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P-2 Co-ordinate Geometry — Maths STD 10 — Question

Maharashtra BoardEnglish MediumSTD 10MathsP-2 Co-ordinate Geometry4 Marks
Question
In the following examples, can the segment joining the given points form atriangle? If triangle is formed, state the type of the triangle considering sides ofthe triangle.
(1) L(6,4), M(-5,-3) , N(-6,8)
(2) P(-2,-6), Q(-4,-2), R(-5,0)
(3) $A (\sqrt{2}, \sqrt{2}), B (-\sqrt{2},-\sqrt{2}), C (-\sqrt{6}, \sqrt{6})$
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Answer

According to the distance formula, the distance 'd' between two points $(a, b)$ and $(c, d)$ is given by
$d=\sqrt[2]{(a-c)^2+(b-d)^2} \ldots . .(1)$
$\text { 1. } LM =\sqrt{(6+5)^2+(4+3)^2}=\sqrt{170}$
$M N=\sqrt{(-6+5)^2+(8+3)^2}=\sqrt{122}$
$NL =\sqrt{(-6-6)^2+(8-4)^2}=\sqrt{160}$
As sum of any two sides are greater than the third side,
The following points form a scalene triangle.
$\text { 2. } P Q=\sqrt{(-4+2)^2+(-2+6)^2}=\sqrt{20}$
$Q R=\sqrt{(-5+4)^2+(0+2)^2}=\sqrt{5}$
$R P=\sqrt{(-5+2)^2+(0+6)^2}=\sqrt{45}$
As $P Q+Q R$ The following points donot form a triangle.
$\text { 3. } A B=\sqrt{\left.((-\sqrt{2})-(\sqrt{2}))^2+(-(\sqrt{2})-(\sqrt{2}))^2\right)}=4$
$B C==\sqrt{\left.((-\sqrt{6})-(-\sqrt{2}))^2+((\sqrt{6})-(-\sqrt{2}))^2\right)}=4$
$A C=\sqrt{\left.((-\sqrt{6})-(\sqrt{2}))^2+((\sqrt{6})-(\sqrt{2}))^2\right)}=4$
As $A B=B C=A C$
The following points form an equilateral triangle.

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