Question

The distance between the points $\left( {a,b} \right)$ and $\left( { - a, - b} \right)$ is:
(a) ${a^2} + {b^2}$
(b) $\sqrt {{a^2} + {b^2}}$
(c) 0
(d) $2\sqrt {{a^2} + {b^2}}$

Answer 1

Here, we will use the formula for distance between two points, and simplify the expression to find the distance between the points $\left( {a,b} \right)$ and $\left( { - a, - b} \right)$, and then choose the correct option.
Formula Used: We will use the following formulas:

1. Distance formula: The distance $d$ between two points $\left( {{x_1},{y_1}} \right)$ and $\left( {{x_2},{y_2}} \right)$ is given by $d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}}$.
2. Rule of exponents: If two or more numbers with the same base and different exponents are multiplied, the product can be written as ${a^b} \times {a^c} = {a^{b + c}}$.

Complete step by step solution:
We will use the distance formula to find the distance between the points $\left( {a,b} \right)$ and $\left( { - a, - b} \right)$.
Let $d$ be the distance between the points $\left( {a,b} \right)$ and $\left( { - a, - b} \right)$.
Substituting ${x_1} = a$, ${y_1} = b$, ${x_2} = - a$, and ${y_2} = - b$ in the distance formula, $d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}}$, we get
$\Rightarrow d = \sqrt {{{\left( { - a - a} \right)}^2} + {{\left( { - b - b} \right)}^2}}$
Subtracting the like terms in the parentheses, we get
$\Rightarrow d = \sqrt {{{\left( { - 2a} \right)}^2} + {{\left( { - 2b} \right)}^2}}$
Rewriting the expression ${\left( { - 2a} \right)^2}$, we get
$\begin{array}{l} \Rightarrow {\left( { - 2a} \right)^2} = \left( { - 2a} \right) \times \left( { - 2a} \right)\\\ \Rightarrow {\left( { - 2a} \right)^2} = \left( { - 1} \right) \times 2{a^1} \times \left( { - 1} \right) \times 2{a^1}\end{array}$
Rearranging the terms, we get
$\Rightarrow {\left( { - 2a} \right)^2} = {\left( { - 1} \right)^2} \times 2 \times 2 \times {a^1} \times {a^1}$
We know that ${\left( { - 1} \right)^n}$ is equal to 1 if $n$ is an even number, and is equal to $- 1$ if $n$ is an odd number.
Thus, we get
${\left( { - 1} \right)^2} = 1$
Therefore, we get
$\Rightarrow {\left( { - 2a} \right)^2} = 1 \times 2 \times 2 \times {a^1} \times {a^1}$
Now by applying rules of exponent, the equation becomes
$\begin{array}{l} \Rightarrow {\left( { - 2a} \right)^2} = 1 \times 2 \times 2 \times {a^{1 + 1}}\\\ \Rightarrow {\left( { - 2a} \right)^2} = 1 \times 2 \times 2 \times {a^2}\end{array}$
Multiplying the terms of the expression, we get
$\Rightarrow {\left( { - 2a} \right)^2} = 4{a^2}$
Rewriting the expression ${\left( { - 2b} \right)^2}$, we get
$\begin{array}{l} \Rightarrow {\left( { - 2b} \right)^2} = \left( { - 2b} \right) \times \left( { - 2b} \right)\\\ \Rightarrow {\left( { - 2b} \right)^2} = \left( { - 1} \right) \times 2{b^1} \times \left( { - 1} \right) \times 2{b^1}\end{array}$
Rearranging the terms, we get
$\Rightarrow {\left( { - 2b} \right)^2} = {\left( { - 1} \right)^2} \times 2 \times 2 \times {b^1} \times {b^1}$
Simplifying the expression, we get
$\Rightarrow {\left( { - 2b} \right)^2} = 1 \times 2 \times 2 \times {b^1} \times {b^1}$
Rewriting using the rule of exponent ${a^b} \times {a^c} = {a^{b + c}}$, we get
$\begin{array}{l} \Rightarrow {\left( { - 2b} \right)^2} = 1 \times 2 \times 2 \times {b^{1 + 1}}\\\ \Rightarrow {\left( { - 2b} \right)^2} = 1 \times 2 \times 2 \times {b^2}\end{array}$
Multiplying the terms of the expression, we get
$\Rightarrow {\left( { - 2b} \right)^2} = 4{b^2}$
Now, substituting ${\left( { - 2a} \right)^2} = 4{a^2}$ and ${\left( { - 2b} \right)^2} = 4{b^2}$ in the equation $d = \sqrt {{{\left( { - 2a} \right)}^2} + {{\left( { - 2b} \right)}^2}}$, we get
$\Rightarrow d = \sqrt {4{a^2} + 4{b^2}}$
Factoring out 4 from the terms, we get
$\Rightarrow d = \sqrt {4\left( {{a^2} + {b^2}} \right)}$
Rewriting the expression as a product of square roots, we get
$\Rightarrow d = \sqrt 4 \sqrt {{a^2} + {b^2}}$
$\Rightarrow d = 2\sqrt {{a^2} + {b^2}}$
Therefore, the distance between the points $\left( {a,b} \right)$ and $\left( { - a, - b} \right)$ is $2\sqrt {{a^2} + {b^2}}$.

The correct option is option (d).

Note:
We rewrote $\sqrt {4\left( {{a^2} + {b^2}} \right)}$ as $\sqrt 4 \sqrt {{a^2} + {b^2}}$. This is because if two square roots are multiplied, then the result is equal to the square of the product of the number inside the root. This means that if $\sqrt x$ and $\sqrt y$ are multiplied, then the result is equal to $\sqrt {xy}$.