Past Papers’ Solutions | Assessment & Qualification Alliance (AQA) | AS & A level | Mathematics 6360 | Pure Core 1 (6360-MPC1) | Year 2007 | January | Q#2
The line AB has equation and the point A has coordinates .
i. Find the gradient of AB.
ii. Hence find an equation of the straight line which is perpendicular to AB and which passes through A.
b. The line AB intersects the line with equation at the point B. Find the coordinates of B.
c. The point C has coordinates and the distance from A to C is 5 . Find the two possible values of the constant k.
Slope-Intercept form of the equation of the line;
Where is the slope of the line.
For the given case, we have;
We can rearrange the given equation in slope-intercept form;
Comparing with standard slope-intercept form of the equation, we can see that;
Hence gradient (slope) of the equation is .
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 required to find the equation of the line which passes through point . Hence we have coordinates of the point and now we need slope of the line to write its equation.
We are given that required line is perpendicular to the line AB with equation;
If two lines are perpendicular (normal) to each other, then product of their slopes and is;
We know from a(i) that
Substituting the value of ;
Point-Slope form of the equation of the line is;
Now we can write equation of the line with slope and passing through point .
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 AB, from (a:i) is;
Equation of the other line is;
We can also rearrange this equation into slope-intercept form as follows;
Equating both equations;
Single value of x indicates that there is only one intersection point.
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.
Hence coordinates of .
Expression to find distance between two given points and is:
We are given that distance between two points and is 5.
We can substitute given data into expression of distance between two points;
Now we have two options.