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miss Akunina [59]
3 years ago
9

Find the equation of a line parallel to the line 2x-y-9=0 and passing through the point of intersection of the lines 5x+y+4=0 an

d 2x+3y=1
Mathematics
1 answer:
Cerrena [4.2K]3 years ago
3 0

Answer:

y=2x-3

Step-by-step explanation:

1. Using the point of intersections, we use the substitution method to find the coordinates of the line parallel to 2x-y-9=0

5x+y+4=0

y= -4-5x

substituting the value of y in 2x+3y=1:

2x +3(-4-5x)=1\\\\2x-12-15x-1=0\\-13x=13x\\x= \frac{13}{-13}= -1

substituting x=-1 in y= -4-5x:

y= 1 (upon solving, you should get this)

(x,y)= (-1,1)

2. Using y=mx+c and making y the subject of the formula 2x-y-9=0 and using the coordinate we found earlier, we will find the equation of the parallel line. (We make y the subject of the formula to find the gradient)

2x-y-9=0\\2x-9=y\\y= 2x-9

y= mx+c\\

-1= 2 x 1 +c

-3=c

  • y= 2x-3

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egoroff_w [7]

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Step-by-step explanation:

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