A straight line meets the axes at A and B such that the centroid of `DeltaOAB` is (a,a). The equation of the line AB is

Options

A. `x+y=a`
B. `x-y=3a`
C. `x+y=2a`
D. `x+y=3a`

Answer

To find the equation of the line AB that meets the axes at points A and B, given that the centroid of triangle OAB is (a, a), we can follow these steps:

### Step 1: Define the Points A and B
Let point A be on the x-axis at (P, 0) and point B be on the y-axis at (0, Q).

### Step 2: Find the Centroid of Triangle OAB
The centroid (G) of triangle OAB, where O is the origin (0, 0), A is (P, 0), and B is (0, Q), is given by the formula:
\[ G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) \]
Substituting the coordinates of points O, A, and B:
\[ G = \left( \frac{0 + P + 0}{3}, \frac{0 + 0 + Q}{3} \right) = \left( \frac{P}{3}, \frac{Q}{3} \right) \]

### Step 3: Set the Centroid Equal to (a, a)
According to the problem, the centroid is given as (a, a). Therefore, we can set up the equations:
\[ \frac{P}{3} = a \quad \text{and} \quad \frac{Q}{3} = a \]
From these equations, we can solve for P and Q:
\[ P = 3a \quad \text{and} \quad Q = 3a \]

### Step 4: Write the Equation of the Line AB
The equation of the line passing through points A (P, 0) and B (0, Q) can be derived using the two-point form of the line equation:
\[ \frac{x}{P} + \frac{y}{Q} = 1 \]
Substituting the values of P and Q:
\[ \frac{x}{3a} + \frac{y}{3a} = 1 \]

### Step 5: Simplify the Equation
Multiplying through by 3a to eliminate the denominators:
\[ x + y = 3a \]

### Final Answer
Thus, the equation of the line AB is:
\[ x + y = 3a \]

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