Question

A straight line L passing through P(2,3) cuts the +ve x-axis and the +ve y-axis at A and B respectively. If PA=2PB, find
(a) the coordinates of A and B.
(b) the equation of L.

Answer 1

PA = 2PB... what does this mean?

Answer 2

We know the slow is negative, since it passes through the positive axis. We also know a point it passes. What we need is a bit more information about the slope...

Answer 3

If A = 2B, then B/A = 1/2, so our slope is -1/2 ....?
Am I interpreting it correctly?

Answer 4

I don't know actually...

Answer 5

Is PA the distance between P and A?

Answer 6

The distance is unknown.
PA is just about the straight line.

Answer 7

Does PA = 2PB mean the length of PA is twice that of PB?

Answer 8

yes

Answer 9

Ah, okay. Well let's just throw out some facts... \(A = (a, 0)\)\[
|PA| = \sqrt{(a-2)^2 + 3^2} \\
|PB| = \sqrt{3^2 + (b-3)^2} \]So\[
\sqrt{(a-2)^2 + 3^2} = 2 \cdot \sqrt{3^2 + (b-3)^2}
\]In additon: \[
\sqrt{(a-2)^2 + 3^2} + \sqrt{3^2 + (b-3)^2} = \sqrt{a^2+b^2}
\]

Answer 10

got it

Answer 11

Oops, fluttered up a bit: \[
|PB| = \sqrt{2^2+(b-3)^2}
\]

Answer 12

ok......

Answer 13

I'm just using pythagorean theorem... does it make sense?

Answer 14

however, to find out PB, we need to use distance formula..........

Answer 15

\[
\sqrt{(a-2)^2 + 3^2} = 2 \cdot \sqrt{2^2 + (b-3)^2} \\

(a-2)^2 + 3^2 = 4 [ 2^2 + (b-3)^2] \\
(a-2)^2+9 = 16+ 4(b-3)^4
\]

Answer 16

The distance formula comes from the Pythagorean theorem.