Question

The distance between $\left( a,b \right),\left( -a,-b \right)$ is

(a) $2\sqrt{{{a}^{2}}+{{b}^{2}}}$

(b) $3\sqrt{{{a}^{2}}+{{b}^{2}}}$

(c) $2{{a}^{2}}$

(d) $2{{b}^{2}}$

Answer 1

Consider the two points given as P (a, b) and Q (-a, -b). We need to find the distance PQ. As the values are already given, substitute them in the distance formula and simplify it.

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Complete step-by-step answer:**

To calculate the distance between 2 points in a plane, we have to use the distance formula as per described in coordinate geometry. We have been given 2 points (a, b) and (-a, -b). Let us consider points as P and Q. Thus the 2 points becomes P (a, b) and Q (-a, -b). We were asked to find the distance between them. Thus we need to find the distance PQ. The distance formula is given by.

Distance = $\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$.

Thus let us take, $P\left( a,b \right)=\left( {{x}_{1}},{{y}_{1}} \right)$

Similarly, $Q\left( -a,-b \right)=\left( {{x}_{2}},{{y}_{2}} \right)$.

Now let us apply these values in the distance formula,

Distance, $PQ=\sqrt{{{\left( -a-a \right)}^{2}}+{{\left( -b-b \right)}^{2}}}$

$=\sqrt{{{\left( -2a \right)}^{2}}+{{\left( -2b \right)}^{2}}}$

$=\sqrt{4{{a}^{2}}+4{{b}^{2}}}=\sqrt{4\left( {{a}^{2}}+{{b}^{2}} \right)}=\sqrt{4}.\sqrt{{{a}^{2}}+{{b}^{2}}}$

$=2\sqrt{{{a}^{2}}+{{b}^{2}}}$

Distance, $PQ=2\sqrt{{{a}^{2}}+{{b}^{2}}}$

Thus the distance between the two points (a, b) and (-a, -b) is $2\sqrt{{{a}^{2}}+{{b}^{2}}}$.

$\therefore$ Option (a) is the correct answer.

Note: In the Distance formula, we can observe that ${{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}$ is the square of the difference in x – coordinates of P and Q and is always positive. The same can be said about ${{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}$ as well. Thus the formula remains the same for any coordinates of P and Q, in any quadrant.