Question

Let M be the mid-point of the side AB of the equilateral triangle ABC having side length 2 units. P is a mid-point on BC such that 'AP+PM' is minimum. If the value of 'AP+PM' is $\sqrt{b}$ then the value of b-7 is equal to _____

Answer 1

0

Answer 2

To find the minimum value of AP + PM, where P is a point on the line segment BC, we use the reflection principle.

1. Reflection Principle: Reflect point M across the line BC to get M'. The sum AP + PM will be minimum when A, P, and M' are collinear, and the minimum value will be the straight-line distance AM'.

2. Set up Coordinate System: Let C be the origin (0,0). Since the triangle ABC is equilateral with side length 2 units, the coordinates of B will be (2,0). The coordinates of A can be found using trigonometry: A = (side length * cos(60°), side length * sin(60°)) A = (2 * 1/2, 2 * $\sqrt{3}/2$) A = (1, $\sqrt{3}$)

3. Find Coordinates of M: M is the mid-point of side AB. A = (1, $\sqrt{3}$) B = (2, 0) M = $\left(\frac{1+2}{2}, \frac{\sqrt{3}+0}{2}\right)$ M = $\left(\frac{3}{2}, \frac{\sqrt{3}}{2}\right)$

4. Find Coordinates of M' (Reflection of M across BC): The line BC lies on the x-axis (y=0). When a point (x,y) is reflected across the x-axis, its reflection is (x,-y). So, M' = $\left(\frac{3}{2}, -\frac{\sqrt{3}}{2}\right)$

5. Calculate the Minimum Distance AM': The minimum value of AP + PM is the distance between A(1, $\sqrt{3}$) and M'($\frac{3}{2}$, $-\frac{\sqrt{3}}{2}$). Using the distance formula: $AM' = \sqrt{\left(\frac{3}{2} - 1\right)^2 + \left(-\frac{\sqrt{3}}{2} - \sqrt{3}\right)^2}$ $AM' = \sqrt{\left(\frac{1}{2}\right)^2 + \left(-\frac{3\sqrt{3}}{2}\right)^2}$ $AM' = \sqrt{\frac{1}{4} + \frac{9 \times 3}{4}}$ $AM' = \sqrt{\frac{1}{4} + \frac{27}{4}}$ $AM' = \sqrt{\frac{28}{4}}$ $AM' = \sqrt{7}$

6. Determine the value of b and then b-7: We are given that the minimum value of AP + PM is $\sqrt{b}$. So, $\sqrt{b} = \sqrt{7}$. This implies b = 7.

   The value of b-7 is $7 - 7 = 0$.