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CH-06-001 · Line–parabola intersections and distance

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A line has equation y=−xy=-x. A curve has equation y=x2−7x−16y=x^{2} - 7x - 16. They meet at A and B. Find both intersection coordinates algebraically. Then calculate AB to 2 decimal places.

Simultaneous Equations (Quadratic & Linear) + Length of a Line Booklet p. 28 & p. 80
Method
1) Substitute the linear equation into the quadratic
2) Rearrange to zero
3) Factorise with a grid
4) Solve and substitute back to pair the coordinates
Worked Example (CH-06-001)
1)
Substitute the linear expression for y into the quadratic: x2−7x−16=−x+0x^2 - 7x - 16=-x + 0.
2)
Rearrange to zero: x2−6x−16=0x^2 - 6x - 16=0.
3)
Multiply a×c=1×(−16)=−16a\times c=1\times(-16)=-16. Choose the signed factor pair 2, −82,\ -8: product −16-16, sum −6-6.
4)
×xx22
xxx2x^22x2x
−8-8−8x-8x−16-16
5)
Read the factors outside the grid: (x−8)(x+2)(x - 8)(x + 2).
6)
x−8=0⇒x=8x - 8=0\quad\Rightarrow\quad x=8
7)
x+2=0⇒x=−2x + 2=0\quad\Rightarrow\quad x=-2
8)
Substitute x=8x=8 into the line: y=(−1)(8)+0=−8y=(-1)(8) + 0=-8; point (8,−8)(8,-8).
9)
Substitute x=−2x=-2 into the line: y=(−1)(−2)+0=2y=(-1)(-2) + 0=2; point (−2,2)(-2,2).
10)
Distance formula (p. 80): AB=(−10)2+(10)2=14.14AB=\sqrt{(-10)^2+(10)^2}=14.14.
Ans = A(-2, 2) and B(8, -8); AB = 14.14 units.