First slide

Equation of line in Straight lines

Question

Suppose that $A (1, 2)$ and $B (3, 4)$ be two points and P be a point on y=x such that $|PA + PB|$ is minimum then the point P

Difficult

A

$(- 15, 5)$
B

$(- 10, 0)$
C

$(\frac{5}{2}, \frac{5}{2})$
D

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

Solution

Suppose that the images of $A (1, 2)$ and $B (3, 4)$ in the line x=y are C and D respectively.
The point P is the point of intersection of the lines AD and x=y , it is coincide with the point of intersection of the lines BC and x=y
The image of $B (3, 4)$ in the line y=x is $D (4, 3)$
The equation of the line joining $D (4, 3)$ an $A (1, 2)$ is $y - 2 = \frac{2 - 3}{1 - 4} (x - 1)$
This can be simplified as
$\begin{matrix} y - 2 = \frac{1}{3} (x - 1) \\ 3 y - 6 = x - 1 \\ x - 3 y + 5 = 0 \end{matrix}$
Plug in x=y to get the point of intersection of the above line and the line x=y.
It implies that
$\begin{matrix} x - 3 x + 5 = 0 \\ 2 x = 5 \\ x = \frac{5}{2} \end{matrix}$
Therefore, $P = (\frac{5}{2}, \frac{5}{2})$

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