Question

What is an equation of the line that is perpendicular to y-3 = -4(x + 2) and

passes through the point (-5,7)?
O A. Y+7=-=(x-5)
O B. 7-7 - }(x+5)
O
O
c. y - 7 = -4(x + 5)
D. y+ 7 = 4(x - 5)

Answer 1

`\large\boxed{y-7=(1)/(4)(x+5)}`

Explanation:

`\text{The point-slope form of an equation of a line:}\\\\y-y_1=m(x-x_1)\\\\m-slope\\\\\text{Let a line}\ k\ \text{has a slope}\ m_1\ \text{and a line}\ l\ \text{has a slope}\ m_2,\ \text{then}\\\\l\ \perp\ k\iff m_1m_2=-1\to m_2=-(1)/(m_1)\\\\l\ \parallel\ k\iff m_1=m_2\\================================`

`\text{We have the equation:}\ y-3=-4(x+2)\to m_1=-4.\\\\\text{Therefore, the slope of the line perpendicular to the given line is}\\\\m_2=-(1)/(-4)=(1)/(4).\\\\\text{Put it and the coordinates of the point (-5, 7) to the equation of a line:}\\\\y-7=(1)/(4)(x-(-5))\\\\y-7=(1)/(4)(x+5)`