# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2016 | Feb-Mar | (P1-9709/12) | Q#5

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Question

Two points have coordinates A(5,7) and B(9,-1).

i.       Find the equation of the perpendicular bisector of AB.

The line through C(1,2) parallel to AB meets the perpendicular bisector of AB at the point X.

ii.       Find, by calculation, the distance BX.

Solution

a.

We are required to write equation of perpendicular bisector of AB with points A(5,7) and B(9,-1).

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).

Since we are required to find the equation of perpendicular bisector we need coordinates of a point  on it and we know that point is mid-point of AB.

Let’s find coordinates of mid-point (M) of AB.

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 Therefore, coordinates of M mid-point of AB with points A(5,7) and B(9,-1);

x-coordinate of mid-point of the line y-coordinate of mid-point of the line Hence, coordinates of mid-point M are (7,3).

Next, we need slope of perpendicular bisector of AB to write its equation.

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

Let’s find the slope of line AB.

Expression for slope of a line joining points and ; Therefore, slope of line AB having points A(5,7) and B(9,-1);   Now we can find slope of perpendicular bisector of AB.    Finally, with coordinates of point M(7,3)on perpendicular bisector of AB and its slope we can write equation of perpendicular bisector of AB.

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

We are required to calculate distance BX.

Expression to find distance between two given points and is: We are given the coordinates of point B(9,-1) but we do not have coordinates of point X.

Let’s find coordinates of point X.

We are given that point X is the point of intersection of line passing through C(1,2) and parallel to  AB, say , and the perpendicular bisector of AB.

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).

From (a) we have equation of the perpendicular bisector of AB; We need to find equation of line .

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 are given that point on the line has coordinates C(1,2). Therefore, we need slope of line to  write its equation.

We are given the line is parallel to AB.

If two lines are parallel to each other, then their slopes and are equal; Therefore; From (a) we have, , therefore; Now we can write equation of line , with coordinates of point C(1,2) on and slope of line as  .

Point-Slope form of the equation of the line is;      Now we can equate equations of perpendicular bisector of AB and line to find the coordinates of  their point of intersection X;         Single value of x indicates that there is only one intersection point.

Therefore, x-coordinate of point X is .

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 equation of line ;   Hence, coordinates of .

Finally, we can calculate distance BX from point and ;      