If the incentre and the circumcentre of the t | vedclass

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(B) The lines are $L_1: x=2$,$L_2: y=3$,and $L_3: 4x+3y+7=0$.The vertices are found by intersection:$A = L_2 \cap L_3: y=3$ $\Rightarrow 4x+9+7=0$ $\R

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$\sqrt{5}$

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(B) The lines are $L_1: x=2$,$L_2: y=3$,and $L_3: 4x+3y+7=0$.The vertices are found by intersection:$A = L_2 \cap L_3: y=3$ $\Rightarrow 4x+9+7=0$ $\Rightarrow 4x=-16$ $\Rightarrow x=-4$. So $A=(-4, 3)$.$B = L_1 \cap L_2: x=2, y=3$. So $B=(2, 3)$.$C = L_1 \cap L_3: x=2$ $\Rightarrow 8+3y+7=0$ $\Rightarrow 3y=-15$ $\Rightarrow y=-5$. So $C=(2, -5)$.The side lengths are $c = AB = \sqrt{(2-(-4))^2 + (3-3)^2} = 6$,$a = BC = \sqrt{(2-2)^2 + (-5-3)^2} = 8$,and $b = AC = \sqrt{(2-(-4))^2 + (-5-3)^2} = \sqrt{6^2 + (-8)^2} = 10$.The incentre $I = \left(\frac{ax_1+bx_2+cx_3}{a+b+c}, \frac{ay_1+by_2+cy_3}{a+b+c}ight) = \left(\frac{8(-4)+10(2)+6(2)}{8+10+6}, \frac{8(3)+10(3)+6(-5)}{8+10+6}ight) = \left(\frac{-32+20+12}{24}, \frac{24+30-30}{24}ight) = (0, 1)$.Since $	riangle ABC$ is a right-angled triangle at $B(2, 3)$,the circumcentre $S$ is the midpoint of the hypotenuse $AC$.$S = \left(\frac{-4+2}{2}, \frac{3-5}{2}ight) = (-1, -1)$.The distance $IS = \sqrt{(0-(-1))^2 + (1-(-1))^2} = \sqrt{1^2 + 2^2} = \sqrt{5}$.

If the incentre and the circumcentre of the triangle formed by the lines $x=2$,$4x+3y+7=0$ and $y=3$ are $I$ and $S$ respectively,then $IS=$

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