Question

Determine whether the points (1,-3) &c(-4,7)are follinear OR mothers

Answer 1

Answer:

$\bf{correct\:Questuon}$

Determine whether the points A(1 , -3 ) B(2, -5) and C(-4 , 7) are volunteering or not

$\bf\red{\underline{\underline{Solution\mapsto :}}}$

$\bf {by\: distant \: formula}$

$\bf{d(A ,B) = \sqrt{(2 - 1)^2+ [(- 5 - (-3)]^2}}$

$\bf\implies\sqrt{1^2 + ( - 5 + 3)^2}$

$\bf\implies\sqrt{1 + ( - 2)^2}$

$\bf\implies\sqrt{1 + 4}$

$\bf\implies\sqrt{5}$..............(1)

$\bf{d(B ,C) = \sqrt{( - 4 - 2)^2 + 1 [7 - ( - 5 )]^2}}$

$\bf\implies\sqrt{( -6 )^2 + (12)^2}$

$\bf\implies\sqrt{36 + 144}$

$\bf\implies\sqrt{180}$

$\bf\implies\sqrt{2\times{2\times{3\times{3\times5}}}}$

$\bf\implies{6\sqrt{5}}$............(2)

$\bf{d( A, C) = \sqrt{(- 4 - 1 )^2 +[ 7 - ( - 3 )]^2}}$

$\bf\implies\sqrt{( - 5 )^2 + (10)^2}$

$\bf\implies\sqrt{25 + 100}= \sqrt{125}$

$\bf\implies\sqrt{5\times{5\times{\times5}}}$

$\bf\implies{5\sqrt{5}}$.......(3)

Now,

Adding (1) and (3)

d(A, B) + d(A ,C) = $\bf\sqrt{5} + 5\sqrt{5} = 6 \sqrt{5}$

$\bf{d(A,B) +d(A,C) = d(B,C)}$

Point A, B and C are collinear.

$\bf\red{\underline{\underline{MuskaŊ}}}$

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Answer 2

Answer:

Determine whether the points A(1 , -3 ) B(2, -5) and C(-4 , 7) are volunteering or not

\bf\red{\underline{\underline{Solution\mapsto :}}}

Solution↦:

\bf {by\: distant \: formula}bydistantformula

\bf{d(A ,B) = \sqrt{(2 - 1)^2+ [(- 5 - (-3)]^2}}d(A,B)=

(2−1)

2

+[(−5−(−3)]

2

\bf\implies\sqrt{1^2 + ( - 5 + 3)^2}⟹

1

2

+(−5+3)

2

\bf\implies\sqrt{1 + ( - 2)^2}⟹

1+(−2)

2

\bf\implies\sqrt{1 + 4}⟹

1+4

\bf\implies\sqrt{5}⟹

5

..............(1)

\bf{d(B ,C) = \sqrt{( - 4 - 2)^2 + 1 [7 - ( - 5 )]^2}}d(B,C)=

(−4−2)

2

+1[7−(−5)]

2

\bf\implies\sqrt{( -6 )^2 + (12)^2}⟹

(−6)

2

+(12)

2

\bf\implies\sqrt{36 + 144}⟹

36+144

\bf\implies\sqrt{180}⟹

180

\bf\implies\sqrt{2\times{2\times{3\times{3\times5}}}}⟹

2×2×3×3×5

\bf\implies{6\sqrt{5}}⟹6

5

............(2)

\bf{d( A, C) = \sqrt{(- 4 - 1 )^2 +[ 7 - ( - 3 )]^2}}d(A,C)=

(−4−1)

2

+[7−(−3)]

2

\bf\implies\sqrt{( - 5 )^2 + (10)^2}⟹

(−5)

2

+(10)

2

\bf\implies\sqrt{25 + 100}= \sqrt{125}⟹

25+100

=

125

\bf\implies\sqrt{5\times{5\times{\times5}}}⟹

5×5××5

\bf\implies{5\sqrt{5}}⟹5

5

.......(3)

Now,

Adding (1) and (3)

d(A, B) + d(A ,C) = \bf\sqrt{5} + 5\sqrt{5} = 6 \sqrt{5}

5

+5

5

=6

5

\bf{d(A,B) +d(A,C) = d(B,C)}d(A,B)+d(A,C)=d(B,C)

Point A, B and C are collinear