Question

A line passes through the point (-6,-9) and has a slope of 5/2.

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

Answer 1

`(\stackrel{x_1}{-6}~,~\stackrel{y_1}{-9})\hspace{10em} \stackrel{slope}{m} ~=~ \cfrac{5}{2} \\\\\\ \begin{array} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-9)}=\stackrel{m}{\cfrac{5}{2}}(x-\stackrel{x_1}{(-6)})\implies y+9=\cfrac{5}{2}(x+6) \\\\\\ y+9=\cfrac{5}{2}x+15\implies y=\cfrac{5}{2}x+6`

Answer 2

y = 5/2x +6

Explanation:

The slope-intercept form of the equation of a line is ...

y = mx +b . . . . line with slope m and y-intercept b

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The above equation can be solved for b to give ...

b = y -mx

Using the given values m = 5/2 and (x, y) = (-6, -9), we find b to be ...

b = -9 -(5/2)(-6) = -9 +15 = +6

Then the slope-intercept equation is ...

y = 5/2x +6