If a variable line drawn through the intersection of the lines $\frac{x}{3} + \frac{y}{4} = 1$ and $\frac{x}{4} + \frac{y}{3} = 1$ , meets the coordinate axes at $A$ and $B$ , $(A \neq B)$ , then the locus of the midpoint of $AB$ is :

$L_{1} : 4x+3y-12=0$ $L_{2} : 3x+4y-12=0$ $L_{1}+\lambda L_{2}=0$ $\left(4x+3y-12\right)+\lambda\left(3x+4y-12\right)=0$ $x\left(4+3\lambda-12\right)+y\left(3+4\lambda\right)-12\left(1+\lambda\right)=0$ Point $A\left(\frac{12\left(1+\lambda\right)}{4+3\lambda},0\right)$ Point $B\left(0, \frac{12\left(1+\lambda\right)}{3+4\lambda}\right)$ mid point $\Rightarrow h=\frac{6\left(1+\lambda\right)}{4+3\lambda}......\left(i\right)$ $k=\frac{6\left(1+\lambda\right)}{3+4\lambda} ......\left(ii\right)$ Eliminate ? from (i) and (ii) then $6\left(h+k\right)=>hk$ $6\left(x+y\right)=>xy$

If a variable line drawn through the intersection

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Straight lines

A straight line is a line having the shortest distance between two points.

A straight line can be represented as an equation in various forms, as show in the image below:

The following are the many forms of the equation of the line that are presented in straight line-

1. Slope – Point Form

Assume P0(x0, y0) is a fixed point on a non-vertical line L with m as its slope. If P (x, y) is an arbitrary point on L, then the point (x, y) lies on the line with slope m through the fixed point (x0, y0) if and only if its coordinates fulfil the equation below.

y – y₀ = m (x – x₀)

2. Two – Point Form

Let's look at the line. L crosses between two places. P₁(x₁, y₁) and P₂(x₂, y₂) are general points on L, while P (x, y) is a general point on L. As a result, the three points P₁, P₂, and P are collinear, and it becomes

The slope of P₂P = The slope of P₁P₂, i.e.

$\frac{y-y_1}{x-x_1} = \frac{y_2-y_1}{x_2-x_1}$

Hence, the equation becomes:

y - y₁ =$ \frac{y_2-y_1}{x_2-x_1} (x-x1)$

3. Slope-Intercept Form

Assume that a line L with slope m intersects the y-axis at a distance c from the origin, and that the distance c is referred to as the line L's y-intercept. As a result, the coordinates of the spot on the y-axis where the line intersects are (0, c). As a result, the slope of the line L is m, and it passes through a fixed point (0, c). The equation of the line L thus obtained from the slope – point form is given by

y – c =m( x - 0 )

As a result, the point (x, y) on the line with slope m and y-intercept c lies on the line, if and only if

y = m x +c

If a variable line drawn through the intersection of the lines $\frac{x}{3} + \frac{y}{4} = 1$ and $\frac{x}{4} + \frac{y}{3} = 1$, meets the coordinate axes at $A$ and $B$, $(A \neq B)$, then the locus of the midpoint of $AB$ is :

The Correct Option is C

Solution and Explanation

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