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What is the equation of the line with a slope of -1/2 that passes through the point (6,-6)?

Sagot :

konrad509
[tex]y-y_1=m(x-x_1)\\\\ y-(-6)=-\dfrac{1}{2}(x-6)\\ y+6=-\dfrac{1}{2}x+3\\ y=-\dfrac{1}{2}x-3[/tex]
dln
[tex]You\ can\ just\ plug\ this\ into\ point-slope\ equation: \\ (y-y1)=m(x-x1)\ \ \ \ m\ is\ slope \\ \\ (y-(-6)= \frac{-1}{2}(x-6) \\ \boxed{y+6= \frac{-1}{2}(x-6) } \\ \\ If\ you\ want\ in\ slope-intercept\ form: \\ y+6= \frac{-1}{2}(x-6) \\ y+6= \frac{-1}{2}x+3 \\ \boxed{ y=- \frac{1}{2}x-3 }[/tex]
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