Question

The abscissa of a point A is equal to its ordinate, and its distance from the point B(1, 3)
is 10 units, What are the coordinates of A?

Answer 1

"Given

abscissa = ordinate

let x = y = k. then A (x,y) = (k,k)

B (1,3)

condition is

distance of AB = 10

squaring on both sides

(AB)^2 = 100

{(k - 1)}^{2}  +  {(k - 3)}^{2}  = 100 \\  {k}^{2}  + 1 - 2k +  {k}^{2}  + 9 - 6k = 100 \\ 2 {k}^{2}  - 8k + 10 = 100 \\  {k}^{2} -  4k + 5 = 50 \\  {k}^{2}  - 4k - 45 = 0 \\  {k}^{2}   + 5k  - 9k - 45 = 0 \\  k(k + 5) - 9(k + 5) = 0 \\ (k + 5)(k - 9) = 0 \\ k =  - 5 \: or \: 9

therefore k = -5 or 9

so A(x,y) can be (-5,-5) or (9,9)"

Answer 2

Answer:

Possible coordinates are (-5, 5) or(9,9)

Step-by-step explanation:

Let a be the abscissa of point A.

So, the coordinate of A is (a, a)

distance of A from B is 10 units.

By distance formula, we get

10  = $\sqrt{(1 -a)^{2}  + (3-a)^{2} }$

Square both sides, we get

100  = $(1 - a)^{2}  + (3 - a)^{2}$

or,  $100  = ( 1  + a^{2} - 2a)  +   ( 9 + a^{2} - 6a)\\ 100 = 10 - 8a  + 2a^{2} \\or,   2a^{2} - 8a - 90 = 0 \\simplifying,    a^{2}  -4a  - 45 = 0$

Here, solving for a , we get

(a-9)(a+5) = 0

it implies either a = 9 or a =-5

So the possible coordinates are (-5, -5) or (9,9)