Question

A point P divides the line segment joining the points A(-3, 5) and B(7, -4) in a certain ratio. If the point P lies on the line y = 2x, then find the ratio AP : PB and coordinates of point P.

Hint:

Using \(k : 1\) as the ratio is always easier than using \(m_1 : m_2\) because it reduces the problem to solving a single-variable equation.

Answer 1

Step 1: Find the equation of line $AB$ first, instead of parametrising $P$ with a ratio straight away.
The slope of line $AB$ through $A(-3,5)$ and $B(7,-4)$ is:
\[ m = \frac{-4-5}{7-(-3)} = \frac{-9}{10} \]
Using point $A$, the equation of line $AB$ is:
\[ y - 5 = -\frac{9}{10}(x+3) \]
\[ y = 5 - \frac{9}{10}x - \frac{27}{10} = \frac{23}{10} - \frac{9}{10}x \]

Step 2: Intersect line $AB$ with the given line $y=2x$.
Since $P$ lies on both lines, set the two expressions for $y$ equal:
\[ 2x = \frac{23}{10} - \frac{9}{10}x \]
\[ 2x + \frac{9}{10}x = \frac{23}{10} \]
\[ \frac{29}{10}x = \frac{23}{10} \]
\[ x = \frac{23}{29} \]
Then:
\[ y = 2x = \frac{46}{29} \]
So $P = \left(\dfrac{23}{29}, \dfrac{46}{29}\right)$.

Step 3: Work backward to find the ratio $AP:PB$ from the coordinates of $P$.
If $P$ divides $AB$ in the ratio $k:1$, then by the section formula, $x = \dfrac{x-x_1}{x_2-x}$ gives $k$ once rearranged:
\[ k = \frac{x - x_1}{x_2 - x} = \frac{\frac{23}{29} - (-3)}{7 - \frac{23}{29}} = \frac{\frac{23}{29}+\frac{87}{29}}{\frac{203}{29}-\frac{23}{29}} = \frac{\frac{110}{29}}{\frac{180}{29}} = \frac{110}{180} = \frac{11}{18} \]

Final Answer:
The ratio $AP:PB$ is $11:18$, and the coordinates of $P$ are $\left(\dfrac{23}{29}, \dfrac{46}{29}\right)$.
\[ \boxed{AP:PB = 11:18,\quad P = \left(\tfrac{23}{29}, \tfrac{46}{29}\right)} \]