Question

A straight line L with negative slope passes through the point (8,2) and positive coordinate axes at points P and Q. The absolute minimum value of OP + OQ as L varies where O is the origin is ?

Answer 1

Hint: Start by drawing a representative diagram, considering a general equation of line, and getting an equation in two variables a and b by putting the point (8,2). Now find the value of OP+OQ in terms of a, and b. Finally, using both the equations minimise the value of OP+OQ.

Complete step-by-step answer:

Let us consider a general line $\dfrac{x}{a}+\dfrac{y}{b}=1$ and as it is in intercept form, we know that the x-intercept of the line is a, and the y-intercept of the line is b.
Now, drawing the diagram of the above line.

Now satisfying the point (8,2) in the equation of the line, we get

$\dfrac{8}{a}+\dfrac{2}{b}=1$

 $\Rightarrow \dfrac{8}{a}=1-\dfrac{2}{b}$

$\Rightarrow \dfrac{8}{a}=\dfrac{b-2}{b}$

$\Rightarrow \dfrac{8b}{b-2}=a............(i)$

Now from the figure:

OP+OQ = a + b

Now let us substitute the value of a from equation (i). On doing so, we get

$OP+OQ=\dfrac{8b}{b-2}+b$

$\Rightarrow OP+OQ=\dfrac{8b+{{b}^{2}}-2b}{b-2}$

$\Rightarrow OP+OQ=\dfrac{6b+{{b}^{2}}}{b-2}.............(ii)$

Now to minimise the value of OP and OB, let’s differentiate both sides of the equation with respect to b.

$\dfrac{d\left( OP+OQ \right)}{db}=\dfrac{\left( 6+2b \right)\left( b-2
\right)-{{b}^{2}}-6b}{{{\left( b-2 \right)}^{2}}}$

For maximum and minimum we know $\dfrac{d\left( OP+OQ \right)}{db}$ is zero.

$\dfrac{6b-12+2{{b}^{2}}-4b-{{b}^{2}}-6b}{{{\left( b-2 \right)}^{2}}}=0$

$\Rightarrow {{b}^{2}}-4b-12=0$

$\Rightarrow {{b}^{2}}-6b+2b-12=0$

$\Rightarrow \left( b+2 \right)\left( b-6 \right)=0$

Therefore, the values of b are -2, and 6. But according to question b can only be positive, so b=6 is the acceptable value.

Now put b=6 in equation (ii) to minimise it.

$\Rightarrow OP+OQ=\dfrac{6\times 6+{{6}^{2}}}{6-2}=\dfrac{72}{4}=18$

So, the answer to the above question is 18.


Note: In such questions, the form of the equation of the line you take decides the complexity of the calculations and the problem. For example: in the above question, we took the intercept form of the line, which helped us to directly write the value of OP + OQ as the sum of the intercepts.