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

What is the equation of the line that passes through the point (−6,−2) (-6,-2) and has a slope of −12 -\frac{1}{2} ?\newlineAnswer:

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Q. What is the equation of the line that passes through the point (−6,−2) (-6,-2) and has a slope of −12 -\frac{1}{2} ?\newlineAnswer:
  1. Identify slope and point: Identify the slope mm and the point (x1,y1)(x_1, y_1) through which the line passes.\newlineThe slope mm is given as −12-\frac{1}{2}, and the point is (−6,−2)(-6, -2).
  2. Use point-slope form: Use the point-slope form of the equation of a line to plug in the values of the slope and the point.\newlineThe point-slope form is y−y1=m(x−x1)y - y_1 = m(x - x_1).\newlineSubstitute m=−12m = -\frac{1}{2}, x1=−6x_1 = -6, and y1=−2y_1 = -2 into the equation to get y−(−2)=−12(x−(−6))y - (-2) = -\frac{1}{2}(x - (-6)).
  3. Simplify the equation: Simplify the equation from Step 22.\newliney+2=−12(x+6)y + 2 = -\frac{1}{2}(x + 6).\newlineNow distribute the slope −12-\frac{1}{2} across (x+6)(x + 6) to get y+2=−12x−3y + 2 = -\frac{1}{2}x - 3.
  4. Solve for y: Solve for y to put the equation in slope-intercept form y=mx+by = mx + b.\newlineSubtract 22 from both sides of the equation to isolate yy.\newliney=−12x−3−2y = -\frac{1}{2}x - 3 - 2.\newliney=−12x−5y = -\frac{1}{2}x - 5.

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