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Home  >>  CBSE XI  >>  Math  >>  Straight Lines
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Find the equation of the line passing through the point of intersection of the lines $4x + 7y – 3 = 0$ and $2x – 3y + 1 = 0$ that has equal intercepts on the axes.

$\begin {array} {1 1} (A)\;13x-13y=6 & \quad (B)\;13x+13y=6 \\ (C)\;6x+6y=13 & \quad (D)\;6x-6y=13 \end {array}$

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  • Equation of the line having intercepts $a$ and $b$ on the axes is $ \large\frac{x}{a}$$+ \large\frac{y}{b}$$=1$
Let the equation of the line having equal intercepts on the axes is
$ \large\frac{x}{a}$$+ \large\frac{y}{a}$$=1$
(i.e.,) $x+y=a$----------(1)
Given equation of the line is
$ 4x+7y-3=0\: and \: 2x-3y+1=0$
Let us solve the above equations for $x$ and $y$.
$ \qquad 4x+7y=3$-------(2)
$ ( \times 2) 2x-3y=-1$--------(3)
$ \qquad 4x+7y=3$
$ \qquad 4x-6y=-2$
$ \qquad \qquad 13y=5 \Rightarrow y = \large\frac{5}{13}$
$4x+7 \bigg( \large\frac{5}{13} \bigg)$$ = 3$
$ \Rightarrow 4x=3-\large\frac{35}{13}$
$ \Rightarrow x = \large\frac{1}{13}$
Hence the point of intersection is $ \bigg( \large\frac{1}{13}$$, \large\frac{5}{13} \bigg)$
Since equation (1) passes through the point $ \bigg( \large\frac{1}{13}$$, \large\frac{5}{13} \bigg)$.
Substituting these values for $x$ and $y$ in equation (1) we get,
$ \large\frac{1}{13}$$+ \large\frac{5}{13}$$=a$
$ \Rightarrow \large\frac{6}{13}$$=a$ Substituting the value of $ 'a'$ in equation (1)
$x+y=\large\frac{6}{13}$
(i.e.,) $13x+13y=6$
answered May 21, 2014 by thanvigandhi_1
 

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