Find the distance d between the points P_{1} and P_{2} P_{1}=(1.2,\

Step 1
The question is taken from the Analytical Geometry in which we have to find the distance between the two points. The given points are P1 and P2 which are (1.2, 2.3) and (-0.3, 1.1) respectively.
Given that distance between the two points is d
Now we have to solve the question by using the distance formula
We move to the next step -2 to solve the question
Step 2
The distance formula is given below so that we can calculate the distance between the two points then
${\displaystyle d=\sqrt{{(x2-x1)}^2+{(y2-y1)}^2}}$
now the points are given that P1 and P2 which are (1.2, 2.3) and (-0.3, 1.1) respectively.
and ${\displaystyle \left(x1y1\right)=(1.2,2.3),\left(x2y2\right)=(-0.3,1.1)}$
so ${\displaystyle x1=1.2,y1=2.3,x2=-0.3,y2=1.1}$
now put the value into the distance formula
${\displaystyle d=\sqrt{{(x2-x1)}^2+{(y2-y1)}^2}}$
${\displaystyle x1=1.2,y1=2.3,x2=-0.3,y2=1.1}$
${\displaystyle d=\sqrt{{(-0.3-1.2)}^2+{(1.1-2.3)}^2}}$
${\displaystyle d=\sqrt{{(-1.5)}^2+{(-1.2)}^2}}$
${\displaystyle d=\sqrt{{\left(1.5\right)}^2+{\left(1.2\right)}^2}}$
${\displaystyle d=\sqrt{2.25+1.44}}$
${\displaystyle d=\sqrt{3.69}}$
${\displaystyle d=1.920937}$
the result is shown in the step-3
Step 3 Result
Result: From the above solution we got the distance between two points P1 and P2 are shown below
${\displaystyle d=1.920937}$ Answer

$d=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$
and $(x_1{y}_1)=(1.2,2.3),(x_2{y}_2)=(-0.3,1.1)$
so $x_1=1.2,y_1=2.3,x_2=-0.3,y_2=1.1$
$\begin{array}{}d=\sqrt{(x2-x1{)}^2+(y2-y1{)}^2}\\ x_1=1.2,y_1=2.3,x_2=-0.3,y_2=1.1\\ d=\sqrt{(-0.3-1.2{)}^2+(1.1-2.3{)}^2}\\ d=\sqrt{(-1.5{)}^2+(-1.2{)}^2}\\ d=\sqrt{(1.5{)}^2+(1.2{)}^2}\\ d=\sqrt{2.25+1.44}\\ d=\sqrt{3.69}\\ d=1.920937\end{array}$