To find the points of intersection, solve the equations simultaneously by substitution.

Substitute the equation of the line into the equation of the curve

$x^{2} + x (\frac{2}{3} x - 2) = 9$

[1]

Expanding the brackets gives

$x^{2} + \frac{2}{3} x^{2} - 2 x = 9$

Collecting like terms and setting the quadratic equal to 0

$\frac{5}{3} x^{2} - 2 x - 9 = 0$

Multiplying through by 3 gives

$5 x^{2} - 6 x - 27 = 0$

[1]

To factorise, we look for a pair of numbers that has a product of -135 (from doing $5 \times - 27$ ) and a sum of -6. The pair of numbers would be -15 and 9 so we split the middle term

$5 x^{2} - 15 x + 9 x - 27 = 0$

Factorising each half of the equation

$5 x (x - 3) + 9 (x - 3) = 0$

We have a common factor of $x - 3$ and the remaining terms create the other factor

$(5 x + 9) (x - 3) = 0$

[1]

Solving each bracket gives the x co-ordinates

$x = - \frac{9}{5}, x = 3$

Substituting these into either equation will give the y co-ordinates

$y = \frac{2}{3} (- \frac{9}{5}) - 2 = - \frac{16}{5}$

$y = \frac{2}{3} (3) - 2 = 0$

$(- \frac{9}{5}, - \frac{16}{5})$ [1]

$(3, 0)$ [1]

Simultaneous Equations (Cambridge O Level Additional Maths)

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