First slide

Equation of line in Straight lines

Question

Suppose that $A (5, 6)$ and $B (3, 4)$ be two points and P be a point on x- axis such that $|PA + PB|$ is minimum then the point P is

Difficult

A

$(- 15, 5)$
B

$(- 10, 0)$
C

$(- 10, 10)$
D

$(- \frac{19}{5}, 0)$

Solution

Suppose that the images of $A (5, 6)$ and $B (3, 4)$ in the X- axis are C and D respectively.
The point P is the point of intersection of the lines AD and x- axis, it is coincide with the point of intersection of the lines BC and x- axis
The image of $B (3, 4)$ in the x- axis is $D (3, - 4)$
The equation of the line joining $D (3, - 4)$ an $A (5, 6)$ is $y - 6 = \frac{6 + 4}{5 - 3} (x - 5)$
This can be simplified as
$\begin{matrix} y - 6 = \frac{6 + 4}{5 - 3} (x - 5) \\ y - 6 = 5 (x - 5) \\ y - 6 = 5 x - 25 \\ 5 x - y - 19 = 0 \end{matrix}$
Plug in x=0 to get the point of intersection of the above line and axis.
It implies that
$\begin{matrix} 5 x = 19 \\ x = \frac{19}{5} \end{matrix}$
Therefore, $P = (\frac{19}{5}, 0)$

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