Question

Find the equation for the line that passes through the point (-1, -3), and that is perpendicular to the line with the equation

-2/3x- 2y = -2​

Answer 1

`\large\boxed{y=3x}`

Explanation:

`\text{The slope-intercept form of an equation of a line:}\\\\y=mx+b\\\\m\ -\ slope\\b\ -\ y-intercept`

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

`\text{We have:}\\\\k:-(2)/(3)x-2y=-2\\\\\text{Convert to the slope-intercept form}:\\\\-(2)/(3)x-2y=-2\qquad\text{multiply both sides by 3}\\\\-2x-6y=-6\qquad\text{divide both sides by (-2)}\\\\x+3y=3\qquad\text{subtract}\ x\ \text{from both sides}\\\\3y=-x+3\qquad\text{divide both sides by 3}\\\\y=-(1)/(3)x+1\to m_1=-(1)/(3)\\\\\text{Calculate the other slope:}\\\\m_1=-(1)/(-(1)/(3))=3\\\\\text{Put it to the equation of a line:}\\\\y=3x+b`

`\text{Put the coordinates of the given point (-1, -3) to the equation:}\\\\-3=3(-1)+b\\-3=-3+b\qquad\text{add 3 to both sides}\\0=b\to b=0\\\\\text{Finally:}\\\\y=3x+0\to y=3x`