Question

Please help! find the distance of AG if A (4,4) and G (-1,-1)

Answer 1

• The distance between two points
  `\rm{A(x_1,y_1)}` and
  `\rm{B(x_2,y_2)}` is given by the formula,

`\rm{AB= √((x_2-x_1)^2+(y_2-y_1)^2) }`

Proof:

`\rm{Let \: X'OX \: and \: YOY' \: be \: the \: z-axis \: and \: y-axis \: respectively. Then, O \: is \: the \: origin.}`

`\rm{Let \: A(x_1,y_1) \: and \: B(x_2,y_2) \: be \: the \: given \: points.}`

`\rm{Draw \: AL \perp \: OX, BM \perp \: OX \: and \: AN \perp BM }`

Now,

`\rm{OL=x_1,OM=x_2,AL=y_1 \: and \: BM=y_2}`

`\rm\therefore{AN=LM=(OM-OL)=(x_2-x_1)}`

`\: \: \: \: \rm{BN=(BM-NM)=(BM-AL)=(y_2-y_1)}`

`\rm{In \: right \: angled \: \triangle ANB, by \: Pythagorean \: theorem,}`

We have,

`\: \: \: \: \rm{AB^2=AN^2+BN^2}`

`\rm{or,AB^2=(x_2-x_1)^2+(y_2-y_1)^2}`

`\rm\therefore AB= √((x_2-x_1)^2+(y_2-y_1)^2)`

The Given question,

Find the distance of AG if A (4,4) and G (-1,-1).

Solution,

The given points are A(4,4) and G(-1,-1).

Then,

`\rm{(x_1=4,y_1=4) and (x_2=-1,y_2=-1)}`

We know that,

The distance formula,

`\rm{AG= √((x_2-x_1)^2+(y_2-y_1)^2) }`

`\: \: \: \: \: \: \: \: = \sqrt{ {( - 1 - 4)}^(2) + {( - 1 - 4)}^(2) }`

`\: \: \: \: \: \: \: \: = \sqrt{ {( - 5)}^(2) + {( - 5)}^(2) }`

`\: \: \: \: \: \: \: \: = √(25 + 25)`

`\: \: \: \: \: \: \: \: = √(50)`

`\: \: \: \: \: \: \: \: = √((2)(25))`

`\: \: \: \: \: \: \: \: = √((2)(5)(5))`

`\: \: \: \: \: \: \: \: = \sqrt{(2)( {5}^(2)) }`

`\rm \: \: \: \: \: \: \: \: = 5 √(2) \: units`

Answered by:

`\frak{\red{moonlight123429}}`

Answer 2

`{\qquad\qquad\huge\underline{{\sf Answer}}}`

Let's solve ~

By using distance formula :

`\qquad \sf  \dashrightarrow \: \sqrt{(x_2 - x_1) {}^(2) + (y_2 - y_1) {}^(2) }`

`\qquad \sf  \dashrightarrow \: \sqrt{(4 - ( - 1)) {}^(2) + (4 - ( - 1)) {}^(2) }`

`\qquad \sf  \dashrightarrow \: \sqrt{(4 + 1) {}^(2) + (4 + 1) {}^(2) }`

`\qquad \sf  \dashrightarrow \: \sqrt{2(5) {}^(2) }`

`\qquad \sf  \dashrightarrow \: 5 √(2 )\: \: units`