Question

if (-2.1) (4.6) (6-3) are the vertices of a
triangle. find the perimeter of a triangle.( this is in co ordinate geometry).​

Answer 1

Question :-

if (-2,1) , (4,6) and (6,3,) are the vertices of a triangle .Find the perimeter of this triangle .

Answer :-

→ Perimeter = √(61) + √(85) + √(80)

To find :-

→ Perimeter of triangle

Formula used :-

→ Perimeter of a triangle = sum of all sides

→ Distance between points ,

$\star \: \sqrt{ {( \: x_2 - x_1)}^{2} + \: \: {( \:y _2\: -y _1) }^{2} }$

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Step - by - step explanation :-

Given that,

Given points are ,

A ( -2,1) ,B (4,6) and C ( 6,3)

If we find perimeter of this triangle ,

Then firstly find the distance of AB , BC and CA ,

Therefore ,

If these are the sides of a triangle then perimeter of triangle = AB+ BC+ CA

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Solution :-

Firstly find distance between A and B ,

Let points are,

$\: x_1 = - 2 \: , \: \: x_2 \: = 4\\ \\ y_1 = 1 ,\: y_2 = 6$

Then,

$AB \: = \sqrt{ {(4 - ( - 2))}^{2} + {(6 - 1)}^{2} } \\ \\ AB \: = \sqrt{36 + 25} \\ \\ \boxed{ AB \: = \sqrt{61} } \: ......(1)$

Now find distance between B and C ,

$BC \: = \sqrt{ {(6 - 4)}^{2} + {( - 3 - 6)}^{2} } \\ \\ BC \: = \sqrt{4 + 81} \\ \\ \boxed{ BC \: = \sqrt{85} } \: \: \: .......(2)$

Now find distance between C and A

$CA \: = \sqrt{ { \big(6 - ( - 2) \big)}^{2} + {( - 3 - 1)}^{2} } \\ \\ CA\: = \sqrt{64 + 16} \\ \\ \boxed{CA \: = \sqrt{80} } \: \: \: .... ....(3)$

Now ,From eq.(1) ,(2) and (3)

We get ,

Perimeter of triangle

→ √(61) + √(85) +√(80)

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Answer 2

Answer:

Given that vertices of a triangle are A(-2,1) ; B(4,6) and C(6,-3) and we have to calculate the perimeter of the triangle .

As, we know that perimeter of the triangle is equal to sum of sum of length of all the three sides .

So,

$perimeter \: of \: \triangle \\ = AB + BC + AC$

Now, using distance formula

$AB = \sqrt{ { {(4 + 2) }^{2} + (6 - 1)}^{2} } \\ \\ AB = \sqrt{36 + 25} \\ \\ AB = \sqrt{61}$

And

$BC = \sqrt{ {(6 - 4)}^{2} + {( - 3 - 6)}^{2} } \\ \\ BC = \sqrt{4 + 81} \\ \\ BC = \sqrt{85}$

And

$AC = \sqrt{ {( - 2 - 6)}^{2} + {(1 + 3)}^{2} } \\ \\AC = \sqrt{64 + 16} \\ \\AC = \sqrt{80} \\ \\ AC = 2 \sqrt{20}$

So,

$perimeter \: of \: \triangle \\ = \sqrt{61} + \sqrt{85} + 2 \sqrt{20}$