Question

A line is drawn through the point (1,2) to meet the coordinate axes at P and Q such that it forms a triangle OPQ, where O is the origin. If the area of the triangle OPQ is least, then the slope of the line PQ is
(a) -0.25
(b) -4
(c) -2
(d) -0.5

Answer 1

Hint:First use fact that the equation of the line which intersects both the coordinate axes with x - intercept equal to a and y – intercept equal to b is given by the equation: $\dfrac{x}{a}+\dfrac{y}{b}=1$. Substitute (1, 2) in this equation to find $b=\dfrac{2a}{a-1}$. Put this in Area (OPQ) formula = $\dfrac{1}{2}\times a\times b$ and then find $\dfrac{dA}{da}=0$ to calculate a and b,then find the slope of the equation using the slope of the line $\dfrac{x}{a}+\dfrac{y}{b}=1$ is $\dfrac{-b}{a}$.

Complete step-by-step answer:
In this question, we are given that A line is drawn through the point (1,2) to meet the coordinate axes at P and Q such that it forms a triangle OPQ, where O is the origin. The area of the triangle OPQ is least.
We need to find the slope of the line PQ.
We already know that the equation of the line which intersects both the coordinate axes with x - intercept equal to a and y – intercept equal to b is given by the following:
$\dfrac{x}{a}+\dfrac{y}{b}=1$
In this question, let OP be the x – intercept and let OQ be the y – intercept.
So, OP = a and OQ = b.
Now, we are also given that the line passes through the point (1, 2). Putting this in the given equation, we will get the following:
$\dfrac{1}{a}+\dfrac{2}{b}=1$
$b=\dfrac{2a}{a-1}$ …(1)
Now, the area of the triangle formed OPQ will be:
Area (OPQ) = $\dfrac{1}{2}\times OP\times OQ$
Area (OPQ) = $\dfrac{1}{2}\times a\times b$
Substituting equation (1) in the above expression, we will get the following:
Area (OPQ) = $\dfrac{1}{2}\times a\times b$
Area (OPQ) = $\dfrac{1}{2}\times a\times \dfrac{2a}{a-1}=\dfrac{{{a}^{2}}}{a-1}$
Now, this area needs to be minimum.
So, the derivative needs to be zero. i.e. $\dfrac{dA}{da}=0$
$\dfrac{d\left( \dfrac{{{a}^{2}}}{a-1} \right)}{da}=0$
We know the formula for derivative that $\dfrac{d\left( \dfrac{f}{g} \right)}{dx}=\dfrac{g\cdot \dfrac{df}{dx}-f\cdot \dfrac{dg}{dx}}{{{g}^{2}}}$.
Also, $\dfrac{d\left( {{a}^{2}} \right)}{da}=2a$ and $\dfrac{d\left( a-1 \right)}{da}=1-0=1$.
Using these, we will get the following:
$\dfrac{2{{a}^{2}}-2a-{{a}^{2}}}{{{\left( a-1 \right)}^{2}}}=0$
$\dfrac{{{a}^{2}}-2a}{{{\left( a-1 \right)}^{2}}}=0$
${{a}^{2}}-2a=0$
$a=0,2$.
For a = 0, the Area (OPQ) = 0 which is not possible.
So, a = 2
Now, $b=\dfrac{2a}{a-1}$
$b=\dfrac{2\times 2}{2-1}=4$
So, a = 2 and b = 4
Now, we know that the slope of the line $\dfrac{x}{a}+\dfrac{y}{b}=1$ is $\dfrac{-b}{a}$
So, the slope of the line PQ is $\dfrac{-4}{2}=-2$.
Hence, option (c) is correct.

Note: In this question, it is very important to know the following: the equation of the line which intersects both the coordinate axes with x - intercept equal to a and y – intercept equal to b is given by the following: $\dfrac{x}{a}+\dfrac{y}{b}=1$ and that for area to be minimum, the derivative needs to be zero. i.e. $\dfrac{dA}{da}=0$.