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I need get General Form of Equation of a Line for $M'P'$ by point $K(9, 2)$ and $M'P' \parallel MP$ $M(1, 4); P(7, 8)$.

I calc: $\overrightarrow{MP} $ {$7-1=6; ~ 8-4=4$}. Next, I use formula: $Ax+By+C=0; ~6x+4y+C=0; ~6 \cdot 9 + 4 \cdot 2 + C = 0; ~C = -62;$, =>, $3x-2y-31=0$. But, answer: $2x-3y-12=0$. Where did I go wrong?

P. S. Sorry if this is not correct. But it difficult for me.

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    Perhaps, I understood my mistake. – 2017-01-23

2 Answers 2

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I needed to find $\vec{n} \perp \vec{AB} $, then $\vec{n}${$4; -6$}. After, I calc: $4*9-6*2+C=0$; $C=-24$. And, result: $2x-3y-12=0$.

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you can use the vector form $$[x,y]=[9,2]+s[6,4]$$ or need you an arithmetic form? from here you will get $$x=9+6s$$ $$y=2+4s$$ by eliminating the real number $s$ you will get your equation

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    Arithmetic form, please. – 2017-01-23
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    My solution contains many mistakes? – 2017-01-23
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    you can't plugging a vector into the equation – 2017-01-23