Question

The points $\left( {3,3} \right)$, $\left( {h,0} \right)$ and $\left( {0,k} \right)$ are collinear, if $\dfrac{1}{h} + \dfrac{1}{k}$ is equal to:
A. $\dfrac{1}{2}$
B. $\dfrac{1}{3}$
C. $1$
D. $2$

Answer 1

In the given question, we have to find the value of $\dfrac{1}{h} + \dfrac{1}{k}$ such that the points $\left( {3,3} \right)$, $\left( {h,0} \right)$ and $\left( {0,k} \right)$ are collinear. For this, first we must know what collinear points are. So, collinear points are those which lie on the same line. Now, if three points are collinear, the area of the triangle formed by the points has to be zero. So, equating the area of the triangle as zero, we get a relation between h and k. We manipulate the obtained relation to find the value of the expression.

Complete step by step answer:
The points given to us in the question are:
$\left( {3,3} \right)$, $\left( {h,0} \right)$ and $\left( {0,k} \right)$
Now, for the points to be collinear, they must satisfy the condition that the area of the triangle formed by the three points as the vertices of the triangle should be zero. Now, we know that the formula for the area of triangle $ABC$ with coordinates of vertices as,
$\left( {{x_1},{y_1}} \right)$, $\left( {{x_2},{y_2}} \right)$ and $\left( {{x_3},{y_3}} \right)$ is $\dfrac{1}{2}\left| {{x_1}\left( {{y_2} - {y_3}} \right) + {x_2}\left( {{y_3} - {y_1}} \right) + {x_3}\left( {{y_1} - {y_2}} \right)} \right|$

Substituting the values of coordinates of the three points, we get,
$\Rightarrow \dfrac{1}{2}\left| {3\left( {0 - k} \right) + h\left( {k - 3} \right) + 0\left( {3 - 1} \right)} \right|$
$\Rightarrow \dfrac{1}{2}\left| { - 3k + hk - 3h + 0} \right|$
Simplifying the expression we get,
$\Rightarrow \dfrac{1}{2}\left| { - 3k + hk - 3h} \right|$
Now, the area of the triangle must be equal to zero.
Hence, we get,
$\Rightarrow \dfrac{1}{2}\left| { - 3k + hk - 3h} \right| = 0$
We know that if the modulus of a quantity is zero, then that quantity must be equal to zero as zero is neither negative nor positive.Multiplying both sides of the equation by $2$ and removing the modulus function, we get,
$\Rightarrow - 3k + hk - 3h = 0$

Dividing both sides of equation by hk, we get,
$\Rightarrow \dfrac{{ - 3k + hk - 3h}}{{hk}} = 0$
$\Rightarrow \left( {\dfrac{{ - 3k}}{{hk}}} \right) + \left( {\dfrac{{hk}}{{hk}}} \right) + \left( {\dfrac{{ - 3h}}{{hk}}} \right) = 0$
Cancelling common factors in numerator and denominator, we get,
$\Rightarrow \left( {\dfrac{{ - 3}}{h}} \right) + 1 + \left( {\dfrac{{ - 3}}{k}} \right) = 0$
Shifting the terms consisting h and k to the right side of the equation. So, we get,
$\Rightarrow 1 = \dfrac{3}{h} + \dfrac{3}{k}$
Dividing both sides of equation by $3$, we get,
$\Rightarrow \dfrac{1}{h} + \dfrac{1}{k} = \dfrac{1}{3}$
So, the value of $\dfrac{1}{h} + \dfrac{1}{k}$ is $\left( {\dfrac{1}{3}} \right)$.

Hence, option B is the correct answer.

Note: We must remember the formula of area of triangle given the vertices of the triangle to solve such questions. We must take care of the calculations as it can change the final answer of the given problem. We must also know the basics and properties of a modulus function in order to tackle such problems.