Question

The equation of the incircle of the triangle formed by the axes and the line 4x+3y=6 is

a)x2+y2-6x-6y+9=0

b)4(x2+y2-x-y)+1=0

c)4(x2+y2+x+y)+1

d)none of these

Answer 1

The equation of the in circle of the triangle formed by the axes and the line 4x+3y=6 is 4[x^2 + y^2 - x -y ] + 1 = 0

Option (b) is correct.

Step-by-step explanation:

[ 4r + 3r - 6 / 5 ] = r

( 7r - 6)^2 = 25r^2

48r^2 - 84 r + 36 = 25 r^2

4r^2 - 14r + 6 = 0

By taking 2 commom

2r^2 - 7r +3 = 0

+7 ± √ 49 - 4.4.3 / 2.2

=> + 7 ± √ 25 / 4

= + 7 ± 5 / 4

r = 3 , r = 1/3

4x + 3y - 6 = 0

12 + 9 -6 =  15

0 + 0 -6 => - 6

( x - 1/2)^2 + ( y - 1/2)^2 = 1/4

x^2 + 1/4 -x + y^2 + 1/4 - y = 1/4

4[x^2 + y^2 - x -y ] + 1 = 0

Thus The equation of the in circle of the triangle formed by the axes and the line 4x+3y=6 is 4[x^2 + y^2 - x -y ] + 1 = 0

Answer 2

Answer:

 4(x²+y²−x−y)+1=0