Thread: Help me on all these problems

1. Help me on all these problems

Find the points that are symmetric to given point (a) across the x axis, (b) across the y axis, and (c) across the origin

1. (1,4)
2. (2,-3)

3. Find equations for the vertical and horizontal lines through the point (1,3)

8. Given the point, P(6,0) and the line, 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

2. Originally Posted by yitsongg
Help me on all these problems
Help me help you. Tell me where it is that you are having problems. Show some work or something, man.

3. Originally Posted by yitsongg
Find the points that are symmetric to given point (a) across the x axis, (b) across the y axis, and (c) across the origin

1. (1,4)
2. (2,-3)

3. Find equations for the vertical and horizontal lines through the point (1,3)

8. Given the point, P(6,0) and the line, 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
Hi yitsongg,

1. Graph this point and let the x and y axes be the lines of symmetry.

(1, 4) reflected across the x axis is (1, -4) and reflected across the y-axis is (-1, 4). With respect to the origin, it would be (-1, -4).

2. See if you can handle this one based on the previous one.

3. The equation of the vertical line through (1, 3) is x = 1.

The equlation of the horizontal line through (1, 3) is y =3.

8. Recall that parallel lines have the same slope. Given line:

$2x-y=-2$
$y=2x+2$

The slope is 2.

To find the equation of a line through (6, 0) with slope 2, use the slope intercept form of the equation and find the y-intercept.

$y=mx+b$
$0=2(6)+b$
$b=-12$

Therefore, the equation of a line through (6, 0) and parallel to the line whose equation is 2x - y = -2 is y = 2x -12.

To find the equation of a line through (6, 0) and perpendicular to 2x - y = -2, use the negative reciprocal of the slope of this line (which is -1/2) and perform the same steps as above to find the y-intercept and ultimately, the equation in slope-intercept form.

$y=mx+b$
$0=-\frac{1}{2}(6)+b$

and so on.