# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2016 | May-Jun | (P1-9709/13) | Q#11

Question

Triangle ABC has vertices at A(2,1), B(4,6) and C(6,3).

i.
Show that triangle ABC is isosceles and find the exact area of this triangle.

ii.    The point D is the point on AB such that CD is perpendicular to AB. Calculate the x-coordinate of  D.

Solution

i.

An isosceles triangle is a triangle with (at least) two equal sides.

To show that triangle ABC with vertices at A(2,1), B(4,6) and C(6,3) is isosceles, we need to show any two sides equal.

Let’s find the lengths of two sides of the triangle.

Expression to find distance between two given points and is: Therefore, with A(2,1) and B(4,6) we can find AB as follows;  Also, with B(4,6) and C(6,3) we can find BC as follows;  Since two sides of triangle ABC are equal , it is isosceles triangle.

Next, we are required to find the area of triangle ABC.

Expression for the area of the triangle is; We have seen above that triangle ABC is isosceles triangle and we have also found lengths of two  equal sides.

We can draw the triangle as shown below. We know that . It is evident that AC is the base of the triangle ABC and BM is the  height of the triangle ABC which also means BM is perpendicular bisector of AC. There are many ways to find the length of BM like Pythagoras Theorem and distance formula. We  will use distance formula.

Expression to find distance between two given points and is: Therefore, we need coordinates of points B and M to find BM.

To find distance BM we have coordinates of point B(4,6) but we need coordinates of point M. Since  BM is perpendicular bisector of AC, the point M is the mid-point of AC.

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 We have A(2,1) and C(6,3), therefore;

x-coordinate of mid-point of the line y-coordinate of mid-point of the line Hence, M(2,-2).

Now we can find distance BM from coordinates of B(4,6) and M(2,-2);   Finally we need length of AC to find the area of triangle ABC.

Expression to find distance between two given points and is: We have A(2,1) and C(6,3), therefore;   Hence;   ii.

We are given that point D is the point on AB such that CD is perpendicular to AB. We can draw  according to given condition. It is evident that point D is intersection point of lines AB and CD. Therefore, we are required to find  the coordinates of a point of intersection.

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).

Therefore, we need equations of both lines AB and CD. Let’s find these equations one by one.

First we find the equation of line AB.

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).

We have coordinates of both points A(2,1) and B(4,6).

Two-Point form of the equation of the line is; Therefore;         Now we find equation of line CD.

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).

We have coordinates of one point on the line CD and that is C(6,3). Therefore, we need slope of  line CD to write its equation.

We are given that lines AB and CD are perpendicular.

If two lines are perpendicular (normal) to each other, then product of their slopes and is;  We can find slope of line CD if we know slope of line AB.

Slope-Intercept form of the equation of the line; Where is the slope of the line.

We have found equation of line in slope-intercept form above as; Therefore, slope of line AB is Hence, we can find slope of line CD as;    Point-Slope form of the equation of the line is; Now, with coordinates of a point C(6,3) and slope of line CD at hand, we can write equation of CD.         Now we have equations of both intersecting lines, so we can find coordinates of point of  intersection.

Equation of the line AB is; Equation of the line CD is; Equating both equations;             Single value of x indicates that there is only one intersection point.

Hence, x-coordinate of point D is .