Sample Question

 

The following question is a typical example of a Leaving Cert question on the line

  1.  Show that the point a=(-3,4)  is on the line L:2x+5y=14 and write down the slope

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  1. p=(-1,6)  q=(-3,2)

  1. Find the midpoint of [pq]

  2. Find the slope of [pq]

  3. Show that the equation of the line K, the perpendicular bisector of [pq] is x+2y=6

  4. Find the point where the line K intersects the x-axis

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  1. Using the lines L and K find the point t such that t = KL

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  1. O is the origin and s(3,1).  Find r such that osrt is a parallelogram & find the area.

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Answer

 

(A)    Show that the point a=(-3,4)  is on the line L:2x+5y=14 and write down the slope

 

To show a point is on a line we sub it into the equation:

  L: 2x  +  5y   = 14

    (-3) + 5(4) = 14

      -6  +  20  = 14

               14 = 14

therefore the point (-3,4) is on the line L:2x + 5y = 14

 

 

To find the slope we need to use the formula y=mx+c as we are given the equation of the line.     

» write out equation   
» write out the line
» get y on the left side     
» get y on its own on the left
» slope is value in front of x

                   

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(B)   p=(-1,6)  q=(-3,2)

  1. Find the midpoint of [pq]

  2. Find the slope of [pq]

  3. Show that the equation of the line K, the perpendicular bisector of [pq] is x+2y=6

  4. Find the point where the line K intersects the x-axis

 

1.    Midpoint of [pq]      

p=(-1,6)  q=(-3,2) » write down values
    (x1,y1)   (x2,y2)      of (x1,y1) & (x2,y2)
» write out the formula         
» sub in the value
» work it out
» Midpoint = (-2,4)

                 

2.    Slope

    Since we are given two points to find the slope we use the formula

  

p=(-1,6)  q=(-3,2) » write down values
   (x1,y1)     (x2,y2)      of (x1,y1) & (x2,y2)     
» write out the formula
       » sub in the values
           » work it out
        » slope  

           

3.    Perpendicular bisector

   To do this part we must first find the equation of the line [pq].  We need a point and the slope in order to find the equation of the line.  We have a point and we have just worked out that the slope.  

Slope = 2            (x1,y1) = p(-1,6)

        y - y1 = m(x-x1) » write out the formula
        y - 6 = 2(x-(-1)) » sub in the values
        y - 6 = 2x+2 » work it out 
        y - 2x -4 = 0   » bring y & x to one side
[pq]:y-2x-4 = 0 » equation of the line [pq]

                  

We are asked to show that K:x+2y=6 is the perpendicular bisector of [pq].  In order to prove this we need to find the slopes of both lines and compare them.  We must remember that multiplying their slopes should result in -1.  to find the slope we can use the formula y=mx+c as we have the two equations of the lines.

       

[pq]:  y - 2x - 4 = 0   K:   x + 2y = 6
          y = mx + c             y = mx + c
           y =  2x + 4             2y =  -x  + 6 
              m1 = 2                 y = x +

     y = -½x + 3    

 

 

            m2 = -½            

Now that we have the slope we can multiply them by each other and we should get -1

       m1 * m2 = 2 * -½  =  -1

therefore K:x+2y=6 is the perpendicular bisector of [pq] : y-2x-4 = 0  

 

4.    Intersects x-axis

 If K intersects the x-axis then y=0
 K:    x + 2y   = 6
       x + 2(0) = 6
       x + 0= 6

          x = 6        Point of intersection: (6,0)

 

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(C)    Using the lines L and K find the point t such that t = KL

         L: 2x  + 5y  = 14              K:    x + 2y = 6     

t is the point of intersection of these two lines.  To find the point of intersection of two lines we use simultaneous equations.

L:     2x + 5y = 14 » write down both lines
K:      x + 2y =  6         under each other     
L:     2x + 5y = 14      » elimnate x by multiplying
K:    -2x - 4y = -12    line K by -2
                   y = 2 » value
   

                       

Sub the y-value back into the original equation to find the x-value.

L:    2x + 5(2) = 14 

       2x + 10 = 14
       2x = 14 - 10

       2x = 4
         x = 2

Point of Intersection:        t(2,2)

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(D)    O is the origin and s(3,1).  Find r such that osrt is a parallelogram & find the area.

 

 The origin is the point (0,0).  We have also worked out that t=(2,2).  Since we know three points of the parallelogram the best thing to do is draw a graph and fill in the information you know.

NOTE:  It is important to remember that when we are dealing with parallelograms opposite lengths of the parallelogram are equal.

This means that    |os| = |tr|                   

To find the point r we use the translations method.

o(0,0) --> s(3,1)        

t(2,2)  -->  r(?,?)        

o(0,0)    s(3,1) » we add three to the x value & 1 to the y value
t(2,2)    r(5,3)  » we add three to the x value & 1 to the y value to get r

 

Area of a parallelogram:

In order to find the area of a parallelogram we must firstly find the area of triangle within the parallelogram and then multiply it by two. 

Since we know that the three points o(0,0), t(2,2) & s(3,1) are correct we will find the area of this triangle.  

o(0,0)  t(2,2)  s(3,1)

 » write down values
  (0,0) (x1,y1) (x2,y2)      of (x1,y1) & (x2,y2) 
½|x1 y2-x2 y1| » write out the formula
½|2. 1-3.2| » sub in the values
½|2-6| » work it out
½|-4| »|-4| means we take the positive value
½|4|  = 2 » Area of triangle tos
2 * 2  = 4 » Area of parallelogram (Area of a Triangle * 2)

      

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