Question

find points on the y axis which are at a distance of 10 units from the point (8,8).​

Answer 1

So this is a geometry question.

To solve this question ,

We need to the distance formula i.e. ,

$\sqrt{(x-a)^{2} +(y-b)^{2}}$ = d

where x , y , a , b are the coordinates and d is distance between them. [(x , y) and (a , b)]

We need to find the points on y -axis which are at a distance of 10 units from the point (8 , 8).

The points of y-axis will be of the form of = (0 , y)

Now,

Using distance formula,

$\sqrt{(x-a)^{2} +(y-b)^{2}}$ = d

where ,

x=0 , y=y , a=8 and b=8

$\sqrt{(0-8)^{2}+(y-8)^{2}}$ = 10

Squaring both sides,

$(-8)^{2}$ + $(y-8)^{2}$ = 100

64 + $y^{2}$ + 64 - 16y = 100

$y^{2}$ - 16y + 128 = 100

$y^{2}$ -16y + 28 = 0

Here , either you can factorize the equation or you can use quadratic formula i.e. (-b ± $\sqrt{b^{2} - 4ac}$)/ 2a

On solving , we get

y = 14 or 2

∴ the points on y-axis which are at a distance of 10 units from (8 , 8) are :

(0 , 2) and (0 , 14)

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I hope you understand.

Gracias.

Answer 2

Answer:

you're welcome to ....