Question

find the equations of line which contain the points (4,1) & X intercept is twice its Y intercept

Answer 1

hey here is your solution

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$so \: here \\ for \: a \: certain \: line \\ given \: points \: are \: (4,1) \\ so \: let \: then \\ (x1,y1) = (4,1) \\ \\ moreover \: let \: \\ its \: \: x - intercept \: be \: a \: \: and \\ \: its \: \: y - intercept \: be \: b \\ \\ according \: to \: given \: condition \\ a = 2b \: \: \: \: \: (1) \\ \\ so \: we \: know \: that \\ double \: intercept \: form \: of \: straight \: line \\ is \: given \: by \\ \\ \frac{x}{a} + \frac{y}{b} = 1 \\ \\ \frac{4}{2b} + \frac{1}{b} = 1 \\ \\ \frac{2}{b} + \frac{1}{b} = 1 \\ \\ \frac{2 + 1}{b} = 1 \\ \\ b = 3 \\ substitute \: value \: of \: b \: in \: (1) \\ we \: get \\ a = 6$

$so \: again \: applying \\ \\ \frac{x}{a} + \frac{y}{b} = 1 \\ \\ \frac{x}{6} + \frac{y}{3} = 1 \\ \\ \frac{3x + 6y}{(6 \times 3)} = 1 \\ \\ \frac{3x + 6y}{18} = 1 \\ \\ 3x + 6y = 18 \\ 3(x + 2y) = 18 \\ \\ x + 2y = 6 \\ or \\ x - 2y - 6 = 0 \\ \\ hence \: the \: equation \: of \: line \: which \: contain \: the \: \\ points \: (4,1) \: and \: \: X - intercept \\ is \: twice \: its \: Y - intercept \: is \\ x + 2y = 6$