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
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.
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)