$\begin {array} {1 1} (A)\;2x + 3y = 12 & \quad (B)\;3x + 2y = 12 \\ (C)\;4x – 3y = 6 & \quad (D)\;5x – 2y = 10 \end {array}$

- The coordinates of the midpoint of the line joining the points $(x_1, y_1)$ and $(x_2, y_2)$ is $ \bigg( \large\frac{x_1+x_2}{2}$$, \large\frac{y_1+y_2}{2} \bigg)$
- Equation of a line in its intercept form is $ \large\frac{x}{a}$$+\large\frac{y}{b}$$=1$ , where $a$ and $b$ are the intercepts in the coordinate axes.

Step 1 :

It is given that the mid point of a line intercepted between the coordinate axes is (3, 2)

Let the equation of the line be

$ \large\frac{x}{a}$$+ \large\frac{y}{b}$$=1$

Let the points A be $(a,o)$ and B be (0, b) and $p(3, 2)$

Since $p$ is the mid point of AB

$ \large\frac{a+0}{2}$$=3 \Rightarrow a = 6$ and

$ \large\frac{b+0}{2}=2 \Rightarrow b = 4$

$ \therefore $ The equation of the line is

$ \large\frac{x}{6}$$+ \large\frac{y}{4}$$=1$

$ \Rightarrow 2x+3y=12$

Hence 'A' is the correct option.

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