SOLUTION: #38 Find an equation in slope-intercept form (where possible) for each line Use slopes to show that the square with vertices at (-2,5),(4,5),(4,-1),and (-2,-1) has diagonals

Algebra ->  Linear-equations -> SOLUTION: #38 Find an equation in slope-intercept form (where possible) for each line Use slopes to show that the square with vertices at (-2,5),(4,5),(4,-1),and (-2,-1) has diagonals      Log On


   

Question 83173: #38
Find an equation in slope-intercept form (where possible) for each line
Use slopes to show that the square with vertices at (-2,5),(4,5),(4,-1),and
(-2,-1)
has diagonals that are perpendicular.
#32
Find an equation in slope-intercept form (where possible) for each line
Through (-2,6),perpendicular to 2x -3y =5.
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
#38


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:

Solved by pluggable solver: Finding the slope


Slope of the line through the points (-2, 5) and (4, -1)



m+=+%28y%5B2%5D+-+y%5B1%5D%29%2F%28x%5B2%5D+-+x%5B1%5D%29


m+=+%28y%5B2%5D+-+y%5B1%5D%29%2F%28x%5B2%5D+-+%28x%5B1%5D%29%29


m+=+%28-1+-+5%29%2F%284+-+%28-2%29%29


m+=+%28-1+-+5%29%2F%284+%2B+2%29


m+=+%28-6%29%2F%286%29


m+=+-1



Answer: Slope is m+=+-1




Now lets find the slope of the line going through (-2,-1) and (4,5)

Solved by pluggable solver: Finding the slope


Slope of the line through the points (-2, -1) and (4, 5)



m+=+%28y%5B2%5D+-+%28y%5B1%5D%29%29%2F%28x%5B2%5D+-+%28x%5B1%5D%29%29


m+=+%285+-+%28-1%29%29%2F%284+-+%28-2%29%29


m+=+%285+%2B+1%29%2F%284+%2B+2%29


m+=+%286%29%2F%286%29


m+=+1



Answer: Slope is m+=+1




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.

#32
Lets solve 2x+-3y+=5 first

Solved by pluggable solver: Graphing Linear Equations


2%2Ax-3%2Ay=5Start with the given equation



-3%2Ay=5-2%2Ax Subtract 2%2Ax from both sides

y=%28-1%2F3%29%285-2%2Ax%29 Multiply both sides by -1%2F3

y=%28-1%2F3%29%285%29%2B%281%2F3%29%282%29x%29 Distribute -1%2F3

y=-5%2F3%2B%282%2F3%29x Multiply

y=%282%2F3%29%2Ax-5%2F3 Rearrange the terms

y=%282%2F3%29%2Ax-5%2F3 Reduce any fractions

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)

So to graph this equation lets plug in some points

Plug in x=-8

y=%282%2F3%29%2A%28-8%29-5%2F3

y=-16%2F3-5%2F3 Multiply

y=-21%2F3 Add

y=-7 Reduce

So here's one point (-8,-7)





Now lets find another point

Plug in x=-5

y=%282%2F3%29%2A%28-5%29-5%2F3

y=-10%2F3-5%2F3 Multiply

y=-15%2F3 Add

y=-5 Reduce

So here's another point (-5,-5). Add this to our graph





Now draw a line through these points

So this is the graph of y=%282%2F3%29%2Ax-5%2F3 through the points (-8,-7) and (-5,-5)


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.


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).


So we have one point (0,-5%2F3)






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


and to the right 3 units to get to our next point



Now draw a line through those points to graph y=%282%2F3%29%2Ax-5%2F3


So this is the graph of y=%282%2F3%29%2Ax-5%2F3 through the points (0,-1.66666666666667) and (3,0.333333333333333)



So now we have the equation

y=%282%2F3%29x-5%2F3

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)

Solved by pluggable solver: FIND a line by slope and one point

What we know about the line whose equation we are trying to find out:

  • it goes through point (-2, 6)

  • it has a slope of -1.5



First, let's draw a diagram of the coordinate system with point (-2, 6) plotted with a little blue dot:



Write this down: the formula for the equation, given point x%5B1%5D%2C+y%5B1%5D and intercept a, is

y=ax+%2B+%28y%5B1%5D-a%2Ax%5B1%5D%29 (see a paragraph below explaining why this formula is correct)

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:

y=-1.5%2Ax+%2B+3

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+.

Here's the graph:





So the equation is y=%28-3%2F2%29x%2B3