What is the equation for a line that passes through (5,-1) and is parallel to the line 2x+5y=15?

Answer 1

2x+5y = 15
subtract 2x both sides
5y = 15-2x
divide 5 both sides
y= 3 -(2/5)x

slope is -2/5

cuz parallel then new line will have the same slope

standart form y= mx+ b, m is slope, b= y-intercept
we have m= -2/5 , x= 5, y = -1 above
-1= (-2/5)*5 +b
-1 = -2 +b
add 2 both sides
1=b
therefore new line is
y= (-2/5) x +1

or y-1 = (-2/5)x

Answer 2

Parallel = the same slop ..!
then find the slop of the given line
d/dx[2x+5y=15]
2+5dy/dx=0 => dy/dx=-2/5 (the slop)
The general equation of any line is [ y=ax+b] where a is the slop and b is the intersection point.
then a=dy/dx=-2/5
then we have y=-2/5x+b
To find b substitute for the point(5,-1):
-1=-2/5*(5)+b => b=+1
The eq. of the line is
y=-2/5x+1

Answer 3

2x + 5y = 15
5y = - 2x + 15
y = (- 2 / 5) x + 3
m = ( - 2 / 5 )
y + 1 = (- 2 / 5) (x - 5)
y = (- 2 / 5) x + 2 - 1
y = (- 2 / 5) x + 1 is required equation.