Question

If x,y ∈ C ,

statement 1 : |$\frac{2x-4y}{6\overline{y}-3\overline{x}}$| = $\frac{2}{3}$

statement 2 : If z ∈ C, then |z|= |$\overline{z}$| = |-z| = |$\overline{-z}$|

• A. Statement 1 is true & statement 2 is correct explanation of statement 1
• B. Statement 1 is false & statement 2 is true
• C. Statement 1 is true & statement 2 is not correct explanation of Statement2
• D. Statement 1 is false & Statement 2 is false

Answer 1

The correct option is A

Statement 1 is true & statement 2 is correct explanation of statement 1

|$\frac{2x-4y}{6\overline{y}-3\overline{x}}$| = |$\frac{2(x-2y)}{3(2\overline{y}-\overline{x})}$| = $\frac{2}{3}$ $\frac{|x-2y|}{|-(\overline{x}-2\overline{y})|}$ = $\frac{2}{3}$ $\frac{|x-2y|}{|-\left(\overline{(x-2y)}\right)|}$

[ ∵ ${\overline{Z}}_1$-${\overline{Z}}_2$ = $\overline{Z_1-Z_2}$ ]

Let x-2y = z

Now, $\frac{2}{3}$$\frac{|x-2y|}{|-\left(\overline{x-2y}\right)|}$

= $\frac{2}{3}$ \(\frac{|z|}{-\overline z}

= $\frac{2}{3}$ $[\because |z|=|-\overline{z}|]$

Hence, Statement 1 is true & Statement 2 is correct explanation of statement 1.