Question

Find the distance of the point (1,2) from the mid-point of the line segment joining the points (6,8) and (2,4).

Answer 1

Hint : To solve the question given above, we will first find out what are mid-points between any two points. Then, we will apply the formula of midpoint between two points given by: $P\left( x,y \right)=\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)$ . With the help of this formula, we will find the midpoint of (6,8) and (2,4). Then, we will find the distance between the obtained midpoint and (1,2) with the help of the distance formula given by:
$AB=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$

Complete step-by-step answer :
Before we start to solve the question given above, we must know what is a midpoint. A midpoint of two points in a plane is the point which is equidistant from both the points i.e. its distance from both the points is same and it lies on a line joining the two points. If two points $\left( {{x}_{1}},{{y}_{1}} \right)\ and\ \left( {{x}_{2}},{{y}_{2}} \right)$ are given then the coordinates of their mid-point (x,y) is given by the formula shown:
$\left( x,y \right)=\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2} \right)$

Let the mid points of the points (6,8) and (2,4) is denoted by P and has coordinates P(a,b), then we can say that:
$\begin{aligned} & P\left( a,b \right)=\left( \dfrac{6+2}{2},\dfrac{8+4}{2} \right) \\\ & P\left( a,b \right)=\left( \dfrac{8}{2},\dfrac{12}{2} \right) \\\ & P\left( a,b \right)=\left( 4,6 \right) \\\ \end{aligned}$
Thus, the mid-point of (6,8) and (2,4) is P(4,6).
Let us assume that the point (1,2) is denoted by Q. Thus, we have to find the distance PQ. We know that the distance between two points $A\left( {{x}_{1}},{{y}_{1}} \right)\ and\ B\left( {{x}_{2}},{{y}_{2}} \right)$ can be obtained with the help of distance formula as shown:
$AB=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$
Thus, we can say that:
$\begin{aligned} & PQ=\sqrt{{{\left( 4-1 \right)}^{2}}+{{\left( 6-2 \right)}^{2}}}units \\\ & \Rightarrow PQ=\sqrt{{{\left( 3 \right)}^{2}}+{{\left( 4 \right)}^{2}}}units \\\ & \Rightarrow PQ=\sqrt{9+16}units \\\ & \Rightarrow PQ=\sqrt{25}units \\\ & \Rightarrow PQ=5units \\\ \end{aligned}$
Thus, the distance between the midpoint of (6,8) and (2,4) and the point (1,2) is 5 units.

Note : The coordinates of midpoints of A(6,8) and B(2,4) can also be calculated by:
We know that P(x,y) is mid-point of A and B so A,P and B will lie on the same line. Thus, the slope of line formed by any two points would be the same. Therefore,
$\begin{aligned} & {{m}_{AB}}={{m}_{AP}} \\\ & \Rightarrow \dfrac{8-4}{6-2}=\dfrac{y-8}{x-6} \\\ & \Rightarrow \dfrac{4}{4}=\dfrac{y-8}{x-6} \\\ & \Rightarrow 1=\dfrac{y-8}{x-6} \\\ & \Rightarrow y-8=x-6 \\\ & \Rightarrow y=x-6+8 \\\ & \Rightarrow y=x+2.................\left( 1 \right) \\\ \end{aligned}$
Also, the distance PA will be equal to PB. Thus, we have:
$\begin{aligned} & PA=PB \\\ & \Rightarrow \sqrt{{{\left( x-6 \right)}^{2}}+{{\left( y-8 \right)}^{2}}}=\sqrt{{{\left( x-2 \right)}^{2}}+{{\left( y-4 \right)}^{2}}} \\\ \end{aligned}$
On squaring both sides, we will get:
$\begin{aligned} & {{\left( x-6 \right)}^{2}}+{{\left( y-8 \right)}^{2}}={{\left( x-2 \right)}^{2}}+{{\left( y-4 \right)}^{2}} \\\ & {{x}^{2}}-12x+36+{{y}^{2}}-16y+64={{x}^{2}}-4x+4+{{y}^{2}}-8y+16 \\\ & -12x-16y+100=-4x-8y+20 \\\ & 8x+8y=80 \\\ & \Rightarrow x+y=10...............\left( 2 \right) \\\ \end{aligned}$
From (1) and (2)
$\begin{aligned} & \Rightarrow x+x+2=10 \\\ & \Rightarrow 2x+2=10 \\\ & \Rightarrow 2x=8 \\\ & \Rightarrow x=4 \\\ \end{aligned}$
On putting the value of x = 4 in (1), we will get:
$\begin{aligned} & \Rightarrow y=4+2 \\\ & \Rightarrow y=6 \\\ \end{aligned}$
Thus, mid-point is P(4,6).