The second day of the eighth day, about the knowledge of linear function, help the little sister. thank you

1) If the coordinates of point A are directly substituted into the linear equation, you can get m=3-2= 1.

Then substitute it into hyperbolic equation to get k=3× 1=3.

2) There are three such Qs.

First of all, OA=√(3? + 1? )=√ 10

(1) Let OQ=OA=√ 10, then the coordinate of Q point is (√ 10,0).

(--√ 10/0,0) is ignored because q is on the positive semi-axis of the X axis.

② Let OA=AQ. According to the properties of the three lines of an isosceles triangle, it is easy to get the coordinate of point Q as (6,0).

③ Let OQ=AQ, and let the coordinates of Q point be (m, 0), m>0.

So m=√[(3-m)? + 1? ]

The solution is m=5/3.

That is, the coordinate of point Q is (5/3,0).