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Question: Find the distance between the following pair of points: \(\left( a,b \right);\left( c,b \right)\)....

Find the distance between the following pair of points: (a,b);(c,b)\left( a,b \right);\left( c,b \right).

Explanation

Solution

We first use the formula of distance between two points (x,y)\left( x,y \right) and (m,n)\left( m,n \right) as d=(xm)2+(yn)2d=\sqrt{{{\left( x-m \right)}^{2}}+{{\left( y-n \right)}^{2}}} for the general form. We then place the values for points (a,b)\left( a,b \right) and (c,b)\left( c,b \right) to find the approximate value of the distance.

Complete step by step answer:
We have to find the distance between points (a,b)\left( a,b \right) and (c,b)\left( c,b \right).
We first find the general formula for distance between two arbitrary points.
We take two points (x,y)\left( x,y \right) and (m,n)\left( m,n \right).
The formula for distance between those two points will be d=(xm)2+(yn)2d=\sqrt{{{\left( x-m \right)}^{2}}+{{\left( y-n \right)}^{2}}}.
For our given points (a,b)\left( a,b \right) and (c,b)\left( c,b \right), we put the values for x=a,y=bx=a,y=b and m=c,n=bm=c,n=b.
Therefore, the distance between those two points is d=(ac)2+(bb)2d=\sqrt{{{\left( a-c \right)}^{2}}+{{\left( b-b \right)}^{2}}}.
Simplifying we get d=(ac)2+0=(ac)2=acd=\sqrt{{{\left( a-c \right)}^{2}}+0}=\sqrt{{{\left( a-c \right)}^{2}}}=\left| a-c \right| units.
Therefore, the distance between the following pair of points: (a,b);(c,b)\left( a,b \right);\left( c,b \right) is ac\left| a-c \right| units.

We take an actual example to simplify the explanation.
Let’s assume we have to find the distance between points (7,11)\left( -7,11 \right) and (8,11)\left( 8,11 \right).
For our given points (7,2)\left( -7,2 \right) and (11,5)\left( 11,-5 \right), we put the values for a=7,b=11a=-7,b=11 and c=8,b=11c=8,b=11.
Therefore, the distance between those two points is d=(78)2+(1111)2=(78)2d=\sqrt{{{\left( -7-8 \right)}^{2}}+{{\left( 11-11 \right)}^{2}}}=\sqrt{{{\left( -7-8 \right)}^{2}}}
Simplifying we get d=78=15d=\left| -7-8 \right|=15 units.

Note: We need to remember that when we are taking the modulus value because distance cannot be negative as it is a scalar value. The negative sign indicates the direction only. So, the modulus value changes it to its value to give the distance.