Question

find the distance between
(a,b) (-a,-b)​

Answer 1

GIVEN

• Points (a,b) and (-a, -b).

TO FIND :-

• Distance between them.

SOLUTION :-

☯ $\\ \underline{\boldsymbol{According\: to \:the\: Question\:now :}}$

$\\ : \implies \displaystyle \sf \sqrt{(x _{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} } \ \ \ \ \ \ \ \bigg \lgroup Distance Formula \bigg \rgroup$

$: \implies \displaystyle \sf \sqrt{ \{a - ( - a) \} ^{2} + \{b - ( - b) \} ^{2} }$

$: \implies \displaystyle \sf \sqrt{(a + a) ^{2} + (b + b) ^{2} }$

$\\ : \implies \displaystyle \sf \sqrt{(a ^{2} + a ^{2} + 2 \times a \times a) + (b ^{2} + b ^{2} + 2 \times b \times b)}$

$: \implies \displaystyle \sf \sqrt{(2a ^{2} + 2a ^{2}) + (2b ^{2} + 2b ^{2} )}$

$: \implies \displaystyle \sf \sqrt{4a ^{2} + 4b ^{2} }$

$: \implies \underline{\boxed { \displaystyle \sf 2\sqrt{a^{2}+ b^{2}}}}$

$\therefore \underline{\displaystyle \sf Distance \ between \ (a,b) \ and \ (-a, -b) \ is \ 2\sqrt{a^{2}+ b^{2}}}$

________________

Identity used,

$\\ \dashrightarrow \displaystyle \sf (a + b)^{2} = a^{2} + b^{2} + 2ab$

Answer 2

Given:

• Points are (a,b) and (-a,-b)

Find:

• Distance between the given points ?

Solution:

Let, points be P(a,b) and Q(-a,-b)

Now,

we, know that

$\underline{ \boxed{ \pink{ \sf PQ = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2} }}}}$

where,

• $\sf x_1$ = a
• $\sf x_2$ = -a
• $\sf y_1$ = b
• $\sf y_2$ = -b

Now,

$\sf \longrightarrow PQ = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}$

$\sf \longrightarrow PQ = \sqrt{{( - a - a)}^{2} + {( - b- b)}^{2}}$

$\sf \longrightarrow PQ = \sqrt{{( - 2a)}^{2} + {( - 2b)}^{2}}$

$\sf \longrightarrow PQ = \sqrt{4 {a}^{2} + 4 {b}^{2} }$

$\sf \longrightarrow PQ = \sqrt{ 4({a}^{2} + {b}^{2})}$

$\underline{ \boxed{ \sf \longrightarrow PQ = 2\sqrt{ {a}^{2} + {b}^{2}} \: units}}$

Hence, the distance between the points P(a,b) and Q(-a,-b) will be $\sf 2\sqrt{ {a}^{2} + {b}^{2}} \: units$