darkfenrir
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I can't seem to think of what to do here... anyone able to help?
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Ruse 2009 trial Q7 part iii) onwards, help plz? (1 Viewer)
- Thread starter darkfenrir
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Carrotsticks
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(iii): Since we have two different equations for the same line L, we can equate the coefficients of x, y and the constant term. That is how (iii) is acquired. Same line have the same ratio between coefficients.
For example, suppose the following two are actually the same line.
Ax+By+C=0
Px+Qy+R=0
A/P = B/Q = C/R
Because they are the same line.
(iv): Since we have the identities from (iii), we can substitute them into Q to get the coordinates of Q in terms of t,a,b and c.
Now, to prove the identity, substitute the above into the equation in (ii).
(v): Similar to previous parts.
(vi): If PQRS is a rhombus, then the diagonals PR and QS are perpendicular. So get the gradient of PR, multiply with the gradient of QS, and let it be equal to -1 and it should cancel out nicely to give b^2=a^2.
For example, suppose the following two are actually the same line.
Ax+By+C=0
Px+Qy+R=0
A/P = B/Q = C/R
Because they are the same line.
(iv): Since we have the identities from (iii), we can substitute them into Q to get the coordinates of Q in terms of t,a,b and c.
Now, to prove the identity, substitute the above into the equation in (ii).
(v): Similar to previous parts.
(vi): If PQRS is a rhombus, then the diagonals PR and QS are perpendicular. So get the gradient of PR, multiply with the gradient of QS, and let it be equal to -1 and it should cancel out nicely to give b^2=a^2.