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Using the distance formula, show that the given points are collinear: (iii) $(-1,-1),(2,3)$ and $(8,11)$ (iv) $(-2,5),(0,1)$ and $(2,-3)$
Using the distance formula, show that the given points are collinear: (iii) $(-1,-1),(2,3)$ and $(8,11)$ (iv) $(-2,5),(0,1)$ and $(2,-3)$
CBSE | Class 10 | Coordinate Geometry | Exercise 16.1 | Maths | RS Aggarwal
Using the distance formula, show that the given points are collinear: (iii) $(-1,-1),(2,3)$ and $(8,11)$ (iv) $(-2,5),(0,1)$ and $(2,-3)$

(iii) Let $\mathrm{A}(-1,-1), \mathrm{B}(2,3)$ and $\mathrm{C}(8,11)$ be the give points. Then

$$
\begin{aligned}
&\mathrm{AB}=\sqrt{(2+1)^{2}+(3+1)^{2}}=\sqrt{(3)^{2}+(4)^{2}}=\sqrt{25}=5 \text { units } \\
&\mathrm{BC}=\sqrt{(8-2)^{2}+(11-3)^{2}}=\sqrt{(6)^{2}+(8)^{2}}=\sqrt{100}=10 \text { units } \\
&\mathrm{AC}=\sqrt{(8+1)^{2}+(11+1)^{2}}=\sqrt{(9)^{2}+(12)^{2}}=\sqrt{225}=15 \text { units } \\
&\therefore \mathrm{AB}+\mathrm{BC}=(5+10) \text { units }=15 \text { units }=\mathrm{AC}
\end{aligned}
$$

Therefore, the given points are collinear

(iv) Let $\mathrm{A}(-2,5), \mathrm{B}(0,1)$ and $\mathrm{C}(2,-3)$ be the give points. Then

$$
\begin{aligned}
&\mathrm{AB}=\sqrt{(0+2)^{2}+(1-5)^{2}}=\sqrt{(2)^{2}+(-4)^{2}}=\sqrt{20}=2 \sqrt{5} \text { units } \\
&\mathrm{BC}=\sqrt{(2-0)^{2}+(-3-1)^{2}}=\sqrt{(2)^{2}+(-4)^{2}}=\sqrt{20}=2 \sqrt{5} \text { units } \\
&\mathrm{AC}=\sqrt{(2+2)^{2}+(-3-5)^{2}}=\sqrt{(4)^{2}+(-8)^{2}}=\sqrt{80}=4 \sqrt{5} \text { units } \\
&\therefore \mathrm{AB}+\mathrm{BC}=(2 \sqrt{5}+2 \sqrt{5}) \text { units }=4 \sqrt{5} \text { units }=\mathrm{AC}
\end{aligned}
$$

Therefore, the given points are collinear

i

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