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)# 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]}