SOLUTION: Write the slope-intercept equation for the line that passes through (70, 92) and is (a) parallel and (B) perpendicular to 7x + 10y = -24. Please help!

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Question 533950: Write the slope-intercept equation for the line that passes through (70, 92) and is (a) parallel and (B) perpendicular to 7x + 10y = -24. Please help!
Answer by bucky(2189) (Show Source):
You can put this solution on YOUR website!
The slope intercept form of an equation is:
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In this form, m (the multiplier of the x) is the slope, and b is the value of the point on the y-axis where the graph crosses the y-axis.
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Given that, lets convert the given equation of:
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into slope-intercept form. Begin by subtracting 7x from both sides to get:
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Now solve for y by dividing all terms on both sides by 10 to get:
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Notice that this is now in slope-intercept form. The slope of the graph of this equation is -7/10 (or - 0.7) which is the multiplier of the x term. And the point where the graph crosses the y-axis is -24/10 (or -2.4) on the y-axis.
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Let's work on Part A of the problem. We are to find the slope-intercept form of the equation whose graph is parallel to the given equation:
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and goes through point (70,92)
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Start with the slope-intercept form:
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In order to be parallel to the given graph, the graph of this new equation needs to have the same slope as the given graph. We found that the slope of the given equation is (-7/10) which for simplification we'll write as -0.7. So our slope-intercept form becomes:
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Next we know that the graph has to go through the point (70,92) in which 70 is the x value when 92 is the y value. So let's substitute 70 for x and 92 for y in the equation that we are developing. When we do that we can solve for the value of b as follows:
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Do the multiplication on the right side and you have:
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Add 49 to both sides and you have:
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Substitute this value for b in the slope intercept equation we are developing and we end up with:
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That's the answer to Part A. Now let's go to Part B. We need to develop a slope intercept equation for a graph that will be perpendicular to the given equation of:
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and also goes through the point (70, 92)
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The critical part of this problem is to know that a perpendicular will have a slope that is the negative inverse of the slope that it is to be perpendicular to. In this case the negative inverse slope is:
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Rather than dividing this out to get a long decimal term, let's just multiply the numerator and denominator by 10 and get that the slope of the perpendicular is:
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Substitute this value into the slope-intercept equation and we get:
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Just as we did previously let's substitute the x and y values of the point that this graph is to go through and solve for b:
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Do the multiplication on the right side and you get:
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Solve for b by subtracting 100 from both sides and you get:
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Substitute this value of b into the slope intercept equation and the equation:
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becomes:
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That should be the answer to Part B.
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Please check my math. It's pretty late and I might be prone to making dumb math errors. The methodology is correct.
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Hope this helps you understand the problem better. The two key things to remember are that graphed lines are parallel if they have the same slope and perpendicular if one of them has the negative inverse slope of the other one.
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