Q 56. - Calculus

Question

Prove Theorem $9.10$ That is, show that if a graph in the plane has any two of three types of symmetry, namely, symmetry about the $x-$axis, symmetry about the $y-$axis, and symmetry about the origin, then it has the third type of symmetry as well.

Step-by-Step Solution

Answer

Proof: $c$ is symmetric to $a$ with respect to the origin.

Step 1: Given information

It has the third type of symmetry as well.

Step 2: Calculation

Take a point $a$ in the first quadrant.

Draw a line $l_1$ from the orlgin to $a$ and let the angle formed by $l_1$ and the $x$ axis.

Now reflect the point $a$ about $x$ axis and call it as point $b$

Draw a line $L_2$ from the origin to $b$ and call $\beta$ the angle formed by $L_2$ and the $x$ axis.

And call $\rho$ the angle formed by $l_2$ and the $y$ axis.

Then $\theta =\beta$ and $l_1=l_2$

Step 3: Calculation

Now reflect $b$ about $y$ axis and call that paint $c$

Draw $l_3$ from the origin to $c$ and call $\varphi$ the angle formed by $l_3$ and $y-$axis.

Then $I_2=I_3$ and therefore $I_3=l_1$

Since $x, y$ axes are orthogonal, $\theta$ and $\beta$ are complements of $\varphi$ and $\rho$

Therefore, $\theta +\beta +\varphi +\beta =180,l_1+l_3$ the diameter of the circle with the origin as the center. Therefore, $c$ is symmetric to $a$ with respect to origin.

Hence the proof.