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

2 Answers
Oct 12, 2017

Let's see.

Explanation:

Let the coordinates of two points PP & QQ be (x_1,y_1)(x1,y1) & (x_2,y_2)(x2,y2) respectively.

Therefore the distance between the two points is rarr

PQ=sqrt((y_2-y_1)^2+(x_2-x_1)^2)PQ=(y2y1)2+(x2x1)2

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

d=sqrt((8-3)^2+(7+2)^2)d=(83)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:)

Oct 12, 2017

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")

Tony BTony B

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