document.write( "Question 83173: #38\r
\n" ); document.write( "\n" ); document.write( "Find an equation in slope-intercept form (where possible) for each line
\n" ); document.write( "Use slopes to show that the square with vertices at (-2,5),(4,5),(4,-1),and
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\n" ); document.write( "Find an equation in slope-intercept form (where possible) for each line
\n" ); document.write( "Through (-2,6),perpendicular to 2x -3y =5.
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Algebra.Com's Answer #59724 by jim_thompson5910(35256)\"\" \"About 
You can put this solution on YOUR website!
#38\r
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\n" ); document.write( "\n" ); document.write( "If we construct the figure that the problem is describing, we get a square with the given vertices and 2 lines. We need to find the slopes of the 2 lines to see if they are perpendicular. So the first line goes through the points (-2,5) and (4,-1) (it is the line that is sloping downward). So lets find the slope of this line:\r
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Solved by pluggable solver: Finding the slope

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\n" ); document.write( " Slope of the line through the points (-2, 5) and (4, -1)
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\n" ); document.write( " \"m+=+%28y%5B2%5D+-+y%5B1%5D%29%2F%28x%5B2%5D+-+x%5B1%5D%29\"
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\n" ); document.write( " \"m+=+%28y%5B2%5D+-+y%5B1%5D%29%2F%28x%5B2%5D+-+%28x%5B1%5D%29%29\"
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\n" ); document.write( " \"m+=+%28-1+-+5%29%2F%284+-+%28-2%29%29\"
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\n" ); document.write( " \"m+=+%28-1+-+5%29%2F%284+%2B+2%29\"
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\n" ); document.write( " \"m+=+%28-6%29%2F%286%29\"
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\n" ); document.write( " \"m+=+-1\"
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\n" ); document.write( " Answer: Slope is \"m+=+-1\"
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\n" ); document.write( "\n" ); document.write( "Now lets find the slope of the line going through (-2,-1) and (4,5)\r
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Solved by pluggable solver: Finding the slope

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\n" ); document.write( " Slope of the line through the points (-2, -1) and (4, 5)
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\n" ); document.write( " \"m+=+%28y%5B2%5D+-+%28y%5B1%5D%29%29%2F%28x%5B2%5D+-+%28x%5B1%5D%29%29\"
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\n" ); document.write( " \"m+=+%285+-+%28-1%29%29%2F%284+-+%28-2%29%29\"
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\n" ); document.write( " \"m+=+%285+%2B+1%29%2F%284+%2B+2%29\"
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\n" ); document.write( " \"m+=+%286%29%2F%286%29\"
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\n" ); document.write( " \"m+=+1\"
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\n" ); document.write( " Answer: Slope is \"m+=+1\"
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\n" ); document.write( "\n" ); document.write( "So we have one slope of \"-1\" and another of \"1\". Since these slopes are negative reciprocals of each other (ie \"1=-%28-1%2F1%29\" and \"-1=-%281%2F1%29%29\", these two lines are perpendicular.\r
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\n" ); document.write( "\n" ); document.write( "Lets solve \"2x+-3y+=5\" first\r
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Solved by pluggable solver: Graphing Linear Equations

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\n" ); document.write( " \"2%2Ax-3%2Ay=5\"Start with the given equation
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\n" ); document.write( " \"-3%2Ay=5-2%2Ax\" Subtract \"2%2Ax\" from both sides
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\n" ); document.write( " \"y=%28-1%2F3%29%285-2%2Ax%29\" Multiply both sides by \"-1%2F3\"
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\n" ); document.write( " \"y=%28-1%2F3%29%285%29%2B%281%2F3%29%282%29x%29\" Distribute \"-1%2F3\"
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\n" ); document.write( " \"y=-5%2F3%2B%282%2F3%29x\" Multiply
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\n" ); document.write( " \"y=%282%2F3%29%2Ax-5%2F3\" Rearrange the terms
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\n" ); document.write( " \"y=%282%2F3%29%2Ax-5%2F3\" Reduce any fractions
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\n" ); document.write( " So the equation is now in slope-intercept form (\"y=mx%2Bb\") where \"m=2%2F3\" (the slope) and \"b=-5%2F3\" (the y-intercept)
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\n" ); document.write( " So to graph this equation lets plug in some points
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\n" ); document.write( " Plug in x=-8
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\n" ); document.write( " \"y=%282%2F3%29%2A%28-8%29-5%2F3\"
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\n" ); document.write( " \"y=-16%2F3-5%2F3\" Multiply
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\n" ); document.write( " \"y=-21%2F3\" Add
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\n" ); document.write( " \"y=-7\" Reduce
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\n" ); document.write( " So here's one point (-8,-7)
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\n" ); document.write( " Now lets find another point
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\n" ); document.write( " Plug in x=-5
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\n" ); document.write( " \"y=%282%2F3%29%2A%28-5%29-5%2F3\"
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\n" ); document.write( " \"y=-10%2F3-5%2F3\" Multiply
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\n" ); document.write( " \"y=-15%2F3\" Add
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\n" ); document.write( " \"y=-5\" Reduce
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\n" ); document.write( " So here's another point (-5,-5). Add this to our graph
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\n" ); document.write( " Now draw a line through these points
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\n" ); document.write( " So this is the graph of \"y=%282%2F3%29%2Ax-5%2F3\" through the points (-8,-7) and (-5,-5)
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\n" ); document.write( " So from the graph we can see that the slope is \"2%2F3\" (which tells us that in order to go from point to point we have to start at one point and go up 2 units and to the right 3 units to get to the next point) the y-intercept is (0,\"-1.66666666666667\") ,or (0,\"-5%2F3\"), and the x-intercept is (\"2.5\",0) ,or (\"5%2F2\",0) . So all of this information verifies our graph.
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\n" ); document.write( " We could graph this equation another way. Since \"b=-5%2F3\" this tells us that the y-intercept (the point where the graph intersects with the y-axis) is (0,\"-5%2F3\").
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\n" ); document.write( " So we have one point (0,\"-5%2F3\")
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\n" ); document.write( " Now since the slope is \"2%2F3\", this means that in order to go from point to point we can use the slope to do so. So starting at (0,\"-5%2F3\"), we can go up 2 units
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\n" ); document.write( " and to the right 3 units to get to our next point
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\n" ); document.write( " Now draw a line through those points to graph \"y=%282%2F3%29%2Ax-5%2F3\"
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\n" ); document.write( " So this is the graph of \"y=%282%2F3%29%2Ax-5%2F3\" through the points (0,-1.66666666666667) and (3,0.333333333333333)
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\n" ); document.write( "\n" ); document.write( "So now we have the equation\r
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\n" ); document.write( "\n" ); document.write( "\"y=%282%2F3%29x-5%2F3\"\r
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\n" ); document.write( "\n" ); document.write( "So this means the perpendicular slope is the negative reciprocal of \"2%2F3\". So the negative reciprocal of \"2%2F3\" is \"-3%2F2\". So we have a line with a slope of \"-3%2F2\" and goes through (-2,6)\r
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Solved by pluggable solver: FIND a line by slope and one point

\n" ); document.write( " What we know about the line whose equation we are trying to find out:
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  • it goes through point (-2, 6)

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  • it has a slope of -1.5

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\n" ); document.write( " First, let's draw a diagram of the coordinate system with point (-2, 6) plotted with a little blue dot:
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\n" ); document.write( " Write this down: the formula for the equation, given point \"x%5B1%5D%2C+y%5B1%5D\" and intercept a, is
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\n" ); document.write( " \"y=ax+%2B+%28y%5B1%5D-a%2Ax%5B1%5D%29\" (see a paragraph below explaining why this formula is correct)
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\n" ); document.write( " Given that a=-1.5, and \"system%28+x%5B1%5D+=+-2%2C+y%5B1%5D+=+6+%29+\", we have the equation of the line:
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\n" ); document.write( " \"y=-1.5%2Ax+%2B+3\"
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\n" ); document.write( " Explanation: Why did we use formula \"y=ax+%2B+%28y%5B1%5D+-+a%2Ax%5B1%5D%29\" ? Explanation goes here. We are trying to find equation y=ax+b. The value of slope (a) is already given to us. We need to find b. If a point (\"x%5B1%5D\", \"y%5B1%5D\") lies on the line, it means that it satisfies the equation of the line. So, our equation holds for (\"x%5B1%5D\", \"y%5B1%5D\"): \"y%5B1%5D+=+a%2Ax%5B1%5D%2Bb\" Here, we know a, \"x%5B1%5D\", and \"y%5B1%5D\", and do not know b. It is easy to find out: \"b=y%5B1%5D-a%2Ax%5B1%5D\". So, then, the equation of the line is: \"+y=ax%2B%28y%5B1%5D-a%2Ax%5B1%5D%29+\".
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\n" ); document.write( " Here's the graph:
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\n" ); document.write( "\n" ); document.write( "So the equation is \"y=%28-3%2F2%29x%2B3\"\r
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