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Graph a line that contains the point (4,3)(4,3) and has a slope of 12\frac{1}{2}.

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Full solution

Q. Graph a line that contains the point (4,3)(4,3) and has a slope of 12\frac{1}{2}.
  1. Identify slope and point: Identify the slope and the point through which the line passes.\newlineThe slope mm is given as 12\frac{1}{2}, and the point (x1,y1)(x_1, y_1) is (4,3)(4,3).
  2. Use point-slope form: Use the point-slope form of the equation of a line to start forming the equation.\newlineThe point-slope form is given by (y−y1)=m(x−x1)(y - y_1) = m(x - x_1), where mm is the slope and (x1,y1)(x_1, y_1) is the point the line passes through.
  3. Substitute slope and point: Substitute the slope and the point into the point-slope form equation.\newlineUsing m=12m = \frac{1}{2} and the point (4,3)(4,3), the equation becomes (y−3)=12(x−4)(y - 3) = \frac{1}{2}(x - 4).
  4. Distribute slope: Distribute the slope on the right side of the equation.\newlineThis gives us y−3=12×x−12×4y - 3 = \frac{1}{2} \times x - \frac{1}{2} \times 4, which simplifies to y−3=12×x−2y - 3 = \frac{1}{2} \times x - 2.
  5. Isolate y: Isolate yy to put the equation into slope-intercept form (y=mx+by = mx + b).\newlineAdd 33 to both sides of the equation to get y=12×x−2+3y = \frac{1}{2} \times x - 2 + 3.
  6. Combine like terms: Combine like terms to find the yy-intercept (bb).\newlineThis simplifies to y=12×x+1y = \frac{1}{2} \times x + 1, which is the equation of the line in slope-intercept form.

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