It is known that the distance between points a (a, 2) and B (2, a + 1) is equal to the root sign 5, and the value of a is obtained

It is known that the distance between points a (a, 2) and B (2, a + 1) is equal to the root sign 5, and the value of a is obtained

There are two distance formulas: root (A-2) ^ 2 + (2-a-1) ^ 2 = root 5
(A-2) ^ 2 + (2-a-1) ^ 2 = 5, that is, 2A ^ 2-6a = 0
A = 0 or 3,
The distance formula of two points is: let two points be (x1, Y1) (X2, Y2)
D = root (x1-x2) ^ 2 + (y1-y2) ^ 2

A. Two points of B are on the straight line y = X-1, and the difference of abscissa between a and B is the root sign 2

The slope is 1. The distance is twice the root of root 2
Distance = 2
Draw an isosceles right triangle ABC, AC is the abscissa distance of AB, AC = root 2, AB must be 2

It is known that the distances from two points a (1,2) and B (3,1) to the line L are radical 2, radical 5-radical 2 respectively
Then the number of lines L satisfying the condition

1
reason:
|Ab | = under radical (1 ^ 2 + 2 ^ 2) = radical 5
And the sum of the distances from a and B to the line L = radical 2 + radical 5 - radical 2 = radical 5
So the line l must be perpendicular to AB and between a and B,
There is only one