Question:

Find the equation of the line passing through the point of intersection of the lines 4x+7y3=04x + 7y - 3 = 0 and 2x3y+1=02x - 3y+ 1 = 0  that has equal intercepts on the axes.

Updated On: Oct 21, 2023
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Solution and Explanation

Let the equation of the line having equal intercepts on the axes be  xa+ya=1\frac{x}{a} +\frac{ y}{a} = 1
x+y=a..(1)x + y = a ….. (1)
On solving equations 4x+7y 3=04x + 7y - 3 = 0 and 2x 3y+1=02x - 3y + 1 = 0, we obtain

x=113x =\frac{ 1}{13} and y=513y =\frac{ 5}{13}

(113,513)∴ (\frac{1}{13}, \frac{5}{13})  is the point of intersection of two given lines.
Since equation (1) passes through point (113,513)(\frac{1}{13},\frac{ 5}{13})
113+513=a\frac{1}{13} +\frac{ 5}{13} = a

a=613a = \frac{6}{13}

∴ Equation (1) becomes 
x+y=613x + y =\frac{ 6}{13}
13x+13y=613x + 13y = 6

Thus, the required equation of the line is 13x+13y=6.13x + 13y = 6.

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