Part a. Write the equation of the line that is parallel to the line 3x-y= -2 and contains the point (1,3).?

Question

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Answer 1

The 'standard' way to write a line equation is:

y = m x + c

where
m is the slope (how fast y changes when x changes by 1) and
c is the intercept (the 'y' value when x = 0), the height at which the line crosses the y axis.

Your line is 3x - y = -2
Which can be written (after moving everything around appropriately -- do it)
y = 3x + 2

Step one: the new line is parallel, then it must have the same slope (m = 3)

the new line is
y = 3x + c

Step two, it must contain the point (1,3) -- in other words, when x = 1, y must equal 3

3 = 3(1) + c

then c must be zero

step 3: write it out:

y = mx + c
where m = 3 and c= 0

y = 3x

Answer 2

original equation y = 3x + 2
answer equation y = 3x

y = 3, x = 1 then let equation be y = 3x + z
in order for equation to be true where y = 3 and x = 1
3 = 3 + z, z must be zero

since which ever number in front of x in relation to the number in front of y determines the slope of the line, as long as that remains constant you can make any parallel line by just changing z the which is the intersect point when x = 0.

Answer 3

Let us find the slope of the given line, as this line is parallel to the line for which we have to find the equation.

3x-y= -2
3x+2= y
therefore the slope of this line is 3
so the slope of the required line is also 3.

we know the slope & we have a point so use the formula to find the equation of the line.

formula is
m=(y-y1)/(x-x1)
m=3, (x1,y1)=(1,3)
so substitute and find
3=(y-3)/(x-1)
3(x-1)=(y-3)
3x-3=y-3
3x=y-3+3
3x=y-0
y=3x
is the equation of the required line