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This is based on a parametric equation problem.

We have two ships A and B at $(-2,at +1)$ and $(4, t+10)$ respectively.

I need to show that $d^2 = (1-a)^2t^2 +18(1-a) t +117$

using the distance formula I have got:

$d^2 = (4+2)^2 +((t+10) - (at +1))^2$

Where can I go from here? I have tried expanding the brackets to no avail.

I feel like the fact that $(1-a)^2 = a^2 - 2a +1$ should help down the line but I am struggling.

Please help.

Mathsguy9020
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  • Just expand the brackets again, but carefully... Or perhaps rearrange the inside before expanding... – Martigan Jun 11 '15 at 13:34

2 Answers2

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Using $(A+B)^2=A^2+2AB+B^2,$ we have$$\begin{align}d^2&=(4-(-2))^2+((t+10)-(at+1))^2\\&=6^2+((1-a)t+9)^2\\&=36+(1-a)^2t^2+2\cdot 9\cdot (1-a)t+9^2\\&=(1-a)^2t^2+18(1-a)t+117\end{align}$$

mathlove
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Planning to rewrite the expression $$(4+2)^2 +((t+10) - (at +1))^2$$as a polynomial in $t$, regroup inside the parenthesis, $$6^2 +((1-a)t+9)^2$$ and evaluate the squares $$36+(1-a)^2t^2+2\cdot9(1-a)t+9^2,$$ i.e. $$(1-a)^2t^2+18(1-a)t+117.$$