What is the equation of the line (in slope-intercept form) that passes through the point (5,βˆ’1) and is parallel to the line y=2xβˆ’7

Accepted Solution

Answer:y = 2x - 11Step-by-step explanation:[tex]\text{Let}\\\\k:y=m_1x+b_1\\\\l:y=m_2x+b_2\\\\l\ \perp\ k\iff m_1m_2=-1\to m_2=-\dfrac{1}{m_1}\\\\l\ \parallel\ k\iff m_1=m_2\\\\===============================[/tex][tex]\text{The slope-intercept form of an equation of a line:}\\\\y=mx+b\\\\m-slope\\b-y-intercept\\\\==========================[/tex][tex]\text{We have the equation of a line:}\ y=2x-7\to m_1=2.\\\\\text{The slope of a parallel line:}\ m_2=m_1=2.\\\\\text{We have the equation:}\\\\y=2x+b\\\\\text{Put the coordinates of the point (5, -1) o the equation:}\\\\-1=2(5)+b\\-1=10+b\qquad\tex\text{subtract 10 from both sides}\\-11=b\to b=-11\\\\\text{Finally:}\\\\y=2x-11[/tex]