Question

A straight line passing through a point (3,2) cuts X and Y-axes at the points A and B respectively. If a point P divides AB in the ratio 2:3, then the equation of the locus of point P is

• A. $\frac{9}{x}+\frac{4}{y}=1$
• B. $9x+4y=5xy$
• C. $4x+9y=5xy$
• D. $\frac{4}{x}+\frac{9}{y}=1$

Hint:

When dealing with locus problems involving lines cutting axes, the intercept form of the line ($\frac{x}{a} + \frac{y}{b} = 1$) is often the most convenient starting point. Use the section formula to relate the coordinates of the moving point to the intercepts.

Answer 1

The Correct Option is C

Let the line cut the X-axis at A(a, 0) and the Y-axis at B(0, b).
The equation of the line in the intercept form is $\frac{x}{a} + \frac{y}{b} = 1$.
Since this line passes through the point (3, 2), we have:
$\frac{3}{a} + \frac{2}{b} = 1$. (Eq. 1)
Let the coordinates of the point P be $(h, k)$.
P divides the line segment AB in the ratio 2:3. Using the section formula:
$h = \frac{2(0) + 3(a)}{2+3} = \frac{3a}{5} \implies a = \frac{5h}{3}$.
$k = \frac{2(b) + 3(0)}{2+3} = \frac{2b}{5} \implies b = \frac{5k}{2}$.
Now substitute these expressions for $a$ and $b$ into Eq. 1:
$\frac{3}{(5h/3)} + \frac{2}{(5k/2)} = 1$.
$\frac{9}{5h} + \frac{4}{5k} = 1$.
To clear the denominators, multiply the entire equation by $5hk$:
$9k + 4h = 5hk$.
The locus of P(h, k) is obtained by replacing $h$ with $x$ and $k$ with $y$.
$4x + 9y = 5xy$.