Find the coordinates of a point which is at a distance of 5 from the point (7,9) and 1 from the point (3,5).?

Answer 1

Let the coordinates of derised point be (p,q)

Then you get, (p-7)^2 + (q-9)^2 = 25 and (p-3)^2 +(q-5)^2 = 1

Simplifying, we get 2 quaderatic eqs then subtract to eliminate p^2 and q^2

> p+q = 9 ....... eq (3)

put the value of p (=9-q) in eq (1) and rearrange, we get quaderatic in q as

> q^2 - 11q +30 =0 its roots are q = 6 and 5 then put in eq. (3)

we get p = 3 and 4

so coordinates of desired points (there will be 2 such points):

(3,6) and (4,5)

I was too good at coordinate. Please read my article on modern measures to safeguard a cricket pitch during odi/test using coordinate geomentry on web site : www.cricbuzz.com
cut & paste

http://www.cricbuzz.com/Cricket_Articles/Just_Cricket/Modern_Measures_to_Safeguard_Cricket_Pitch_during_a_Test%10ODI_Match

Answer 2

well i tried this problem for abour an hour....i know 1 thing tht u hav 2 make 2 linear equations n eliminate 1...i got 1 of d linear equation as x + y = 8....i am still tryin 2 find out d second equation when i get i will tell ya...

Answer 3

can i pass this question, i am weak at co-ordinate geometry