# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2014 | Oct-Nov | (P1-9709/12) | Q#9

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Question

The diagram shows a trapezium ABCD in which AB is parallel to DC and angle BAD is . The coordinates of A, B and C are ,  and  respectively.

i.       Find the equation of AD.

ii.       Find, by calculation, the coordinates of D.

The point E is such that ABCE is a parallelogram.

iii.       Find the length of BE.

Solution

i.

To find the equation of the line either we need coordinates of the two points on the line (Two-Point  form of Equation of Line) or coordinates of one point on the line and slope of the line (Point-Slope  form of Equation of Line).

For the given case, line  we have the coordinates of a point on line  i.e . For the slope  we utilize the fact that  is perpendicular to .

If two lines are perpendicular (normal) to each other, then product of their slopes  and  is;

Hence;

To find the slope of .

Expression for slope of a line joining points  and ;

For the given case;

Hence;

Now we can write the equation of the line  using;

Point-Slope form of the equation of the line is;

For the given case;

ii.

To calculate the coordinates of point , we utilize the fact that it is the point of intersection of the  lines  and . Hence we are required to find the coordinates of a point of intersection of two  lines.

If two lines (or a line and a curve) intersect each other at a point then that point lies on both lines i.e.  coordinates of that point have same values on both lines (or on the line and the curve).  Therefore, we can equate  coordinates of both lines i.e. equate equations of both the lines (or the  line and the curve).

For the given case;

Equation of the line  from (i) is;

Now we find the equation of .

To find the equation of the line either we need coordinates of the two points on the line (Two-Point  form of Equation of Line) or coordinates of one point on the line and slope of the line (Point-Slope  form of Equation of Line).

For the given case, line  we have the coordinates of a point on line  i.e . For the slope we utilize the fact that  is parallel to . Hence;

We have found slope of  in (i) as;

Therefore;

Now we can write the equation of the line  using;

Point-Slope form of the equation of the line is;

For the given case;

Equation of the line  is;

We can also manipulate this equation;

Equating both equations of lines  and ;

Single value of x indicates that there is only one intersection point.

With x-coordinate of point of intersection of two lines (or line and the curve) at hand, we can find the  y-coordinate of the point of intersection of two lines (or line and the curve) by substituting value  of x-coordinate of the point of intersection in any of the two equations.

We choose;

Hence coordinates of  point are .

iii.

We are given that ABCE is a parallelogram.

Consider the diagram below. If ABCE is a parallelogram, then BC and AE must be equal and  parallel. We also know that diagonal of a parallelogram bisect each other.

Therefore, for the given case, AC and BE diagonals bisect at point  which is the mid-point of both AC and BE. We can find coordinates of .

To find the mid-point of a line we must have the coordinates of the end-points of the line.

Expressions for coordinates of mid-point of a line joining points  and;

x-coordinate of mid-point  of the line

y-coordinate of mid-point  of the line

Since  is the mid-point of  and ;

x-coordinate of mid-point  of the line

y-coordinate of mid-point  of the line

Therefore,
.

Since  is also mid-point of BE, we can find coordinates of point E.

Therefore, .

Now we can find length of BE.

Expression to find distance between two given points  and is:

For  and  .