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Question

(A) The required line passes through the point of intersection of the lines $3x - y = 5$ and $x + 3y = 1$. Solving these equations: $3(3x - y) = 3(5)

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$5x + 5y - 7 = 0$

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(A) The required line passes through the point of intersection of the lines $3x - y = 5$ and $x + 3y = 1$. Solving these equations: $3(3x - y) = 3(5) \Rightarrow 9x - 3y = 15$ Adding $x + 3y = 1$ to $9x - 3y = 15$,we get $10x = 16 \Rightarrow x = \frac{8}{5}$. Substituting $x = \frac{8}{5}$ in $3x - y = 5$: $3(\frac{8}{5}) - y = 5 \Rightarrow y = \frac{24}{5} - 5 = -\frac{1}{5}$. So,the point of intersection is $(\frac{8}{5}, -\frac{1}{5})$. The equation of a line with equal intercepts is $x + y = a$ or $x - y = a$. Since the line passes through $(\frac{8}{5}, -\frac{1}{5})$,for $x + y = a$: $\frac{8}{5} - \frac{1}{5} = a \Rightarrow a = \frac{7}{5}$. Thus,$x + y = \frac{7}{5} \Rightarrow 5x + 5y - 7 = 0$.

The equation of the line passing through the point of intersection of the lines $3x - y = 5$ and $x + 3y = 1$ and making equal intercepts on the axes is

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