Question

the equation of the line joining the point (3, 5) to the point of intersection of the lines
4x + y - 1 = 0 and 7x - 3y - 35 = 0 ​

Answer 1

Answer:

your solution is as follows

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Step-by-step explanation:

$to \: find : \\ equation \: of \: straight \: line \\ \\ so \: here \\ given \: simultanous \: equations \: are \\ \\ 4x + y = 1 \: \: \: \: \: \: \: \: \: \: \: \: \: \: (1) \\ 7x - 3y = 35 \: \: \: \: \: \: \: \: \: (2) \\ \\ applying \: \: now \\ (1) \times 3 \\ \\ 12x + 3y = 3 \: \: \: \: \: \: \: \: \: \: \: (3) \\ \\ applying \: then \\ (2) + (3) \\ \\ 19x = 38 \\ \\ x = 2 \\ \\ substituting \: value \: of \: \: x \: in \: (1) \\ we \: get \\ \\ y = - 7$

$so \: let \: then \\ (x1,y1) = (3,5) \\ (x2,y2) = (2, - 7) \\ \\ we \: know \: that \\ two \: points \: form \: \: equation \: of \: any \\ straight \: line \: is \\ given \: by : \\ \\ \frac{x - x1}{x1 - x2} = \frac{y - y1}{y1 - y2} \\ \\ \frac{x - 3}{3 - 2} = \frac{y - 5}{5 - ( - 7)} \\ \\ \frac{x - 3}{1} = \frac{y - 5}{5 + 7} \\ \\ x - 3 = \frac{y - 5}{12} \\ \\ 12(x - 3) = y - 5 \\ \\ 12x - 36 = y - 5 \\ \\ 12x - y = 31 \\ \\ or \\ \\ 12x - y - 31 = 0$

$hence \: \: equation \: of \: the \: line \: \\ joining \: the \: point \: (3, 5) \: to \: the \: point \: of \: \\ intersection \: of \: the \: lines \: \: \\ </p><p>4x + y - 1 = 0 \: \: \\ and \: \: \: \: 7x - 3y - 35 = 0 \: \: is \\ \: \: given \: as \\ \\ 12 - y = 31$