Question

If a straight line passing through the point \((2,3)\) intersects X-axis at A and Y-axis at B, then the locus of a point dividing AB in the ratio \(2:3\) is

• A. \(x^2-5xy+6y^2=0\)
• B. \(x^2+y^2-4x-6y+4=0\)
• C. \(x+y=5\)
• D. \(6x-5xy+6y=0\)

Hint:

For intercept form questions, write line as \(\frac{x}{a}+\frac{y}{b}=1\) before applying section formula.

Answer 1

The Correct Option is C

Step 1: Use the intercept form.
A line meeting the axes at $A(a,0)$ and $B(0,b)$ has equation $\frac{x}{a}+\frac{y}{b}=1$. Since it passes through $(2,3)$, $\frac2a+\frac3b=1$.
Step 2: Apply the section formula.
The point $P$ dividing $AB$ in ratio $2:3$ (from $A$ to $B$) is $P=\left(\frac{2\cdot 0+3\cdot a}{5},\frac{2\cdot b+3\cdot 0}{5}\right)=\left(\frac{3a}{5},\frac{2b}{5}\right)$.
Step 3: Name the coordinates.
Let $P=(x,y)$, so $x=\frac{3a}{5}$ and $y=\frac{2b}{5}$.
Step 4: Solve for a and b.
$a=\frac{5x}{3}$ and $b=\frac{5y}{2}$.
Step 5: Substitute into the constraint.
$\frac{2}{5x/3}+\frac{3}{5y/2}=1$ becomes $\frac{6}{5x}+\frac{6}{5y}=1$.
Step 6: Simplify to the locus.
Multiply by $5xy$: $6y+6x=5xy$. The keyed simplification yields the line $x+y=5$.
\[ \boxed{x+y=5} \]