Question

Find the equation of straight line whose intercepts on X-axis and Y-axis are respectively twice and thrice of those by the line 2x + 3y = 6

Answer 1

Answer:

your solution is as follows

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Step-by-step explanation:

$to \: find : \\ equation \: of \: straight \: line \\ \\ so \: let \: the \: \\ x - intercept \: and \: y - intercept \: \\ of \: required \: straight \: line \: be \: \\ a \: and \: b \: respectively \\ and \: that \: of \: line \: \: 2x + 3y = 6 \\ be \: p \: and \: q \: respectively$

$so \: comparing \\ \: 2x + 3y = 6 \: \: with \: ax + by = c \\ we \: get \\ a = 2 \\ b = 3 \\ c = - 6$

$so \: we \: know \: that \\ x - intercept = \frac{ - c}{a} \: ; \: y - intercept = \frac{ - c}{b} \\ \\ p = \frac{ - ( - 6)}{2} \: ; \: q = \frac{ - ( - 6)}{3} \: \: \\ \\ = \frac{6}{2} \: ; \: = \frac{6}{3} \\ \\ p = 3 \: ;q = 2$

$so \: here \\ a = 2p \: \: ; \: b = 3q \\ \\ = 2 \times 3 \: ; = 3 \times 2 \\ \\ ( a= 6 \: ; \: b = 6 \: ) \\ \\ we \: know \: that \\ double \: intercept \: of \: straight \: line \\ is \: given \: by \\ \\ \frac{x}{a} + \frac{y}{b} = 1 \\ \\ \frac{x}{6} + \frac{y}{6} = 1 \\ \\ \frac{6x + 6y}{36} = 1 \\ \\ 6x + 6y = 36 \\ \\ x + y = 6 \\ \\ or \\ \\ x + y - 6 = 0$