Question

find the distance between the following pair of points: (2) P(-5,2), Q L-1,5)


Solve it step by step!​

Answer 1

Answer:

Distance between P(-5, 2) and Q(-1, 5) is 5 units

Explanation:

Given points,

• P(-5, 2)
• Q(-1, 5)

By distance formula :-

$\rm \longrightarrow \: d = \sqrt{(x_2 - x_1)^{2} + (y_2 - y_1)^{2} }$

Where,

• $\rm x_1 \: and \: y_1 \: denotes \: the \: coordinates \: of \: point \: P$
• $\rm x_2 \: and \: y_2 \: denotes \: the \: coordinates \: of \: point \: Q$

Substituing the coordinates,

$\rm \longrightarrow \: d = \sqrt{[( - 1 )- ( - 5)]^{2} + (5 - 2)^{2} }$

$\rm \longrightarrow \: d = \sqrt{[ - 1 + 5]^{2} + (5 - 2)^{2} }$

$\rm \longrightarrow \: d = \sqrt{[ 4]^{2} + (3)^{2} }$

$\rm \longrightarrow \: d = \sqrt{16 + 9}$

$\rm \longrightarrow \: d = \sqrt{25}$

$\rm \longrightarrow \: d = 5$

∴ Distance between the points P and Q is 5 units.

Answer 2

$\sf\huge\bold{Answer:}$

Given :

points

P(-5,2), Q (-1,5)

To find :

Distance between P and Q

Solution :

$\fbox{Distance formula}$

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

where,

$⇒ x_{1} \: and \: y_{1} \: are \:x \: and \: y \: coordintes \: of \: P \: respectively \\ \\ ⇒x_{2} \: and \: y_{2} \: are \: x \:and \: y \: coordintes \: of \: Q\: respectively$

Step 1 : Distinguishing values of x and y for P and Q

P(-5,2)

$x_{1}$= -5

$y_{1}$= 2

Q (-1,5)

$x_{2}$= -1

$y_{2}$= 5

Step 2 : Inserting values in distance formula

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

$⟶PQ= \sqrt{ {( - 1 - ( - 5)) {}^{2}+ ( 5 - 2 ) {}^{2} } }$

$⟶PQ = \sqrt{( - 1 + 5){}^{2} +(3){}^{2} }$

$⟶PQ = \sqrt{(4){}^{2}+9}$

$⟶PQ = \sqrt{16+9}$

$⟶PQ = \sqrt{25}$

$⟶PQ = 5 units$

Hence,distance between points P(-5,2), Q (-1,5) is

$\sqrt{5} units$