}} \right),\left( {{x_2},{y_2}} \right),\left( {{x_3},{y_3}} \right)$ are collinear than $\left[ {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right] = 0$. Another concept of section formula can also be used if these points are collinear than $\left( {{x_3},{y_3}} \right)$ must divide the line segment of the points $\left( {{x_1},{y_1}} \right),\left( {{x_2},{y_2}} \right)$ in ratio 1: 1.