Past Papers’ Solutions | Edexcel | AS & A level | Mathematics | Core Mathematics 1 (C1-6663/01) | Year 2012 | January | Q#6
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Question
The line has equation 2x − 3y +12 = 0.
a. Find the gradient of .
The line crosses the x-axis at the point A and the y-axis at the point B, as shown in Figure.
The line is perpendicular to
and passes through B.
b. Find an equation of .
The line crosses the x-axis at the point C.
c. Find the area of triangle ABC.
Solution
a.
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;
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 neither coordinates of a point on line nor slope of the line
.
First we find coordinates of a point on line .
We are given that line passes through B. We are also given that the line
crosses the y-axis at the point B. Therefore, point B is y-intercept of the 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 are given equation of line ;
We substitute ;
Hence, coordinates of point B are (0,4).
Now we have coordinates of a point on 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(0,4) 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 the area of .
We need coordinates of all three vertexes of triangle to sketch it in order to calculate its area.
We have found in (b) that coordinates of point B are (0,4).
We now find coordinates of point A.
We are given that the line crosses the x-axis at the point A. Therefore, point A 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 are given equation of the line ;
We substitute ;
Therefore, coordinates of point A are (-6,0).
Next, we find coordinates of point C.
We are given that the line crosses the x-axis at the point C. Therefore, 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 ;
Therefore, coordinates of point C are .
We sketch triangle with the help of these coordinates of three points.
Expression for the area of the triangle is;
For ;
We need to find AC and BD.
First we find AC.
Expression for the distance between two given points and
is:
We have coordinates of and
. Therefore;
Next we need to find BD. It is evident from the diagram that it is only the distance of B from origin D(0,0) along y-axis which is 4 since B(4,0).
Therefore;
Hence;
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