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Question

it says given p(6,0) L: 2x-y=-2

a. find an equation for the line through p parallel to l

b. find an equation of the line through p perpendicular to l

can you show it step by step because i dont understand how to do this

Answer 1

2x-y = -2, so y= 2x+2

Thus slope of line L is 2

A line || to L will also have a slope of 2, so its equation is:

y = 2x + b

Since the line passes through (6,0) we have:

0 = 2(6) + b which means b = -12

So equation is y = 2x-12

A line _|_ to line L will have a slope = -1/2 <-- negative reciprocal of 2

Its equation is thus y = -x/2 + b

So 0 = -6/2 +b --> b = 3

So equation is y = -x/2 + 3

Answer 2

First take your line L, and solve it for y:

2x - y = -2

-y = -2x - 2

y = 2x + 2

This is now in slope-intercept form of y = mx + b.  You can simply read off the slope (m = 2).

PART A:

A line parallel to y = 2x + 2 will have the exact same slope.  So it will have m = 2.

y = 2x + b

The only thing you have to figure out, is what value of b will make the line go through the point p(6,0).  Take the values of x = 6, y = 0 and solve for b:

0 = 2(6) + b

0 = 12 + b

b = -12

Answer:

y = 2x - 12

PART B:

Here it asks for a line *perpendicular* to the first line.  A perpendicular line has a slope that is the *negative reciprocal*.  So instead of m = 2, you will have m = -½

Again set-up the line:

y = -½x + b

Plug in the point p(6, 0) and solve for b:

0 = -½(6) + b

0 = -3 + b

b = 3

Answer:

y = -½x + 3