# C1 2017 8 Edexcel

(a) Note that l_1 has gradient \dfrac 4 5.

If x = 5 then:

5y = 20 + 10

So:

y = 6

So P = (5, 6).

Since l_2 is perpendicular to l_1 it has gradient -\dfrac 5 4. So l_2 has equation:

\displaystyle y - 6 = -\dfrac 5 4 (x - 5)

So:

\displaystyle 4y - 24 = -5(x - 5) = -5x + 25

So:

\displaystyle 4y + 5x - 49 = 0

(b)

We want to find T and S. Setting y = 0 in the equation for l_1 we have 4x = -10, so x = -\dfrac 5 2. So \displaystyle S = \left(-\frac 5 2, 0\right)

Setting y = 0 in the equation for l_2 we have 5x = 49, so x = \dfrac {49} 5. So \displaystyle T = \left(\frac {49} 5, 0\right).

Note that the perpendicular height of the triangle \Delta SPT is just 6 (since P = (5,6)) , so the area of the triangle is:

\begin{align*} \frac 1 2 \times 6 \times \left(\frac {49} 5 + \frac 5 2\right) & = 3 \times \left(\frac {98} {10} + \frac {25} {10}\right) \\ & = 3 \times \left(\frac {123} {10}\right) \\ & = \frac {369} {10}\end{align*}

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