JEE MAIN - Mathematics (2004 - No. 41)
The equation of the straight line passing through the point $$(4, 3)$$ and making intercepts on the co-ordinate axes whose sum is $$-1$$ is :
$${x \over 2} - {y \over 3} = 1$$ and $${x \over -2} +{y \over 1} = 1$$
$${x \over 2} - {y \over 3} = -1$$ and $${x \over -2} +{y \over 1} = -1$$
$${x \over 2} + {y \over 3} = 1$$ and $${x \over 2} +{y \over 1} = 1$$
$${x \over 2} + {y \over 3} = -1$$ and $${x \over -2} +{y \over 1} = -1$$
Explanation
Let the required line be $${x \over a} + {y \over b} = 1.......\left( 1 \right)$$
then $$a+b=-1$$ $$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,.........\left( 2 \right)$$
$$(1)$$ passes through $$(4,3), $$ $$ \Rightarrow {4 \over a} + {3 \over b} = 1$$
$$ \Rightarrow 4b + 3a = ab\,\,...............\left( 3 \right)$$
Eliminating $$b$$ from $$(2)$$ and $$(3),$$ we get
$${a^2} - 4 = 0 \Rightarrow a = \pm 2 \Rightarrow b = - 3$$, $$1$$
$$\therefore$$ Equation of straight lines are
$${x \over 2} + {y \over { - 3}} = 1$$
or $${x \over { - 2}} + {y \over 1} = 1$$
then $$a+b=-1$$ $$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,.........\left( 2 \right)$$
$$(1)$$ passes through $$(4,3), $$ $$ \Rightarrow {4 \over a} + {3 \over b} = 1$$
$$ \Rightarrow 4b + 3a = ab\,\,...............\left( 3 \right)$$
Eliminating $$b$$ from $$(2)$$ and $$(3),$$ we get
$${a^2} - 4 = 0 \Rightarrow a = \pm 2 \Rightarrow b = - 3$$, $$1$$
$$\therefore$$ Equation of straight lines are
$${x \over 2} + {y \over { - 3}} = 1$$
or $${x \over { - 2}} + {y \over 1} = 1$$
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