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Find the equation of straight line passing through the mid-point of (1,2) and (3,4) and having slope 3.

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Answer:

[tex] \large{ \tt{❃ \: EXPLANATION}} : [/tex]

  • Let us assume the two points ( 1 , 2 ) and ( 3 , 4 ) be ( x₁ , y₁ ) and ( x₂ , y₂ ) respectively. Now , Find out the midpoint of those points :

[tex] \boxed{\large{ \tt{✧ \: MIDPOINT = ( \frac{x_{1} + x_{2}}{2} } \:, \frac{y_{1} + y_{2}}{2} ) }}[/tex]

[tex] \large{ \tt{↦ \: ( \frac{1 + 3}{2} \: , \frac{2 + 4}{2} )}}[/tex]

[tex] \large{ \tt{↦( \frac{4}{2} \:, \frac{6}{2} })}[/tex]

[tex] \large{ \tt{↦ \underline{( \: 2 \:, 3 \: )}}}[/tex]

  • To find the equation of straight line passing through a point and a slope , we use the equation of straight line in point slope form i.e y - y₁ = m ( x - x₁ ).

  • We have : Slope ( m ) = 3 & assume the midpoint of ( 1 , 2 ) and ( 3 , 4 ) i.e ( 2 , 3 ) be ( x₁ , y₁ ).

[tex] \large{ \tt{✎ \: LET'S \: START}} : [/tex]

[tex] ☯ \: \boxed { \large{\tt{y - y_{1} = m(x - x_{1})}}}[/tex]

[tex] \large{ \tt{↬y - 3 = 3(x - 2)}}[/tex]

[tex] \large{ \tt{↬y - 3 = 3x - 6}}[/tex]

[tex] \large{ \tt{↬3x - 6 = y - 3}}[/tex]

[tex] \large{ \tt{↬3x - y - 6 + 3 = 0}}[/tex]

[tex]\large{ \tt{↬ \boxed{ \tt{3x - y - 3 = 0}}}}[/tex]

  • Hence , The required equation of a straight line is 3x - y - 3 = 0 .

[tex] \tt{✺ \: CHOOSE \: YOUR \: HABITS \: CAREFULLY \: ,THEY \: DECIDE \: YOUR \: FUTURE! \: ♪}[/tex]

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