Past Papers’ Solutions | Edexcel | AS & A level | Mathematics | Core Mathematics 1 (C1-6663A/01) | Year 2014 | January | Q#6

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Question

The straight line  has equation 2y = 3x + 7.

The line  crosses the y-axis at the point A as shown in Figure.

a.    

                    i.       State the gradient of .

                  ii.       Write down the coordinates of the point A.

Another straight line  intersects  at the point B (1, 5) and crosses the x-axis at the point C, as  shown in Figure.

Given that ,

b.   find an equation of  in the form ax + by + c = 0, where a, b and c are integers. 

The rectangle ABCD, shown shaded in Figure, has vertices at the points A, B, C and D.

c.   Find the exact area of rectangle ABCD.

Solution

a.    

                                            i.
  

We are given equation of line ;

We are required to find the gradient of .

Slope-Intercept form of the equation of the line;

Where  is the slope of the line.

Therefore, we can rearrange the given equation of line  in slope-intercept form, as follows, to find  the gradient of the line.

Hence, gradient of the line  is;


ii.
 

We are required to find the coordinates of the point A which is y-intercept of line .

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

We substitute x=0 in equation of the line ;

Hence, coordinates of y-intercept of the line  (point A) are .

b.    

We are required 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 have coordinates of a point B(1,5) on line  but not slope of the line .

Next we need slope of the line  to write its equation. 

We are given that the line  is perpendicular to .

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

Therefore, if we have slope of line , we can find slope of the line .

From (a), we have found that , therefore;

With coordinates of a point on the line  as B(1,5) and its slope  at hand, we can write equation of the line .

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

c.
 

We are required to find area of rectangle ABCD.

Expression for the area of the rectangle with length  and width  is;

It is evident that length of rectangle ABCD is BC and width is AB.

We need to find the distances AB and BC.

Expression for the distance between two given points  and is:

We already have coordinates of points A and B but we need to find coordinates of point C.

It is evident that point C is the x-intercept of the 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). 

We have found equation of the line  in (b);

We substitute y=0;

Hence, coordinates of point C are .

Now we can find the distances AB and BC.

Coordinates of points A, B and C are respectively as follows;

Therefore;

Hence;

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