Question

find co –ordinates of points on y axis which are at a distance of 10 units from the point (-8,4)

Answer 1

The co-ordinates of points on y axis is generally (0,y),

Let a point on y axis be (0,y),

According to the question, Distance between (0,y) and (-8,4) is 10 units,

Generally, Distance between two points (a,b) and (c,d) is

$\sqrt{ (a-c)^{2}+(b-d)^{2} }$,

Now using this formula let us find the co-ordinate, By the formula we can write that,

$\sqrt{ (-8-0)^{2} + (4-y)^{2}}$,

=> 64 + (4-y)² = 100,
=> (4-y)² = 36,
=> (4-y) = either +6 or -6,
=> y = either -2 or 10,

So there are 2 co-ordinates on y-axis whose distance from (-8,4) = 10 units, They are (0,-2) and (0,10),

Hope you understand, Have a great day !

Answer 2

$10 = \sqrt{(x + 8) ^{2} + (y - 4) ^{2} } \\ 10 = \sqrt{ {x}^{2} + 16x + 64 + {y}^{2} - 8y + 16 } \\ 10 = \sqrt{ {0}^{2} + {y}^{2} + 16 \times 0 + 64 + - 8y + 16 } \\ {10}^{2} = {y}^{2} - 8y + 64 \\ 100 - 64 = {y}^{2} - 8y \\ {y}^{2} - 8y - 36 = 0$
find y from given information