SOLUTION: Write an equation of the Line 1 passing through (3,-4) that is parallel to the Line II 5x-3y-9=0. Does the line I pass through the point (6,7)? Does the line II pass through the p

Algebra.Com
Question 149086: Write an equation of the Line 1 passing through (3,-4) that is parallel to the Line II 5x-3y-9=0. Does the line I pass through the point (6,7)? Does the line II pass through the point (6,7)?

Answer by Electrified_Levi(103)   (Show Source): You can put this solution on YOUR website!
Hi, Hope I can help,
.
Write an equation of the Line 1 passing through (3,-4) that is parallel to the Line II 5x-3y-9=0. Does the line I pass through the point (6,7)? Does the line II pass through the point (6,7)?
.
First, we have to put Line II in slope-intercept form (y = Ax+b)("A" is the slope)
.
5x-3y-9=0
.
We will move (-3y) to the right side
.
5x - 3y + 3y - 9 = 0 + 3y
.
5x - 9 = 3y
.
3y = 5x - 9
.
We will now divide each side by "3"
.

.
y = x - 3
.
(y = x - 3) is the slope-intercept form of Line II
.
Since Line I is parallel to Line II, it has the same slope
.
Equation to Line I
.
y = x + b
.
We can solve "b" by replacing "x" and "y" with (3,-4)(x,y)(Since the line contains that point
.

.
We will multiply the right side
.

.
We will move "5" to the left side
.

.

.
b = (-9), we can replace "b" in our equation
.
y = x + b
.
y = x + (-9)
.
y = x - 9
.
(y = x - 9) is the slope-intercept equation to Line I
.
You can check by replacing "x" and "y" with (3, -4)(x,y)
.

.

.

.
(y = x - 9) is the correct equation to Line I
.
y = x - 9( to get a different form of equation, you get rid of the fraction)(We will multiply each side by "3")
.

.
3y = 5x - 27
.
We will move "3y" to the right side
.
3y - 3y = 5x - 3y - 27
0 = 5x - 3y - 27
(5x - 3y - 27 = 0) is another form,( you can move (-27) over, and it will become ( this is the standard form)
.
To see if (6,7) is a point of either of the two lines, you will replace "x" and "y" with (6,7)(x,y)
.
First equation,
.

.

.

.
This is not true, so point (6,7) is not a point on the first line.
.
Second line
.
5x-3y-9=0 ( standard form = (5x - 3y = 9)(moved (-9) to the right side)
.
5x - 3y = 9
.
We can replace "x" and "y" with (6,7)(x,y)
.
5(6) - 3(7) = 9
.
30 - 21 = 9
.
(9 = 9) This is true, (6,7) is a point on Line II
.
Hope I helped, Levi

RELATED QUESTIONS

Hi, can you help me with this? i. Find an equation of the line that passes through the (answered by Maths68)
Write the equation of a line parallel to the given line but passing through the given... (answered by Cromlix)
Write an equation of the line passing through (5, 0) and parallel to the line y = 5x – 2. (answered by checkley75)
Write an equation of the line passing through (5, 0) and parallel to the line y = 5x –... (answered by Alan3354)
How do I solve this? The line through (3, 4) that is parallel to the line 5x + 3y = 9 (answered by mananth)
Can I have some help finding an equation of the line through (0, 9) and parallel to the... (answered by iluvbuilding429)
write an equation of the line through the point (4, -2) that is parallel to the line -5x... (answered by Alan3354)
write an equation of a line that passes through ( -5, 4) and )3, -2) Write and... (answered by solver91311)
The line through (3,4) that is parallel to the line 5x+3y=9. Write the equation of each... (answered by mananth)