# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2010 | May-Jun | (P1-9709/11) | Q#8

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**Question**

The diagram shows a triangle in which is and is . The gradients of , and are , and respectively, where is a positive constant.

**
i. **Find the gradient of and deduce the value of m.

** ii. **Find the coordinates of C.

The perpendicular bisector of meets at .

** iii. **Find the coordinates of D.

**Solution**

i.

Expression for slope (gradient) of a line joining points and ;

We have the coordinates of the two pints of line AB. is and is .

Therefore;

We are given that;

Hence;

ii.

To find the coordinates of point C we recognize that point C is the point of intersection of lines BC & AC.

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 the lines BC & AC.

First we find the equation of BC.

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 both coordinates of point B and slope of the line BC. is and . From (i), , therefore, .

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

Hence, equation of the line BC;

Now we find the equation of AC.

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 both coordinates of point A and slope of the line AC. is and . From (i), , therefore, .

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

Hence, equation of the line AC;

Equating both equations;

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 equation of line BC;

Hence, coordinates of point .

iii.

Perpendicular bisector of AB intersects BC at D. That means perpendicular bisector of AB intersects AB at mid-point of AB i.e. point M. It is evident that point D is the intersection point of BC & MD.

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 the lines BC & MD.

We have equation of line BC from (i);

Now we find the equation of the line MD.

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).

First we find the coordinates of the point M. Since MD is the perpendicular bisector of AB, point M is the mid-point of AB.

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

We have coordinates of both and .

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;

Therefore, coordinates of point .

Now we need slope of MD to write its equation.

We know that MD is perpendicular to AB.

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

From (i), we have . Therefore;

Now we can write equation of the perpendicular bisector MD.

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

Hence, equation of the line MD;

Now we have equations of both lines BC & MD. We can equate both equations to find the coordinates of their point of intersection D.

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 equation of line BC;

Hence, coordinates of point .

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