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

What is the equation of the line that passes through the point (−6,−6) (-6,-6) 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,−6) (-6,-6) 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.\newlineWe have:\newlineSlope mm: 12\frac{1}{2}\newlinePoint (x1,y1)(x_1, y_1): (−6,−6)(-6, -6)\newlineThe slope-intercept form of a line is y=mx+by = mx + b, where mm is the slope and bb is the y-intercept.
  2. Use Point-Slope Form: Use the point-slope form of the equation of a line to find the y-intercept bb. The point-slope form is y−y1=m(x−x1)y - y_1 = m(x - x_1). Substitute the given point and slope into the point-slope form: y−(−6)=12(x−(−6))y - (-6) = \frac{1}{2}(x - (-6)) Simplify and solve for yy: y+6=12(x+6)y + 6 = \frac{1}{2}(x + 6)
  3. Distribute Slope: Distribute the slope (12)(\frac{1}{2}) on the right side of the equation.\newliney+6=12×x+12×6y + 6 = \frac{1}{2} \times x + \frac{1}{2} \times 6\newliney+6=12×x+3y + 6 = \frac{1}{2} \times x + 3
  4. Isolate y: Isolate yy to get the equation in slope-intercept form (y=mx+by = mx + b).\newlineSubtract 66 from both sides of the equation:\newliney=12⋅x+3−6y = \frac{1}{2} \cdot x + 3 - 6\newliney=12⋅x−3y = \frac{1}{2} \cdot x - 3

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