If a variable line drawn through the intersection of the lines [x 3 + y 4 = 1] and [x 4 + …

Question

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

Accepted Answer

L1 : 4x + 3y - 12 = 0

L2 : 3x + 4y - 12 = 0

Equation of line passing through  the intersection of these two lines L1 and L2 is

L1 + $$\lambda $$L2 = 0

$$ \Rightarrow $$ $\,\,\,$(4x + 3y$-$12) +$\lambda $(3x + 4y$-$12) = 0$ \Rightarrow $$ $\,\,\,$$ x(4 + 3$$\lambda $$) + y(3 + 4$$\lambda $$) - 12(1 + $$\lambda $$) = 0

this line meets x coordinate at point A and y coordinate at point B.

$$\therefore\,\,\,$$ Point A = $$\left( {{{12\left( {1 + \lambda } \right)} \over {4 + 3\lambda }},0} \right)$$

and Point B = $$\left( {0,\,\,{{12\left( {1 + \lambda } \right)} \over {3 + 4\lambda }}} \right)$$

Let coordinate of midpoint of line AB is  (h, k).

$$\therefore\,\,\,$$ h = $${{6\left( {1 + \lambda } \right)} \over {4 + 3\lambda }}$$ . . . . . (1)

and k = $${{6\left( {1 + \lambda } \right)} \over {3 + 4\lambda }}$$ . . . . (2)

Eliminate $$\lambda $$ from (1) and (2), then  we get

6(h + k) = 7 hk

$$\therefore\,\,\,$$ Locus of midpoint of line AB is ,

6(x + y) = 7xy

Solutions · 1