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?

1 Answer
Jan 29, 2018

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

Explanation:

If we draw an arc of radius xx 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 dd, then we get one point if x=dx=d, two points ifx>dx>d and no point if x<dx<d.

Here distance is |(0-2+1)/sqrt2|=sqrt202+12=2 and as 5>sqrt25>2, we should get two points.

Now the point (h,k)(h,k) les on line y=x+1y=x+1, its ordinate is 11 unit more than its abscissa and hence k=h+1k=h+1 i.e. we can write point as (h,h+1)(h,h+1).

Its distance from (0,2)(0,2) is 55 units. In other words

sqrt((h-0)^2+(h+1-2)^2)-5(h0)2+(h+12)25

or h^2+(h-1)^2=25h2+(h1)2=25

or 2h^2-2h-24=02h22h24=0

or (2h+6)(h-4)=0(2h+6)(h4)=0

i.e. h=-3h=3 or 44

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

if h=4h=4, k=5k=5 and point is (4,5)(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]}