Question

find the equation of straight line parallel to line joining the points (7,5) (1,3) and passing through the point ( -3,4)

Answer 1

$\bold{Answer:}$

$\bold{x - 3y + 15 = 0}$

$\bold{Step} - \bold{by} - \bold{step\ explanation:}$

$First find the slope of the line passing through points (7, 5) and (1, 3). \\ \\ Slope \ = \frac{5 - 3}{7 - 1} = \frac{2}{6} = \frac{1}{3} \\ \\ As both lines are parallel, both have same slope. \\ \\ Let \ (x, y)\ be a point on the line passing through the point (-3, 4). \\ \\ The slope of this line is also \ \frac{1}{3}.$

$\frac{y - 4}{x - (-3)} = \frac{1}{3} \\ \\ \frac{y - 4}{x + 3} = \frac{1}{3} \\ \\ 3(y - 4) = 1(x + 3) \\ \\ 3y - 12 = x + 3 \\ \\ x + 3 - 3y + 12 = 0 \\ \\ \bold{x - 3y + 15 = 0} \\ \\ \\ \therefore\ \bold{x - 3y + 15 = 0}\ is the equation.$

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Answer 2

slope of line = ( 5-3)/ (7-1) = 2/6 = 1/3
( As parallel line is passing through (7,5)(1,3) so its slope would be equal to required line)

let equation of line be y = mx + c

put m= 1/3

y = x/3 + c

as its passing through (-3,4)

4 = -1 + c

c= 5

So eq is y= x/3 + 5

y = x + 15)/3

3y -x= 15