Question

find the equation of straight line making non-zero equalintersepts on the coordinate axes passing through the point of intersection of line 2x-5y+1=0 and x-3y-4=0.
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Answer 1

Answer:

your solution is as follows

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Step-by-step explanation:

$to \: find : \\ equation \: of \: straight \: line \\ \\ so \: for \: a \: certain \: straight \: line \\ let \: its \\ \: x - intercept \: and \: y - intercept \\ be \: a \: and \: b \: respectively \\ \\ according \: to \: given \: condition \\ a = b ≠0 \\ \\ so \: thus \: then \\ given \: linear \: equations \: are \\ \\ 2x - 5y = - 1 \: \: \: \: \: \: \: (1) \\ \\ x - 3y = - 4 \: \: \: \: \: \: \: \: \: (2)$

$applying \: \: now \\ (2) \times 2 \\ \\ 2x - 6y = 8 \: \: \: \: \: \: (3) \\ \\ now \: using \\ (1) - (3) \\ \\ - 5y - (6y) = - 1 - 8 \\ \\ - 5y + 6y = - 9 \\ \\ y = - 9 \\ \\ substituting \: value \: of \: \: y \: \: \\ in \: equation \: \: (2) \\ \\ we \: get \: \: \\ x = - 23$

$s o\: we \: know \: that \\ double \: intercept \: \: form \: of \: straight \\ line \: is \: given \: \: by : \\ \\ \frac{x}{a} + \frac{y}{b} = 1 \\ \\ \frac{x}{a} + \frac{y}{a} = 1 \\ \\ x + y = a \\ \\ = - 23 - 9 \\ \\ a = - 32 \\ \\ a = b = - 32$

$so \: then \\ \: again \: by \: using \\ double \: intercept \: form \: equation \\ of \: straight \: line \\ we \: have \\ \\ \frac{x}{a} + \frac{y}{b} = 1 \\ \\ \frac{x}{( - 32)} + \frac{y}{( - 32)} = 1 \\ \\ \frac{ - x - y}{32} = 1 \\ \\ - (x + y) = 32 \\ \\ x + y = - 32 \\ \\ or \\ \\ x + y + 32 = 0$

$so \: equation \: of \: required \: straight \\ line \: \: is \: here \\ \\ x + y = - 32$