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Home  >>  CBSE XI  >>  Math  >>  Straight Lines
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Find the equation of the lines which passes through the point (3, 4) and cuts off intercepts from the coordinate axes such that their sum is 14.

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  • Equation of a line which has $a$ and $b$ as the $x$ and $y$ intercepts is $ \large\frac{x}{a}$$+\large\frac{y}{b}$$=1$
Step 1
Equation of the line in the intercept form is
$ \large\frac{x}{a}$$+\large\frac{y}{b}$$=1$-----(1)
It is given that the sum of the intercepts is 14.
(i.e) $a+b=14$
or $ b = 14-a$
Substituting this in equivalent (1) we get.
(i.e) $x(14-a)+y(a)=a(14-a)$
Step 2
This line passes through the point (3, 4)
$ \therefore 3(14-a)+4(a)=a(14-a)$
$ \Rightarrow 42-3a+4a=14a-a^2$
(i.e) $a^2-13a+42=0$
On factorizing this we get,
$(a-6) (a-7)=0$
$ \therefore a = 6 $ or $ a = 7$
Step 3
Case (i)
If $ a = 6$ then $ b = 8$
Hence the equation of the line is
$ \large\frac{x}{6}$$+\large\frac{y}{8}$$=1$
(i.e) $8x+6y=48$
or $4x+3y=24$
Step 4
Case (ii)
If $ a=7$ then $b = 7$
Hence the equation of the line is
$ \large\frac{x}{7}$$+\large\frac{y}{7}$$=1$
$ \Rightarrow x+y=7$
answered Jun 30, 2014 by thanvigandhi_1

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