Question

Find the equation of a straight line which is parallel to the line 3x −7y = 12 and passing through the point (6,4). *


a.)3x −7y +10 = 0 .

b.)3x −7y -10 = 0

c.)3x +7y +10 = 0 .

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Answer 1

EXPLANATION.

→ equation of straight line which is parallel

to the line = 3x - 7y = 12

→ passing through the point ( 6,4)

→ slope of the parallel lines = -a/b

-3/-7 = 3/7

→ equation of the straight line

→ ( y - y¹ ) = m ( x - x¹ )

→ equation passing through the point ( 6,4)

→ ( y - 4 ) = 3/7 ( x - 6 )

→ 7 ( y - 4 ) = 3 ( x - 6 )

→ 7y - 28 = 3x - 18

→ 3x - 7y + 10 = 0 → equation of straight line.

→ More information.

$\sf : \implies \: (1) = distance \: formula \\ \\ \sf : \implies \: \sqrt{( x_{1} - x_{2}) {}^{2} + ( y_{1} - y_{2}) {}^{2} } \\ \\ \sf : \implies \: (2) = section \: formula \\ \\ \sf : \implies \: x \: = \frac{m x_{2} \: \pm \: n x_{1} }{m \: \pm n} \: \: \: \: and \: \: \: \: y = \frac{m {y}^{} _{2} \: \pm \: n y_{1} }{m \: \pm \: n}$

$\sf : \implies \:(3) = slope \: formula \\ \\ \sf : \implies \: line \: joining \: two \: point \: ( x_{1}, y_{1}) \: and \: ( x_{2}, y_{2}) \\ \\ \sf : \implies \: \frac{ y_{2} - y_{1} }{ x_{2} - x_{1} }$

Answer 2

Step-by-step explanation:

Equation of straight line which is parallel

to the line 3x - 7y = 12

Passing through point ( 6 , 4 )

==> 3x - 7y = 12

==> 3x - 12 = 7y

==> 7y = 3x - 12

$==>y = \frac{3x}{7} - \frac{12}{7}$

Compare with y = mx + c

Slope = m

$Therefore \: the \: slope \: is \: \frac{3}{7} , so \: a \: line \: \\ parallel \: to \: this \: have \: slope \: of \: \frac{3}{7} \: also \: .$

Now since the equation of a line is

==> y - y1 = m ( x - x1 )

where m is the slope of the line, therefore with point ( 6 , 4 ) , we have ,

$\implies \: (y - 4) = \frac{3}{7} (x - 6) \\ \implies \: 7(y - 4) = 3(x - 6) \\ \implies \: 7y - 28 = 3x - 18 \\ \implies \: 3x - 18 = 7y - 28 \\ \implies \: 3x - 7y - 18 + 28 = 0 \\ \implies \: 3x - 7y + 10 = 0$

Hence , Option a is correct .