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Find the equation of the line: parallel to 3x−y=11 through (−2, 0).

1 Answer

4 votes

Answer:


\huge\boxed{y=3x+6\to 3x-y=-6}

Explanation:


\text{Let}\ k:y=m_1x+b_1;\ l:y=m_2x+b_2\\\\k\ ||\ l\iff m_1=m_2\\k\ \perp\ l\iff m_1m_2=-1\to m_2=-(1)/(m_1)\\\\===========================


\text{We have the equation of a line in the standard form:}\ 3x-y=11.\\\text{Convert it to the slope-intercept form:}\ y=mx+b\\\\3x-y=11\qquad\text{subtract}\ 3x\ \text{from both sides}\\-y=-3x+11\qquad\text{change the signs}\\y=3x-11\to \boxed{m_1=3}\\\\\text{Let}\ k:y=3x-11\ \text{and}\ l:y=mx+b.\\\\l\ ||\ k\Rightarrow l:y=3x+b\\\\\text{Substitute the coordinates of the point}\ (-2;\ 0)\ \text{to the equation of}\ l:\\\\0=3(-2)+b\\0=-6+b\qquad\text{add 6 to both sides}\\6=b\to\boxed{b=6}


\text{Finally}\ l:y=3x+6\\\\\text{Convert to the standard form:}\\\\y=3x+6\qquad\text{subtract}\ 3x\ \text{from both sides}\\-3x+y=6\qquad\text{change the signs}\\l:\ 3x-y=-6

User Lucas Bernalte
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