# Past Papers’ Solutions | Edexcel | AS & A level | Mathematics | Core Mathematics 1 (C1-6663/01) | Year 2010 | January | Q#3

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Question

The line  has equation 3x+5y-2=0.

a.   Find the gradient of line for .

The line  is perpendicular to  and passes through the point (3,1).

b.   Find the equation of  in the form y=mx+c, where m and c are constants..

Solution

a.

We are given equation of the line  as;

We are required to find the gradient of the line .

Slope-Intercept form of the equation of the line;

Where  is the slope of the line.

Therefore, we can find slope of the line  by rearranging its equation in terms slope-intercept form  as follows.

Therefore, slope of the line  is;

b.

We are required to find equation of .

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 given that line  passes through the point (3,1) and is perpendicular to line .

If a line  is normal to the curve , then product of their slopes  and  at that point (where line  is normal to the curve) is;

Therefore;

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

From (a) we have found that;

Therefore;

Now we can write the equation of line

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

Therefore;