Question

What is the distance between $P$ and $Q$ and the coordinates of the midpoint of the line segment $PQ$ if $P(7,6)$ ,$Q(7,2)$?

Answer 1

Hint: The midpoint of a line segment is the point in the centre of the segment, this means that the midpoint cuts the section in half . It is the centroid of both the segment and the endpoints, and it is equidistant from both.
The distance between the points can be found by using the coordinates of the points.
Formula used:
The distance formula:
$d = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} $Here,
$d$ is the distance between the points.
$(x_1,y_1)$ are the coordinates of the first point and $(x_2,y_2)$ are the coordinates of the second point.
The midpoint formula :
The midpoint of a line segment is given by :
$(\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2})$
Here,
$(x_1,y_1)$ are the coordinates of the first point and $(x_2,y_2)$ are the coordinates of the second point.

Complete step-by-step solution:
Here the coordinates of the two points $P$ and $Q$ are $(7,6)$and $(7,2)$.
This means that,
${x_1} = 7$
${y_1} = 6$
${x_2} = 7$
${y_2} = 2$
To find the distance between $P$ and $Q$, the distance formula can be used which is:
$d = \sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}} $
Substituting the values,
$d = \sqrt {{{(7 - 7)}^2} + {{(6 - 2)}^2}} = \sqrt {({0^2} + {4^{2)}}} = \sqrt {16} = 4$
Thus $d = 4$
To find the midpoint of the line segment we can use the midpoint formula:
The midpoint of a line segment is given by :
$(\dfrac{{{x_1} + {x_2}}}{2},\dfrac{{{y_1} + {y_2}}}{2})$
Thus, the midpoint of $P$ and $Q$ is : $\left( {\dfrac{{7 + 7}}{2},\dfrac{{6 + 2}}{2}} \right) = \left( {\dfrac{{14}}{2},\dfrac{8}{2}} \right) = \left( {7,4} \right)$
Thus, the distance between $P$ and $Q$ is $d = 4$and the midpoint of $P$ and $Q$ is $(7,4)$.

Note: The midpoint formula allows you to determine the exact middle of two points.
The distance formula, which is derived from the Pythagorean Theorem, is used to calculate the distance between two points in the plane.