geometry question

Question

. Choose a pair of coordinate points. On a sheet of paper or in a
graphing utility, graph the line that connects the two points. Write the
slope-intercept form of this line (A). Write the slope-intercept
form of the line (B) perpendicular to line A that passes
through the origin. Write the slope-intercept form of the line (C)
perpendicular to line A that passes through the y-intercept
of line A. Detail your work and tell what the coordinates of all
of the relevant points are.

Answer 1

The slope-intercept form of the equation of a line is ideal for answering this problem.  Let’s consider it first.  Given two points, P₁:(x₁,y₁) and P₂:(x₂,y₂), the slope-intercept form is:

     y = mx + b      where m is the slope [the “rise over run” or (y₂-y₁)/(x₂-x₁) ]

             b    is the y-intercept (the value of y when x=0; this is where the line crosses the y-axis)

     y = [(y₂-y₁)/(x₂-x₁)]x + b

 

With actual points, you can calculate the value of b [(use the slope and either point for (x,y)].

 

This answers (A).

 

Now, for (B), realize that all lines perpendicular to the first line have a slope that is the negative reciprocal of the slope of this line.  Their slope is (-1/m).  They have a different y-intercept:

          y  = (-1/m)x + d

 

The problem states that the line goes through the Origin [the point (0,0), so d=0].

 

For (C), the perpendicular line that passes through the y-intercept of line a has the same y-intercept (note: it “passes through” that point), so b=d and the equation of that line is:

          y = (-1/m)x + b

 

Now, check that all points are identified and that you have included sufficient detail.                                                                                                                                                                                                                                                          

Answer 2

Slope intercept form of a line is of the form y=mx+b where m is the slope of the line and b the y intercept

The problem doesn't specify the initial points.  Let's take points (-4,4) and (6,3).  The slope of a line joining these points is (3-4)/(6+4)=-1/10

 

We have one of the unknowns, namely the slope of the line.  The slope intercept form now becomes

 

y=(-1/10)x+b

 

All points that lie on this line must satisfy this equation.  Plugging in the y and x values from the first point we solve for b as follows

 

4=(-1/10)(-4)+b

 

solving this for b, we have b = 18/5

 

first part is complete.  

 

Any line perpendicular to this has slope of negative reciprocal of this line.  The slope of a line perpendicular to this line will be -1 * (-10) = 10

 

The general slope intercept equation of this line will be y=10x+b

 

This line must pass through the origin which has coordinates of (0,0).  Plugging in zeros for x and y in the above equation, you have 0 = 10*0+b

 

solving for b, we have b=0.  So the equation of the line perpendicular to the first line is y=10x.