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 ${\displaystyle \frac{x}{a}}+{\displaystyle \frac{y}{b}}=1$ 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

${\displaystyle \frac{8}{a}}+{\displaystyle \frac{2}{b}}=1$

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

$\Rightarrow {\displaystyle \frac{8}{a}}={\displaystyle \frac{b-2}{b}}$

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

Now from the figure:

OP+OQ = a + b

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

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

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

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

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

${\displaystyle \frac{d(OP+OQ)}{db}}={\displaystyle \frac{(6+2b)(b-2)-b^2-6b}{{(b-2)}^2}}$

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

${\displaystyle \frac{6b-12+2b^2-4b-b^2-6b}{{(b-2)}^2}}=0$

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

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

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

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.

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

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.