Math, asked by tejasvini16durugwar, 10 months ago

construct a triangle ABC in which angle b is equal to 60 degree ab is equal to 5.4 cm and BC is equal to 5.8 CM ​

Answers

Answered by sumitkokate32
0

Step-by-step explanation:

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 po

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