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

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) (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)

Answer: Slope is

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)

Answer: Slope is

So we have one slope of and another of . Since these slopes are negative reciprocals of each other (ie and , these two lines are perpendicular.

#32
Lets solve first

Solved by pluggable solver: Graphing Linear Equations

Start with the given equation

Subtract from both sides

Multiply both sides by

Distribute

Multiply

Rearrange the terms

Reduce any fractions

So the equation is now in slope-intercept form () where (the slope) and (the y-intercept)

So to graph this equation lets plug in some points

Plug in x=-8

Multiply

Add

Reduce

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

Now lets find another point

Plug in x=-5

Multiply

Add

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 through the points (-8,-7) and (-5,-5)

So from the graph we can see that the slope is (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,) ,or (0,), and the x-intercept is (,0) ,or (,0) . So all of this information verifies our graph.

We could graph this equation another way. Since this tells us that the y-intercept (the point where the graph intersects with the y-axis) is (0,).

So we have one point (0,)

Now since the slope is , this means that in order to go from point to point we can use the slope to do so. So starting at (0,), 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

So this is the graph of through the points (0,-1.66666666666667) and (3,0.333333333333333)

So now we have the equation

So this means the perpendicular slope is the negative reciprocal of . So the negative reciprocal of is . So we have a line with a slope of 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 and intercept a, is

(see a paragraph below explaining why this formula is correct)

Given that a=-1.5, and , we have the equation of the line:

Explanation: Why did we use formula ? 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 (, ) lies on the line, it means that it satisfies the equation of the line. So, our equation holds for (, ): Here, we know a, , and , and do not know b. It is easy to find out: . So, then, the equation of the line is: .

Here's the graph:

So the equation is

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