Straight Line and Pair of Straight Lines 1 Question 16
16. If the line $3 x+4 y-24=0$ intersects the $X$-axis at the point $A$ and the $Y$-axis at the point $B$, then the incentre of the triangle $O A B$, where $O$ is the origin, is
(2019 Main, 10 Jan I)
(a) $(4,3)$
(b) $(3,4)$
(c) $(4,4)$
(d) $(2,2)$
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Solution:
- Given equation of line is
$$ 3 x+4 y-24=0 $$
For intersection with $X$-axis put $y=0$
$$ \begin{aligned} \Rightarrow & & 3 x-24 & =0 \\ \Rightarrow & & x & =8 \end{aligned} $$
For intersection with $Y$-axis, put $x=0$
$$ \Rightarrow \quad 4 y-24=0 \Rightarrow y=6 $$
$\therefore \quad A(8,0)$ and $B(0,6)$
Let $A B=c=\sqrt{8^{2}+6^{2}}=10$
$O B=a=6$
and $O A=b=8$
Also, let incentre is $(h k)$, then
$$ \begin{aligned} h & =\frac{a x _1+b x _2+c x _3}{a+b+c} \quad\left(\text { here, } x _1=8, x _2=0, x _3=0\right) \\ & =\frac{6 \times 8+8 \times 0+10 \times 0}{6+8+10}=\frac{48}{24}=2 \end{aligned} $$
and $\quad k=\frac{a y _1+b y _2+c y _3}{a+b+c} \quad\left(\right.$ here, $\left.y _1=0, y _2=6, y _3=0\right)$
$$ =\frac{6 \times 0+8 \times 6+10 \times 0}{6+8+10}=\frac{48}{24}=2 $$
$\therefore$ Incentre is $(2,2)$.
