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The equation of line ab is y = 5x + 1. write an equation of a line parallel to line ab in slope-intercept form that contains point (4, 5). y = 5x − 15 y = 5x + 15 y = x + y = x −

Respuesta :

ray2me123
METHOD 1:
The equation of line parallel to ax+by=c is,
ax+by=k
Similarly,in this case equation of line parallel to the st line y=5x+1 (i.e, 5x-y=-1) is given by,
5x-y=k......(i)
As equation (i) is the line containing (4,5),
k=5×4-5=15
Hence equation (i) becomes,
5x-y=15
y=5x-15➡required equation (ANS)



METHOD 2:

As slope of AB is 5, the slope of line parallel to AB should also be 5.
Thus, the equation of st line parallel to AB (with slope 5) & passing through (4,5) is:
m=(y-y1)/(x-x1)
5=(y-5)/(x-4)
5x-20=y-5
y=5x-15➡ required eqn



profantoniofonte

Answer:

y = 5x − 15

Step-by-step explanation:

Hi there!

The First equation y=ax+b then y=5x+1

1) Parallel lines have the same slope. Let's call it m' for the first and m" for the second equation. So m' equals 5 and m" equals 5 for the second equation.

2) As for the Second Equation, we know m"=5. Let's write it down the 2nd Linear Equation.

y=m"x+b = y =5x+b

Since (4,5) is an element of this equation let's plug it in:

So let's find the value of x, 5=5(4)+b

5=20+b

5-20=20-20+b

-15=b

b=-15

And that's the Second Equation y=5x-15

3) Testing our work

y=5x-15 Plugging in (4,5)

5=5(4)-15

5=5

In addition to this check both equations, in the graphs below!

Ver imagen profantoniofonte

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