Question

Write the equation in slope-intercept form of the line that is perpendicular to the graph of each equation and passes through the given point. y = -5x + 1; through (2, -1)

Answer 1

The equation is y = 1/5x -7/5

Explanation:

The given equation is in the form of y = mx + b

⇒ m = -5

since the line is perpendicular, therefore perpendicular line has a slope reciprocal of the given slope, that is m = 1/5

By using point slope equation (y-y1) = m(x-x1)

here (y1,x1) = (2,-1) and m = 1/5

(y-(-1)) = (1/5)(x-2)

y+1 = 1/5x -2/5

y+1-1 = 1/5x - 2/5 -1

y = 1/5x -7/5 is the desired equation.

Answer 2

`\large\boxed{y=(1)/(5)x-1(2)/(5)}`

Explanation:

`\text{The slope-intercept form:}\\\\y=mx+b\\\\m-slope\\b-\ y-intercept`

`\text{Let}\ k:y=m_1x+b_1\ \text{and}\ l:y=m_2x+b_2.\\\\l\ \perp\ k\iff m_1m_2=-1\to m_2=-(1)/(m_1)\\\\l\ ||\ k\iff m_2=m_1\\------------------------\\\\\text{We have}\ k:y=-5x+1\to m_1=-5.\\\text{Therefore}\ m_2=-(1)/(-5)=(1)/(5)\\\\l:y=(1)/(5)x+b\\\\\text{The line}\ l\ \text{passes through point (2. -1). Substitute x = 2 and y = -1}\\\text{to the equation of the line}\ l:`

`-1=(1)/(5)(2)+b\\\\-1=(2)/(5)+b\qquad\text{subtract}\ (2)/(5)\ \text{from both sides}\\\\-1-(2)/(5)=(2)/(5)-(2)/(5)+b\\\\-1(2)/(5)=b\to b=-1(2)/(5)\\\\\text{Finally we have:}\\\\l:y=(1)/(5)x-1(2)/(5)`