1) Write the standard form of the equation of the line parallel to the graph of 2y-6=0 and passing through B(4,-1).

Answer: y+1=0

2)Write an equation of the line perpendicular to the graph of 2-2y-6=0 and passing through A(-3,2).

Answer: 2x+y-8=0

Thanks.

assume you mean

2 x - 2 y -6 = 0
slope of this line is one
so slope of perpendicular is -1/1 = -1
2 = -1 (-3) + b
b = -1
so
y = -1 x -1
1 x + 1 y + 1 = 0

To find the standard form of a line parallel to a given line, we need to know that parallel lines have the same slope. The given equation, 2y - 6 = 0, is already in slope-intercept form (y = mx + b) where m is the slope. So, to find the slope of the given line, we can rearrange the equation to isolate y:

2y - 6 = 0
2y = 6
y = 3

Since the slope m is equal to 3 for the given line, any line parallel to it will also have a slope of 3. Now that we have the slope, we can use the point-slope form of a line to find the equation. The point-slope form is:

y - y1 = m(x - x1)

where (x1, y1) is a point on the line and m is the slope. Using the point B(4, -1) and the slope m = 3:

y - (-1) = 3(x - 4)
y + 1 = 3x - 12
y = 3x - 13

To write the equation in standard form, we rearrange it as:
3x - y = 13, which is the standard form of the equation of the line parallel to the graph 2y - 6 = 0 and passing through the point B(4, -1).

Now, let's move on to the second question. To find the equation of a line perpendicular to a given line, we need to find the negative reciprocal of the slope of the given line. The given equation, 2 - 2y - 6 = 0, can be simplified as 2y = -4, then y = -2. The slope m of this line is -2.

To find the negative reciprocal of -2, we invert the fraction and change the sign:
Negative reciprocal of -2 = 1/2.

Now, using the point (-3, 2) and the slope m = 1/2, we can write the equation using the point-slope form:

y - y1 = m(x - x1)
y - 2 = 1/2(x - (-3))
y - 2 = 1/2(x + 3)
y - 2 = 1/2x + 3/2

To write the equation in standard form, we can multiply both sides of the equation by 2 to eliminate the fraction:

2(y - 2) = 2(1/2x + 3/2)
2y - 4 = x + 3
x - 2y + 7 = 0

Therefore, the equation in standard form of the line perpendicular to the graph 2 - 2y - 6 = 0 and passing through the point A(-3, 2) is x - 2y + 7 = 0.

I hope this explanation helps! Let me know if you have any further questions.