Question

line p contains point (6 ,- 5 and is perpendicular to line q the equation for line q is y=3x+5 the slope of line q is 3. the slope of line p is -1/3. Equation for line p in point slope form is y = -1/3x - 3) Use your equation to write an equation for line p in slope-intercept form

Answer 1

Given:

Line p contains the point (6,- 5 and is perpendicular to line q the equation for line q is y=3x+5 the slope of line q is 3. and the slope of line p is -1/3. Equation for line p in point-slope form is y = -1/3x - 3.

Required:

Write an equation for line p in slope-intercept form.

Explanation:

The equation of line q is:

$y=3x+5$

The line p is perpendicular to line q.

The slope of line q = 3

We know that the perpendicular line has negative reciprocal slopes.

So the slope of line p

$=\frac{-1}{3}$

The equation of line has slope m and passes through from one point is given as:

$y-y_1=m(x-x_1)$$(x_1,y_1)=(6,-5)$

Thus the equation of line p is:

$\begin{gathered} y-(-5)=-\frac{1}{3}(x-6) \\ y+5=-\frac{1}{3}x+2 \\ y=-\frac{1}{3}x-3 \end{gathered}$

Final Answer:

The equation of line p is:

$y=-\frac{1}{3}x-3$