Question

The point (h,k) lies on the line y=x+1 and is 5 units from the point (0,2). What is the two equations connecting h and k and hence the possible values of h and k? I have written my answers below, can someone double check them?

1) k= h + 1
2) k = 2h + 5
*I used simultaneous to solve the equation
ANSWERS, h=-4 and k=5
Is this correct?

Answer 1

There are two points - `(-3,-2)` and `(4,5)`.

Explanation:

If we draw an arc of radius `x` units from a point, it cn cut the line in two points, one point or not at all. This depends on the distance of point from line. If this distance is `d`, then we get one point if `x=d`, two points if`x>d` and no point if `x<d`.

Here distance is `|(0-2+1)/sqrt2|=sqrt2` and as `5>sqrt2`, we should get two points.

Now the point `(h,k)` les on line `y=x+1`, its ordinate is `1` unit more than its abscissa and hence `k=h+1` i.e. we can write point as `(h,h+1)`.

Its distance from `(0,2)` is `5` units. In other words

`sqrt((h-0)^2+(h+1-2)^2)-5`

or `h^2+(h-1)^2=25`

or `2h^2-2h-24=0`

or `(2h+6)(h-4)=0`

i.e. `h=-3` or `4`

If `h=-3`, `k=-2` and point is `(-3,-2)` and

if `h=4`, `k=5` and point is `(4,5)`.

graph{((x-4)^2+(y-5)^2-0.03)(x^2+(y-2)^2-0.03)((x+3)^2+(y+2)^2-0.03)(x-y+1)=0 [-9.67, 10.33, -3.56, 6.44]}