Question

1. Write an equation in slope-intercept form of the line that passes through the given point and is parallel to the graph of the given equation. (1,-3; y + 2 – 4(x-1)

2. Are the graphs of the equations y = -2x +3 and 2x +y = 7, parallel, perpendicular, or neither? Explain.

3. A line passes through (5,0) and is perpendicular to the graph of y +1 = 2(x-3). What equation represents the line in slope-intercept form?

4. A bike path is being planned for the park. The bike path will be parallel to Main Street and will pass through the park entrance. What is an equation of the line that represents the bike path? (use image from Ex.4)

5. Determine whether this statement is always, sometimes, or never true. Two line with the same slope and different y-intercepts are perpendicular.

Answer 1

1. the equation y + 2 = -4(x - 1) needs to be in slope-intercept form. it makes it easier to solve. first, distributive property. y + 2 = -4x + 4 now you can subtract 2 from both sides to isolate y. new equation: y = -4x + 2 now you can graph this equation and the point (1, -3). we know the slope of the equation has to be the same so we can go up 4 and 1 to the left from (1, -3). this new point is (0, 1). so the answer is: y = -4x + 1

2. the graphs of the equations y = -2x + 3 and y = -2x + 7 are parallel.
i changed 2x + y = 7 to y = -2x + 7 by subtracting 2 from both sides. (standard form to slope-intercept form) then, i determined they were parallel by graphing them. you can visually see they're parallel to each other.

3. y + 1 = 2(x - 3) <- standard form. to transform this into slope-intercept form, you have to start by using the distributive property. so, do 2 times x and 2 times -3. new equation: y + 1 = 2x - 6 now you add 6 to both sides. new equation: y + 7 = 2x all you have to do now is get this in slope-intercept form. to do this, just subtract 7 from both sides. new equation: y = 2x - 7

4. the bike path is going to be parallel to main street. it is also going to pass through the park entrance (0, 4). the equation of main street is y = 2x - 3
we know that the park entrance is (0, 4) so the equation is going to look like this so far: y = x + 4 we also know the slope of the equation of main street is 2x and for the bike path to be parallel the slope needs to be the same. therefore, the equation of the line that represents the bike path is y = 2x + 4.

5. the statement: two lines with the same slope and different y-intercepts are perpendicular. is never true. this is because two lines with the same slope and different y-intercepts are parallel, not perpendicular.

i know this is a lot to read, but i like explaining things just to make sure you understand everything. the answers are in bold. i hope this helps! 

Answer 2

Answer:

#1) y = -4x + 1; #2) parallel; #3) y = -1/2x + 5/2; #4) y = 2x + 4; #5) Never

Step-by-step explanation:

#1) We want a line that passes through (1, -3) and is parallel to y = 2-4(x-1).  First we simplify our equation using the distributive property:

y = 2-4(x-1) = 2-4(x) -4(-1) = 2-4x--4 = 2-4x + 4 = -4x + 6

Lines that are parallel have the same slope; this means we want a line through (1, -3) with a slope of -4.  Using point-slope form,

$y-y_1=m(x-x_1)\\\\y--3=-4(x-1)\\\\y+3=-4(x)-4(-1)\\\\y+3=-4x+4\\\\y+3-3=-4x+4-3\\\\y=-4x+1$

#2) We must first write our second equation in slope-intercept form by isolating the y term:

2x+y=7

Subtract 2x from each side"

2x+y-2x = 7-2x

y = -2x+7

This means the slope is -2; the slopes are the same, so the lines are parallel.

#3) The slope of our given equation is 2.  To be perpendicular, the second line must have a slope that is a negative reciprocal (flipped and opposite signs); this makes it -1/2.  Using point-slope form,

$y-0=\frac{-1}{2}(x-5)\\\\y=\frac{-1}{2}(x)+\frac{-1}{2}(-5)\\\\y=\frac{-1}{2}x+\frac{5}{2}$

#4) The slope of Main Street on the diagram, found by using rise/run, is 2.  This means the bike path will also have a slope of 2 in order to be parallel.  The park entrance is at (0, 4).  This makes the equation y = 2x+4.

#5) Two lines with the same slope are always parallel.  They are never perpendicular.