Past Papers’ Solutions | Assessment & Qualification Alliance (AQA) | AS & A level | Mathematics 6360 | Pure Core 1 (6360-MPC1) | Year 2006 | January | Q#2

 

Question

The point  has coordinates  and the point  has coordinates .

The line  has equation .

a.    

                           i.       Show that .

                         ii.       Hence find the coordinates of the mid-point of .

b.   Find the gradient of .

c.   Line  is perpendicular to the line .

                           i.       Find the gradient of .

                         ii.       Hence find the equation of the line .

                       iii.       Given that point  lies on the x-axis, find its x-coordinate.

Solution

a.
 

                    i.
 

We are given that point  lies on the line  whose equation is given as;

Since coordinates of a point on the given line must satisfy equation of the lien.

Therefore  must satisfy the equation;

                  ii.
 

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

For the given case we have the point ( and the point . Therefore;

x-coordinate of mid-point  of the line

y-coordinate of mid-point  of the line

Hence mid-point of has coordinates .

b.    

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

For the given case we have the point ( and the point . Therefore;

c.    

                            i.
 

We are given that line  is perpendicular to the line .

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

Therefore;

From (b) we have   ;

                          ii.
 

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).

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

For line AC we have coordinates one point on the line  and also slope from (c:i) as .

Therefore;

 

                         iii.
 

We are given that point  lies on x-axis that means it is x-intercept of line

The point  at which curve (or line) intercepts x-axis, the value of . So we can find the  value of  coordinate by substituting  in the equation of the curve (or line).

Equation of the line , from (c:ii) is;

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