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find the equation of a straight line which is parallel to the line with equation 5x+7y=14 and passes through a point (-2,-3)​

User Mister Why
by
7.7k points

2 Answers

7 votes

Answer:


7y=-5x-11

Explanation:

Since it says the lines are parallel both of their gradients will be same.

so gradient of line 1 =
y=(14-5x)/(7) =
y=(14)/(7)-(5x)/(7)=
-(5)/(7)

so equation of line 2 =
y-y_(1)=m(x-x_(1) )

=
y-(-3)_{}=(-5)/(7) (x-(-2)_{} )

=
y +3=-(5)/(7)x+(10)/(7)

=
y =-(5x+10)/(7)-3

=
7y=-5x-11

User Scud
by
8.0k points
11 votes

Explanation:

basically a line equation typically looks like

y = ax + b

with a being the slope, and b bent the y-intercept (y value when x = 0).

5x + 7y = 14

7y = -5x + 14

y = -5/7 x + 14/7 = -5/7 x + 2

so, wie know the slope of the original line : -5/7 .

any line parallel to it must have the same slope.

our desired line looks like

y = -5/7 x + b

to get b we use the provided point (-2, -3).

-3 = -5/7 × -2 + b = 10/7 + b

-21/7 = 10/7 + b

-31/7 = b

and the full equation is

y = -5/7 x - 31/7

7y = -5x - 31

5x + 7y = -31

User Siegfred
by
8.1k points

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