Current location - Training Enrollment Network - Mathematics courses - The second day of the eighth day, about the knowledge of linear function, help the little sister. thank you
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).