Past Papers’ Solutions  Cambridge International Examinations (CIE)  AS & A level  Mathematics 9709  Pure Mathematics 1 (P19709/01)  Year 2018  OctNov  (P19709/12)  Q#10
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Question
The equation of a curve is and the equation of a line is , where k is a constant.
i. Find the set of values of k for which the line does not meet the curve.
In the case where k = 15, the curve intersects the line at points A and B.
ii. Find the coordinates of A and B.
iii. Find the equation of the perpendicular bisector of the line joining A and B.
Solution
i.
We are given equation of the curve and the line as follows;
We can write equation of the line in slopeintercept form.
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).
Equation of the line is given as;
Equation of the curve is;
Equating both equations;
Therefore, if given line and the given curve do not intersect, there should be no solution of this equation.
It is evident that it is a quadratic equation.
Standard form of quadratic equation is;
Expression for discriminant of a quadratic equation is;
If ; Quadratic equation has two real roots.
If ; Quadratic equation has no real roots.
If ; Quadratic equation has one real root/two equal roots.
Therefore, for the given equation;
To find the set of values of k for which , we solve the following equation to find critical values of ;
Now we have two options;




Hence the critical points on the curve for the given condition are 12 & 12.
Standard form of quadratic equation is;
The graph of quadratic equation is a parabola. If (‘a’ is positive) then parabola opens upwards and its vertex is the minimum point on the graph. If (‘a’ is negative) then parabola opens downwards and its vertex is the maximum point on the graph.
It is evident that is an upward opening parabola.
Therefore, conditions for are;
Hence, line and curve do not intersect for;
ii.
We are given equation of the curve and the line as follows;
We are given that ;
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).
Equation of the line is given as;
Equation of the curve is;
Equating both equations;
Now we have two options.





Two values of x indicate that there are two intersection points.
Corresponding values of y coordinate can be found by substituting values of x in any of the two equation i.e either equation of the line or equation of the curve.
We choose equation of the line;






Hence, line and the curve intersect at points;



iii.
We are required to write equation of perpendicular bisector of AB with points and .
To find the equation of the line either we need coordinates of the two points on the line (TwoPoint form of Equation of Line) or coordinates of one point on the line and slope of the line (PointSlope 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 midpoint of AB.
Let’s find coordinates of midpoint (M) of AB.
To find the midpoint of a line we must have the coordinates of the endpoints of the line.
Expressions for coordinates of midpoint of a line joining points and;
xcoordinate of midpoint of the line
ycoordinate of midpoint of the line
Therefore, coordinates of M midpoint of AB with points A(4,11) and B(1,14);
xcoordinate of midpoint of the line
ycoordinate of midpoint of the line
Hence, coordinates of midpoint M are (,).
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(4,11) and B(1,14);
Now we can find slope of perpendicular bisector of AB.
Finally, with coordinates of point M(,) on perpendicular bisector of AB and its slope we can write equation of perpendicular bisector of AB.
PointSlope form of the equation of the line is;
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