Question

In a plane rectangular coordinate system, points $L, M, N,$ and $O$ are represented by the coordinates $(-5,0), (1,-1), (0,5),$ and $(-1,5)$ respectively. Consider a variable point $P$ in the same plane. The minimum value of $PL + PM + PN + PO$ is:

• A. $1 + \sqrt{37}$
• B. $5\sqrt{2} + 2\sqrt{10}$
• C. $\sqrt{41} + \sqrt{37}$
• D. $\sqrt{41} + 1$
• E. None of these

Hint:

When minimizing sums of distances, check if the expression can be split into pairs. Each pair is minimized when the point lies on the line segment joining the two fixed points. Hence, the solution is often the intersection of diagonals of the quadrilateral formed by the given points.

Answer 1

The Correct Option is B

Step 1: Understand the problem.
We are asked to minimize the sum of distances: \[ S = PL + PM + PN + PO \] where $P$ is a variable point in the plane. This type of problem is usually solved by considering symmetry and properties of line segments (Fermat point or intersection of diagonals of quadrilaterals).

Step 2: Analyze the structure.
- To minimize $(PL + PN)$, point $P$ must lie on the line segment $LN$ (by triangle inequality property). - To minimize $(PM + PO)$, point $P$ must lie on the line segment $MO$. Therefore, the optimal point $P$ is at the

intersection of $LN$ and $MO$, i.e., the diagonals of quadrilateral $LMNO$.

Step 3: Find intersection point $P$.
Equation of $LN$: points $L(-5,0)$ and $N(0,5)$. Slope = $\frac{5-0}{0-(-5)} = \frac{5}{5} = 1$. Equation: $y-0 = 1(x+5) \Rightarrow y = x+5$. Equation of $MO$: points $M(1,-1)$ and $O(-1,5)$. Slope = $\frac{5-(-1)}{-1-1} = \frac{6}{-2} = -3$. Equation: $y+1 = -3(x-1) \Rightarrow y = -3x+2$. Now solve simultaneously: \[ x+5 = -3x+2 \quad \Rightarrow \quad 4x = -3 \quad \Rightarrow \quad x = -\tfrac{3}{4}, \; y = \tfrac{17}{4}. \] So $P = \left(-\tfrac{3}{4}, \tfrac{17}{4}\right)$.

Step 4: Compute $LN$ and $OM$.
At this optimal point, \[ PL + PN = LN, \quad PM + PO = MO \] So the minimum value = $LN + MO$. Distance $LN$: \[ LN = \sqrt{(0 - (-5))^2 + (5-0)^2} = \sqrt{25+25} = \sqrt{50} = 5\sqrt{2}. \] Distance $MO$: \[ MO = \sqrt{(1 - (-1))^2 + (-1-5)^2} = \sqrt{(2)^2 + (-6)^2} = \sqrt{4+36} = \sqrt{40} = 2\sqrt{10}. \]

Step 5: Final Answer.
So the minimum value is: \[ PL+PM+PN+PO = LN+MO = 5\sqrt{2} + 2\sqrt{10} \] \[ \boxed{5\sqrt{2} + 2\sqrt{10}} \]