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We have two points F1, F2. F1-F2 is 21m. We have a point (P) outside the line.

The line from F1-P is called D1. The line from F2-P is called D2.

P is 12m away from F1-F2 on a straight line crossing F1-F2 in (N) dividing the triangle F1-P-F2 into two 90*triangles.

Problem

Task A) Use the Pythagorean theorem to show that $\sqrt{D1^2-12^2} + \sqrt{D2^2-12^2}=21$

Since $K1^2 + K2^2 = H^2$ then $D1^2-12^2=21-N>F2$ and $D2^2-12^2=21-N>F1$ And then $\sqrt{D1^2-12^2} + \sqrt{D2^2-12^2}=21$

Task B)

$D1=D2+7m$

Find D1 and D2.

Halp!! I have tried and tried, isolating square roots and quadrating both sides then repeating the process and using the quadratic equation but i get the wrong answer each time. The answer is supposed to be D1=20m D2=13m but i can't get that answer at all.

Antonio Vargas
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Arne Morten
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1 Answers1

1

This is task B:

Let $x = N-F2$. Then $$ \begin{align} x^2+12^2&=D_2^2 \\ (21-x)^2+12^2&=D_1^2 \\ D_1&=D_2+7 \end{align} $$ Substitute the last equation in the second to get $$ \begin{align} x^2+12^2&=D_2^2 \\ 441-42 x + x^2 +12^2&=D_2^2+14~D_2 + 49 \end{align} $$ Solve for $D_2$ from the first equation and substitute in the second to get $$ 14\,\sqrt{{x}^{2}+144}=42\,x-392 $$ Square both sides and re-arrange $$ 1568\,{x}^{2}-32928\,x+125440=0 $$ Solve to get $$ x = 5, 16$$ Now go back and find $D_2$ and $D_1$. $x=16$ gives negative values for $D_1$ and $D_2$. So the answer is $$ x=5, D1=20, D2=13 $$

Note: For task A, solve for $x$ from the first equation and substitute in the second or just observe that the two square roots are F2 to N and from N to F1, so they add to 21

user44197
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