 QUESTION

# A straight L with negative lines passes through the point (8, 2) and intersects the positive coordinates axes at points P and Q. As L varies the absolute minimum value of OP + OQ isA. 12B. 14C. 18D. 16

Hint: Here, first find the equation of a line with slope –m and $\left( {{x}_{1}},{{y}_{1}} \right)=(8,2)$, by the formula:
$y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$.Then the $x$ and $~y$ intercept so obtained will be OP and OQ respectively. Next, find OP + OQ, afterwards find the min (OP + OQ).

Complete Step-by-Step solution:
Here, we are given that the straight line L with negative line passes through the point (8, 2)and intersects the positive coordinates axes at points =P and Q.
Now, we have to find the minimum value of OP + OQ
We know that the equation of the line passing through the point $\left( {{x}_{1}},{{y}_{1}} \right)$ with slope m is given by the formula:
$y-{{y}_{1}}=m\left( x-{{x}_{1}} \right)$
Here, $\left( {{x}_{1}},{{y}_{1}} \right)=(8,2)$
Since the straight line L is negative, therefore slope = -m.
Hence we will get:
\begin{align} & y-2=-m\left( x-8 \right) \\ & y-2=-mx+8m \\ \end{align}
Next, by taking -2 to the left side it becomes 2 and by taking –mx to the left side it becomes mx. Hence, we obtain:
$y+mx=8m+2$
Now, by cross multiplication, taking 8m to the left side we get:
$\dfrac{y+mx}{8m+2}=1$
Next, by splitting the terms in the left side we get:
$\dfrac{y}{8m+2}+\dfrac{mx}{8m+2}=1$
In the next step, from the second term in the left side, take x to the denominator, we obtain:
$\dfrac{y}{8m+2}+\dfrac{x}{\dfrac{8m+2}{m}}=1$
By cancellation we get:
$\dfrac{y}{8m+2}+\dfrac{x}{8+\dfrac{2}{m}}=1$
This is of the form:
$\dfrac{x}{a}+\dfrac{y}{b}=1$, a is the x-intercept and b is the y- intercept.
Here, the x- intercept is OP and y-intercept is OQ. Hence, we will get:
OP = $8m+2$
OQ = $8+\dfrac{2}{m}$
\begin{align} & OP+OQ=8m+2+8+\dfrac{2}{m} \\ & OP+OQ=10+8m+\dfrac{2}{m} \\ \end{align}
Now, we can make this into a perfect square by adding and subtracting 8. Hence, we will get:
\begin{align} & OP+OQ=10+{{\left( 2\sqrt{2m} \right)}^{2}}+{{\left( \sqrt{\dfrac{2}{m}} \right)}^{2}}+8-8 \\ & OP+OQ={{\left( 2\sqrt{2m} \right)}^{2}}+{{\left( \sqrt{\dfrac{2}{m}} \right)}^{2}}+8+10-8 \\ & OP+OQ={{\left( 2\sqrt{2m} \right)}^{2}}+{{\left( \sqrt{\dfrac{2}{m}} \right)}^{2}}+8+18 \\ \end{align}
Hence, ${{\left( 2\sqrt{2m} \right)}^{2}}+{{\left( \sqrt{\dfrac{2}{m}} \right)}^{2}}+8$ is the expansion of ${{\left( \left( 2\sqrt{2m} \right)+\left( \sqrt{\dfrac{2}{m}} \right) \right)}^{2}}$, we will get:
$OP+OQ={{\left( \left( 2\sqrt{2m} \right)+\left( \sqrt{\dfrac{2}{m}} \right) \right)}^{2}}+18$
Next, we have to find to find the absolute minimum value of OP + OQ
\begin{align} & \min \left( OP+OQ \right)=\min \left( {{\left( \left( 2\sqrt{2m} \right)+\left( \sqrt{\dfrac{2}{m}} \right) \right)}^{2}}+18 \right) \\ & \min \left( OP+OQ \right)=\min {{\left( \left( 2\sqrt{2m} \right)+\left( \sqrt{\dfrac{2}{m}} \right) \right)}^{2}}+\min 18 \\ \end{align}
We know that the minimum value of any square is zero and min 18= 18, therefore we can write:
\begin{align} & \min \left( OP+OQ \right)=0+18 \\ & \min \left( OP+OQ \right)=18 \\ \end{align}
Hence, the absolute minimum value of OP + OQ = 18.
Therefore, the correct answer for this question is option (c).

Note: Here, you can also find the minimum by taking the derivative of OP + OQ with respect to m then equate it to zero and find the value of m. If the second derivative is greater than zero then we can say that the function is minimum at that particular m. Now, substitute the m so obtained in OP+OQ, you will get the minimum value.