<s>[INST] New Maths 9-1 GCSE for 2017 Sample Question My teacher gave me some practice questions for my end of year exam which will be like the new GCSE and this question is very tackling to me. Could any with clear working solve the question and show me how they did it:
A(-2,1), B(6,5), and C(4,k) are the vertices of a right angled triangle ABC.
Angle ABC is the right angle.
Find an equation of the line that passes through A and C.
Give your answer in the form ay+bx=c where a, b and c are integers.
(Total for Question = 5 marks)
 [/INST] Since $\angle ABC$ is the right angle, according to Pythagorean theorem, we have $AB^2+BC^2=AC^2$
$AB^2=(6-(-2))^2+(5-1)^2=80$
$BC^2=(k-5)^2+(4-6)^2=(k-5)^2+4$
$AC^2=(k-1)^2+(4-(-2))^2=(k-1)^2+36$
So we have $80+(k-5)^2+4=(k-1)^2+36 \Rightarrow k=9$
The line passing through A and C has the form $y=\alpha x+\beta$
Calculate the slope $\alpha$:
$\alpha=\frac{k-1}{4-(-2)}=\frac{9-1}{4-(-2)}=\frac{4}{3}$
Plug in $A(-2,1)$ into the equation $y=\frac{4}{3}x+\beta$:
$1=\frac{4}{3}(-2)+\beta \Rightarrow \beta =\frac{11}{3}$
So, $y=\frac{4}{3}x+\frac{11}{3}$. We can rewrite this as $y-\frac{4}{3}x=\frac{11}{3}$, and multiply it by 3 to get $3y-4x=11$
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