Question:
Let a variable line $L$ meet $x$-axis and $y$-axis at points A and B, respectively. Suppose the distance of the line $L$ from the origin is 3 units. Then the equation of the locus of the point C that divides the line segment AB internally in the ratio $2 : 1$ is
Let a variable line $L$ meet $x$-axis and $y$-axis at points A and B, respectively. Suppose the distance of the line $L$ from the origin is 3 units. Then the equation of the locus of the point C that divides the line segment AB internally in the ratio $2 : 1$ is
Show Hint
Be careful with the order of points in the section formula.
Since the ratio is $2 : 1$ from A to B, the coordinate $h$ is associated with $a$ (since $x_B = 0$), and $k$ is associated with $b$ (since $y_A = 0$).
Since the ratio is $2 : 1$ from A to B, the coordinate $h$ is associated with $a$ (since $x_B = 0$), and $k$ is associated with $b$ (since $y_A = 0$).
Updated On: Jun 16, 2026
- $\frac{4}{x^2} + \frac{1}{y^2} = 1$
- $\frac{1}{x^2} + \frac{4}{y^2} = 1$
- $x^2 + 4y^2 = 9$
- $4x^2 + y^2 = 9$
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Question:
A line intersects both axes forming a segment AB. Its distance from the origin is constant.
We need to find the equation representing the locus of a point C dividing this segment in a $2 : 1$ ratio.
Step 2: Key Formula or Approach:
- The intercept form of a line is:
\[ \frac{x}{a} + \frac{y}{b} = 1 \]
- The perpendicular distance from the origin $(0,0)$ is:
\[ d = \frac{1}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}} \]
- The internal section formula for dividing segment joining $(x_1, y_1)$ and $(x_2, y_2)$ in ratio $m : n$ is:
\[ \left( \frac{mx_2 + nx_1}{m + n}, \frac{my_2 + ny_1}{m + n} \right) \]
Step 3: Detailed Explanation:
• Let the coordinates of A and B be $(a, 0)$ and $(0, b)$, respectively.
• The perpendicular distance of the line $\frac{x}{a} + \frac{y}{b} = 1$ from the origin is 3 units:
\[ \frac{1}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}} = 3 \implies \frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{9} \]
• Let $C(h, k)$ be the point that divides the segment AB internally in the ratio $2 : 1$.
Using the section formula with $m = 2$, $n = 1$:
\[ h = \frac{2(0) + 1(a)}{2 + 1} = \frac{a}{3} \implies a = 3h \]
\[ k = \frac{2(b) + 1(0)}{2 + 1} = \frac{2b}{3} \implies b = \frac{3k}{2} \]
• Now substitute $a$ and $b$ into the distance relation:
\[ \frac{1}{(3h)^2} + \frac{1}{\left(\frac{3k}{2}\right)^2} = \frac{1}{9} \]
\[ \frac{1}{9h^2} + \frac{4}{9k^2} = \frac{1}{9} \]
• Multiplying both sides of the equation by 9:
\[ \frac{1}{h^2} + \frac{4}{k^2} = 1 \]
• Replacing $(h, k)$ with general coordinates $(x, y)$ gives the locus of point C:
\[ \frac{1}{x^2} + \frac{4}{y^2} = 1 \]
Step 4: Final Answer:
The equation of the locus of point C is $\frac{1}{x^2} + \frac{4}{y^2} = 1$.
A line intersects both axes forming a segment AB. Its distance from the origin is constant.
We need to find the equation representing the locus of a point C dividing this segment in a $2 : 1$ ratio.
Step 2: Key Formula or Approach:
- The intercept form of a line is:
\[ \frac{x}{a} + \frac{y}{b} = 1 \]
- The perpendicular distance from the origin $(0,0)$ is:
\[ d = \frac{1}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}} \]
- The internal section formula for dividing segment joining $(x_1, y_1)$ and $(x_2, y_2)$ in ratio $m : n$ is:
\[ \left( \frac{mx_2 + nx_1}{m + n}, \frac{my_2 + ny_1}{m + n} \right) \]
Step 3: Detailed Explanation:
• Let the coordinates of A and B be $(a, 0)$ and $(0, b)$, respectively.
• The perpendicular distance of the line $\frac{x}{a} + \frac{y}{b} = 1$ from the origin is 3 units:
\[ \frac{1}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}} = 3 \implies \frac{1}{a^2} + \frac{1}{b^2} = \frac{1}{9} \]
• Let $C(h, k)$ be the point that divides the segment AB internally in the ratio $2 : 1$.
Using the section formula with $m = 2$, $n = 1$:
\[ h = \frac{2(0) + 1(a)}{2 + 1} = \frac{a}{3} \implies a = 3h \]
\[ k = \frac{2(b) + 1(0)}{2 + 1} = \frac{2b}{3} \implies b = \frac{3k}{2} \]
• Now substitute $a$ and $b$ into the distance relation:
\[ \frac{1}{(3h)^2} + \frac{1}{\left(\frac{3k}{2}\right)^2} = \frac{1}{9} \]
\[ \frac{1}{9h^2} + \frac{4}{9k^2} = \frac{1}{9} \]
• Multiplying both sides of the equation by 9:
\[ \frac{1}{h^2} + \frac{4}{k^2} = 1 \]
• Replacing $(h, k)$ with general coordinates $(x, y)$ gives the locus of point C:
\[ \frac{1}{x^2} + \frac{4}{y^2} = 1 \]
Step 4: Final Answer:
The equation of the locus of point C is $\frac{1}{x^2} + \frac{4}{y^2} = 1$.
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