Question

The distance between the two points of intersection of \(x^{2}+y=7\) and \(x+y=7\) (rounded off to two decimal places) is __________________.

Hint:

- Use substitution from the linear equation into the quadratic.
- Distance between \((x_1,y_1)\) and \((x_2,y_2)\): \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\).

Answer 1

Correct Answer: 1.4

Step 1: From \(x+y=7\Rightarrow y=7-x\). Substitute in \(x^2+y=7\): \(x^2+7-x=7\Rightarrow x^2-x=0\Rightarrow x(x-1)=0\).
Step 2: Hence \(x=0\) or \(x=1\). Corresponding \(y\) values: \(y=7\) or \(y=6\). Points are \((0,7)\) and \((1,6)\).
Step 3: Distance \(=\sqrt{(1-0)^2+(6-7)^2}=\sqrt{1+1}=\sqrt{2}\approx 1.414\Rightarrow 1.41\) (2 d.p.).