Question

How do you find the distance between `(7, 8)` and `(-2, 3)`?

Answer 1

Let's see.

Explanation:

Let the coordinates of two points `P` & `Q` be `(x_1,y_1)` & `(x_2,y_2)` respectively.

Therefore the distance between the two points is `rarr`

`PQ=sqrt((y_2-y_1)^2+(x_2-x_1)^2)`

Hence, the distance between `(7,8)` & `(-2,3)` is `rarr`

`d=sqrt((8-3)^2+(7+2)^2)` units.

`:.d=sqrt(5^2+9^2)` units.

`:.d=sqrt(25+81)` units.

`:.d=sqrt(106)` units. (Answer).

Hope it Helps:)

Answer 2

This is why you use the method adopted by Aditya

Distance between points is `sqrt(106)" "` Exact answer
Distance between points is `10.296" "` Approximate answer

Explanation:

Looking at the graph below you will observe that you can use the line between the two points like the hypotenuse of a right triangle. Thus we may use Pythagoras the solve for the required length.

`color(magenta)("Always read a graph left to right on the x-axis")`

Set point 1 as `P_1->(x_1,y_1)=(-2,3)`
Set point 2 as `P_2->(x_2,y_2)=(7,8)`

Let the distance between the points be `d`

So the change horizontally is `x_2-x_1color(white)("d")=color(white)("d")7-(-2)color(white)("d")=color(white)("d")9`

and the change vertically is `y_2-y_1color(white)("d")=color(white)("d")8-3color(white)("d")=color(white)("d")5`

Now comparing to that of Aditya's method

Using Pythagoras

`d=(P_2-P_1)^2=(x_2-x_1)^2+(y_2-y_1)^2`

`d=P_2-P_1=sqrt( (x_2-x_1)^2+(y_2-y_1)^2)`

`d=P_2-P_1=sqrt(color(white)("ddd")9^2color(white)(".ddd")+color(white)("dd")5^2color(white)("dddd"))`

`d = sqrt(106)`

There are only two factors of 106 (other than 1 and 106) and they are prime numbers so we are not able to simplify `sqrt(106)` further.

Distance between points is `sqrt(106)" "` Exact answer
Distance between points is `10.296" "` Approximate answer