JEE MAIN - Mathematics (2016 - 9th April Morning Slot - No. 19)
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 :
6xy = 7(x + y)
4(x + y)2 − 28(x + y) + 49 = 0
7xy = 6(x + y)
14(x + y)2 − 97(x + y) + 168 = 0
Explanation
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
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
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