# Thread: Graph of the equation is symmetric

1. ## Graph of the equation is symmetric

I need help solving this problem, can anyone do a step by step guide on how to do this?

Determine whether the graph of the equation is symmetric with respect to the line Y = -X

1) Y = 2X - 1

2. One way of checking it is by slopes, because both x = -y and y = 2x +1 are straight lines only.
If y = 2x +1 were symmetrical with respect to y = -x, then y = 2x +1 should be perpendicular to y = -x. If not, then there is no symmetry.

y = -x
slope, m1 = -1

y = 2x +1
slope, m2 = 2

To be perpendicular, their slopes must be the negative reciprocals.
m2 = -1/m1
2 = -1 / -1
2 = 1
False, so the two lines are not perpendicular, and so y = 2x +1 is not symmetrical wirh respect to y = -x.

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Another way.

y = -x -------------axis of symmetry.

y = 2x +1 --------(i)

If (i) is symmetrical, when we swap the x and y into (i), we should get an equation similar or equal to (i).

-x = 2(-y) +1
-x = -2y +1
2y = x +1
y = (1/2)(x +1) -------not the same as (i), so, no symmetry.

3. Originally Posted by ticbol
y = -x -------------axis of symmetry.

y = 2x +1 --------(i)

If (i) is symmetrical, when we swap the x and y into (i), we should get an equation similar or equal to (i).

-x = 2(-y) +1
-x = -2y +1
How come the -x remained as a -x. While the 2y turned into a -2y and the -1 turned into a 1?

4. Originally Posted by Brazuca
How come the -x remained as a -x. While the 2y turned into a -2y and the -1 turned into a 1?
what are you talking about? he replaced y with -x and x with -y and solved for y. that's it.

5. Originally Posted by Jhevon
what are you talking about? he replaced y with -x and x with -y and solved for y. that's it.
I don't understand.

If Y = -X then wouldn't the -X have to turn into a X for the Y to turn into a -Y?

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

Wouldn't the problem have to turn into this?
-x = 2y + 1
x = 2(-y) +1
x = -2y + 1

6. Originally Posted by Brazuca
I don't understand.
we have
$\displaystyle y = -x$ which is also saying that $\displaystyle x = -y$, this is line (1)

then we are given $\displaystyle y = 2x - 1$ ........line (2)

to find if $\displaystyle y = 2x - 1$ is symmetric with respect to line one, plug in $\displaystyle y = -x$ and $\displaystyle x = -y$ as line (1) directed. if we simplify and get the original formula for line (2), then line (2) is symmetric with respect to line (1)

$\displaystyle {\color {red}y } = 2 {\color {blue}x} - 1$

Plug in $\displaystyle y = -x$ and $\displaystyle x = -y$, we get:

$\displaystyle {\color {red}-x } = 2({\color{blue}-y}) - 1$

solving for $\displaystyle y$, we get:

$\displaystyle y = \frac 12x - \frac 12$

which is not the original formula for line (2), so line (2) is NOT symmetric with respect to line (1)

7. Originally Posted by Jhevon
we have
$\displaystyle y = -x$ which is also saying that $\displaystyle x = -y$, this is line (1)
Thanks, I had to read everything you posted step by step to understand that y = -x then x = -y.

I always looked at it as y = -x then -x = y. I was not switching the signs.

8. Originally Posted by Brazuca
Thanks, I had to read everything you posted step by step to understand that y = -x then x = -y.

I always looked at it as y = -x then -x = y. I was not switching the signs.
if you multiply both sides of the equation y = -x by -1 you get -y = x