# C1 2016 10 Edexcel

(a)

The gradient of l_1 is:

\displaystyle \frac {7 - 2} 3 = \frac 5 3

Since l_2 is perpendicular to l_1 it has gradient:

-\dfrac 3 5

Since l_2 passes through (3,7), it has equation:

\displaystyle y - 7 = -\frac 3 5 (x - 3)

So:

\displaystyle 5y - 35 = -3(x - 3) = -3x + 9

So:

\displaystyle 5y + 3x - 44 = 0

In particular a = 3, b = 5, c = -44.

(b)

The point R has y = 0, so:

3x = 44

So:

\displaystyle x = \frac {44} 3

So R has coordinates \displaystyle \left(\frac {44} 3, 0\right)

Let S be the intersection of l_2 with the y-axis. This has y coordinate satisfying:

5y - 44 = 0

So:

\displaystyle y = \frac {44} 5

We can then work out the area of the quadrilateral as:

\mathrm{area}(\square ORPQ) = \mathrm{area}(\triangle SOR) - \mathrm{area}(\triangle SQP)

The triangle SOR has height \dfrac {44} 5 and base length \dfrac {44} 3, so has area:

\displaystyle \frac 1 2 \times \frac {44} 5 \times \frac {44} 3 = \frac {22 \times 44} {15} = \frac {968} {15}

The triangle SQP has perpendicular height 3 (the x-coordinate of Q) and base length \displaystyle |PS| = \frac {44} 5 - 2 = \frac {34} 5. So has area:

\displaystyle \frac 1 2 \times 3 \times \frac {34} 5 = \frac {17 \times 3} 5 = \frac {51} 5

So:

\displaystyle \mathrm{area}(\square ORQP) = \frac {968} {15} - \frac {51} 5 = \frac {968 - 153} {15} = \frac {815} {15} = \frac {163} 3

There are loads of other ways to split up the quadrilateral \square ORQP and a direct way using determinants.

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