The line L: 3x + 4Y + 24 = 0 intersects with the x-axis and y-axis at two points a and B respectively, and O is the coordinate origin. The equation of the inscribed circle of the triangle AOB is obtained

The line L: 3x + 4Y + 24 = 0 intersects with the x-axis and y-axis at two points a and B respectively, and O is the coordinate origin. The equation of the inscribed circle of the triangle AOB is obtained

The solution is that the line L: 3x + 4Y + 24 = 0 makes x = 0, i.e. y = - 6 makes y = 0, i.e. x = - 8, i.e. the line L: 3x + 4Y + 24 = 0 intersects with the x-axis, Y-axis intersects a (- 8,0), B (0, - 6) from ∠ AOB = 90 ° i.e. AB & # 178; = OA & # 178; + ob & # 178; i.e. AB = 10, so the inscribed circle radius of Δ ABO is r = (OA + ob-ab) / 2 = (6 + 8-10) / 2 = 2

The line 3x + 4y-12 = 0 intersects with X axis and Y axis at two points a and B respectively. O is the origin coordinate. How to find the perimeter of triangle OAB?

First, find the coordinates of the intersection a and B of the line 3x + 4y-12 = 0 and the X and Y axes,
When x = 0, y = 3 B (0,3)
When y = 0, x = 4, a (4,0)
So, OA length is 4, OB length is 3, according to the formula of oblique side length of triangle
If the square of AB = the square of OA + the square of ob, then AB = 5
So the perimeter of triangle OAB is: 3 + 4 + 5 = 12

If the line 3x + 4y-12 = 0 intersects the x-axis at point a and the y-axis at point B, and O is the coordinate origin, then what is the standard equation of the inscribed circle of the triangle OAB,
Urgent, urgent!

It is easy to get a coordinate (4,0), B coordinate (0,3), ab = 5
Let the radius of the inscribed circle be r
r=(4+3-5)/2=1
So, the coordinates of the center of the circle are (1,1)
The equation is (x-1) ^ 2 + (Y-1) ^ 2 = 1