Question

a point whose abscissa and ordinate are equalis square root of 10 units from (-1,-3). find the points

Answer 1

\[\sqrt{(x+1)^{2}+(y+3)^{2}}=\sqrt{10}\]

Answer 2

do u have any options or u need to solve it completely

Answer 3

?

Answer 4

ahmmm.. i justwant to know HOW to solve that question :))

Answer 5

see as i have the equation...........one way is you can arbitrary values of x and y and check when the left hand side is equal to square root of 10

Answer 6

**arbitrarily substitute

Answer 7

like it can be (2,-2)

Answer 8

errr.. the books says it was 0,0 and -4, -4

Answer 9

ya even both these points satisfy the given equation

Answer 10

i just gave u an example ......

Answer 11

but the condition is both the abscissa and the ordinate are equal.. :))

Answer 12

ok ...............got in this equation put x=y .............you will get a quadratic in x nad then solve to get two values of x and two values of y

Answer 13

(x+1)^2+(x+3)^2=10
\[2x ^{2}+8x+10=10\]
\[2x ^{2}+8x=0\]
2x(x+4)=0
x=0 or x=-4
since x=y,
thus,(0,0) and (-4,-4)

Answer 14

hmmmmmm... i will consider that.. by the way, thanks! i now have a direction on how to solve this stuff.. i will inform you if i solve that stuff :))

Answer 15

see you had two equations::
\[\sqrt{(x+1)^{2}+(y+3)^{2}}=\sqrt{10}\]
and x=y

Answer 16

wiow! :0

Answer 17

wow! now i know wheres my error .. thank you so much! you really help me in this stuff :D

Answer 18

ur welcome......