find the equation of the line through the intersection of the lines 2x+3y-4 = 0 and x-5y = 7 that has its intercept equal to -4.

Here first we have to find the intersection point of these two lines.

So on solving both equation simultaneously we have x = 41/13 and y = -10/13

Let the equation of straight line in intercept form is x/a + y/b = 1

Here a is x-intercept and b = y-intercept and the point x = 41/13 and y = -10/13 will satisfy this line

As x-intercept is a = -4

$\frac{x}{-4}+\frac{y}{b}=1\phantom{\rule{0ex}{0ex}}Soputingthepointisthisequation:\phantom{\rule{0ex}{0ex}}\frac{41}{13\times -4}+\frac{-10}{13b}=1\phantom{\rule{0ex}{0ex}}Sob=\frac{-40}{93}\phantom{\rule{0ex}{0ex}}Sothegeneralequationwillbe\phantom{\rule{0ex}{0ex}}\frac{x}{-4}+\frac{y}{{\displaystyle \frac{-40}{93}}}=1\phantom{\rule{0ex}{0ex}}Or\frac{x}{-4}+\frac{-93y}{40}=1\left(ans\right)$

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