Question

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

Answer 1

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

Let's solve ~

By using distance formula :

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

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

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

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

$\qquad \sf  \dashrightarrow \: 5 \sqrt{2 }\: \: units$

Answer 2

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

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

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,

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

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

$\rm\therefore AB= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

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,

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

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

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

$\: \: \: \: \: \: \: \: = \sqrt{25 + 25}$

$\: \: \: \: \: \: \: \: = \sqrt{50}$

$\: \: \: \: \: \: \: \: = \sqrt{(2)(25)}$

$\: \: \: \: \: \: \: \: = \sqrt{(2)(5)(5)}$

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

$\rm \: \: \: \: \: \: \: \: = 5 \sqrt{2} \: units$

Answered by:

$\frak{\red{moonlight123429}}$