चौरसाकृती मैदानाला चारपदरी तारेचे कुंपण करण्यासाठी २ किमी तार लावली होती मैदान सर्व बाजूंनी ५ मी वाढले तर नवीन मैदानाला चारपदरी तारेचे कुंपण करण्यासाठी आणखी किती तार आवश्यक आहे?
Answers
Answer:
We have to show whether the points P(1,2), Q(2, 8/5) and R (3,6/5) are collinear or not that is they lie on the same straight line by using the concept of slope.
we have 3 points,
P (1,2) , Q(2, \frac{8}{5} ) and R (3, \frac{6}{5})P(1,2),Q(2,
5
8
)andR(3,
5
6
)
Now, for P, Q and R to be collinear the slope m1 of the line from P to Q , the slope m2 of line from point Q to R and the slope m3 of line from point P to R should be equal, therefore
Solving for m1
\begin{lgathered}m_{1} = \frac{y_{2}-y_{1} }{x_{2} -x_{1} } \\= \frac{\frac{8}{5}-2 }{2-1} = \frac{8-10}{5} = -\frac{2}{5}\end{lgathered}
m
1
=
x
2
−x
1
y
2
−y
1
=
2−1
5
8
−2
=
5
8−10
=−
5
2
m_{1} = -\frac{2}{5}m
1
=−
5
2
- (1)
Solving for m2
\begin{lgathered}m_{2} = \frac{y_{2}-y_{1} }{x_{2} -x_{1} } \\= \frac{\frac{6}{5}-\frac{8}{5} }{3-2} = \frac{6-8}{5} = -\frac{2}{5}\end{lgathered}
m
2
=
x
2
−x
1
y
2
−y
1
=
3−2
5
6
−
5
8
=
5
6−8
=−
5
2
m_{2} = -\frac{2}{5}m
2
=−
5
2
- (2)
Solving for m3
\begin{lgathered}m_{2} = \frac{y_{2}-y_{1} }{x_{2} -x_{1} } \\= \frac{\frac{6}{5}-2 }{3-1} = \frac{6-10}{(5)(2)} =\frac{-4}{10} = -\frac{2}{5}\end{lgathered}
m
2
=
x
2
−x
1
y
2
−y
1
=
3−1
5
6
−2
=
(5)(2)
6−10
=
10
−4
=−
5
2
m_{3} = -\frac{2}{5}m
3
=−
5
2
- (3)
From (1), (2) and (3) we have
m_{1} =m_{2} =m_{3} = -\frac{2}{5}m
1
=m
2
=m
3
=−
5
2
Since m1 = m2 = m3 , therefore points P, Q and R are collinear.