Question

the work out to the line x-y+4=0 intersects the curve y=2x^2-4x+1 at points p and q if coordinates of p are 3,7 what are the coordinates of q

Answer 1

Given the equation of a straight line as;

$x-y+4=0$

Let's make y the subject of the equation by subtracting (x+4) from both sides of the equation, we have;

$\begin{gathered} x-y+4-x-4=0-(x+4) \\ -y=-(x+4) \\ y=x+4\ldots\ldots\ldots\ldots\ldots\text{.}\mathrm{}\text{equation 1} \end{gathered}$

Also, the equation of the curve is;

$y=2x^2-4x+1\ldots\ldots\ldots\ldots\ldots\text{.equation 2}$

Then, we equate equation 1 and equation 2, we have;

$\begin{gathered} x+4=2x^2-4x+1 \\ 2x^2-4x+1-x-4=0 \\ 2x^2-5x-3=0 \\ 2x^2-6x+x-3=0 \\ 2x(x-3)+1(x-3)=0 \\ x-3=0 \\ x=3; \\ 2x+1=0 \\ 2x=-1 \\ x=-\frac{1}{2} \\ x=-0.5 \end{gathered}$

At point x=3;

$\begin{gathered} y=x+4 \\ y=3+4 \\ y=7 \end{gathered}$

The coordinate of point P is;

$(3,7)$

At point x=-0.5;

$\begin{gathered} y=-0.5+4 \\ y=3.5 \end{gathered}$

The coordinate of point Q is;

$(-0.5,3.5)$