Question

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

पर्याय

• x² + y² − 6x −6y + 9 = 0 

• 4 (x² + y² − x − y) + 1 = 0

• 4 (x² + y² + x + y) + 1 = 0

• none of these

Answer 1

 4 (x² + y² − x − y) + 1 = 0
The line 4x + 3y = 6 cuts the coordinate axes at \[\left( \frac{3}{2}, 0 \right) \text { and } \left( 0, 2 \right)\]

The coordinates of the incentre is

\[\left( \frac{a x_1 + b x_2 + c x_3}{a + b + c}, \frac{a y_1 + b y_2 + c y_3}{a + b + c} \right)\].

Here,

\[a = \frac{5}{2}, b = \frac{3}{2}, c = 2, x_1 = 0, y_1 = 0, x_2 = 0, y_2 = 2, x_3 = \frac{3}{2}, y_3 = 0\]

Thus, the coordinates of the incentre:

\[\left( \frac{0 + 0 + 3}{6}, \frac{0 + 3 + 0}{6} \right)\]

\[ = \left( \frac{1}{2}, \frac{1}{2} \right)\]

The equation of the incircle:

\[\left( x - \frac{1}{2} \right)^2 + \left( y - \frac{1}{2} \right)^2 = a^2\]

Also, radius of the incircle = \[\frac{\sqrt{s\left( s - a \right)\left( s - b \right)\left( s - c \right)}}{s}\]

Here, \[s = \frac{a + b + c}{2} = \frac{\frac{5}{2} + \frac{3}{2} + 2}{2} = 3\]

∴ Radius of the incircle = \[\frac{\sqrt{3\left( 3 - a \right)\left( 3 - b \right)\left( 3 - c \right)}}{3}\]

\[= \frac{\sqrt{3\left( 3 - \frac{5}{2} \right)\left( 3 - \frac{3}{2} \right)\left( 3 - 2 \right)}}{3}\]

\[ = \frac{\sqrt{3\left( \frac{1}{2} \right)\left( \frac{3}{2} \right)}}{3}\]

\[ = \frac{1}{2}\]

The equation of circle: \[\left( x - \frac{1}{2} \right)^2 + \left( y - \frac{1}{2} \right)^2 = \frac{1}{4}\]

\[\Rightarrow 4\left( x^2 + y^2 - x - y \right) + 1 = 0\]