Section 1 - Topic 9 Parallel and Perpendicular Lines - Part 2 (please help) 1. Write the equation of the line that it is perpendicular to y = 7x - 3 and passes through the origin.

2. Write the equation of the line that it is parallel to y = 2x + 1 and passes through the solution of the following system of equations. (3x - 2y = 10 x + y = 5

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Step-by-step explanation:

1. The slope of the given line is the x-coefficient, 7. The slope of the perpendicular line is the negative reciprocal of this: -1/7. Since the line goes through the origin, the y-intercept is zero.

  y = -1/7x

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2. The first part of this problem is to find the solution to the given system of equations. We can substitute for x to do that:

  3(5 -y)-2y = 10

  15 -5y = 10 . . . simplify

  -3 +y = -2 . . . . divide by -2

  y = 1, x = 4 . . . add 3 to find y; subtract y from 5 to find x

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The second part of this problem requires we note the slope of the parallel line. It is the x-coefficient, 2. The y-intercept can be found from ...

  b = y -mx = 1 -2(4) = -7

So, the desired line in y=mx+b form is ...

  y = 2x -7

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The first attachment shows the perpendicular lines of problem 1.

The second attachment shows the parallel lines of problem 2.

answer: f = 32.2m

step-by-step explanation: that is the answer

for the friend request.

when a figure, such as triangle (xyz) is reflected over the line y = x to create x'y'z' the x-coordinates and y-coordinates of the original image (pre-image) will be reversed for the image. (e.g., (x, y) > (y, x)) this means that the pre-image position of the x- and y-coordinates switch places which in turn provides a mirror like image of the original xyz.  

i notice that when i draw a line segment from x to x', that the points are parallel from each other -- they are exactly across from each other and do not intersect with any other points.  

yes, think i would see the same characteristic if you drew the line segment connecting y with the reflecting line and then y' with the reflecting line because of the essence of reflections. because it is a mirror image, each point was effected the same way which leads me to conclude that they same would apply to y and y'.