Write a slope-intercept equation for a line passing through the point \( (6,13) \) that is parallel to \( y=\frac{1}{2} x+4 \). Then write a second equation for a line passing through the given point that is perpendicular to the given line. Which answer below is correct? A. parallel: \( y=\frac{1}{2} x+13 \) perpendicular: \( y=-2 x+13 \) B. parallel: \( y=\frac{1}{2} x+10 \) perpendicular: \( y=-2 x+25 \) parallel: \( y=\frac{1}{2} x+10 \quad \) perpendicular: \( y=2 x+25 \)
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To find the slope-intercept form of the parallel line, we know it needs to have the same slope as the original line \( y = \frac{1}{2} x + 4 \) which is \( \frac{1}{2} \). Using point-slope form, the equation becomes: \[ y - 13 = \frac{1}{2}(x - 6) \] This simplifies to: \[ y = \frac{1}{2} x + 10 \] Next, for the perpendicular line, the slope will be the negative reciprocal of \( \frac{1}{2} \) which is \( -2 \). So we proceed with: \[ y - 13 = -2(x - 6) \] This simplifies to: \[ y = -2 x + 25 \] Thus, we have the correct equations: the parallel line is \( y=\frac{1}{2} x + 10 \) and the perpendicular line is \( y=-2 x + 25 \). Therefore, the right answer from the choices given is: **B. parallel: \( y=\frac{1}{2} x+10 \) perpendicular: \( y=-2 x+25 \)** You got it! Now, let’s look at how this can tie back into real-world scenarios. Slopes often represent rates—like how quickly something increases or decreases. By analyzing slopes in a graph, whether that's in finance, physics, or engineering, you can make data-driven decisions that resemble the trajectory of the lines we've discussed. When creating equations, one common mistake is forgetting to apply the slope to both parts of the equation properly. Always double-check your calculations when converting from point-slope to slope-intercept forms to avoid small errors that can lead to big mistakes in graphing!
