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.

Asked by Abhisek | 1 year ago |  99

##### Solution :-

Consider the equation of the line having equal intercepts on the axes as

$$\dfrac{x}{a}+\dfrac{y}{a}=1$$

It can be written as

x + y = a ….. (1)

By solving equations 4x + 7y – 3 = 0 and 2x – 3y + 1 = 0 we get

x = $$\dfrac{1}{13}$$ and y = $$\dfrac{5}{13}$$

($$\dfrac{1}{13}$$, $$\dfrac{5}{13}$$) is the point of intersection of two given lines

We know that equation (1) passes through point ($$\dfrac{1}{13}$$,$$\dfrac{5}{13}$$)

$$\dfrac{1}{13}$$ +$$\dfrac{5}{13}$$ = a

a =$$\dfrac{6}{13}$$

So the equation (1) passes through ($$\dfrac{1}{13}$$, $$\dfrac{5}{13}$$)

$$\dfrac{1}{13}$$ +$$\dfrac{5}{13}$$ = a

We get

a = $$\dfrac{6}{13}$$

Here the equation (1) becomes

x + y = $$\dfrac{6}{13}$$

13x + 13y = 6

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

Answered by Pragya Singh | 1 year ago

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