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    A geometric transformation $T:\binom{x}{y} \mapsto\binom{x^{\prime}}{y^{\prime}}$ is defined by $$ T:\binom{x^{\prime}}{y^{\prime}}=\left(\begin{array}{cc} 7 -10 2 -3 \end{array}\right)\binom{x}{y}+\binom{-5}{4} $$

    Question
    HLPaper 1

    A geometric transformation T:(xy)↦(x′y′)T:\binom{x}{y} \mapsto\binom{x^{\prime}}{y^{\prime}}T:(yx​)↦(y′x′​) is defined by T:(x′y′)=(7−102−3)(xy)+(−54)T:\binom{x^{\prime}}{y^{\prime}}=\left(\begin{array}{cc} 7 -10 2 -3 \end{array}\right)\binom{x}{y}+\binom{-5}{4}T:(y′x′​)=(7−102−3​)(yx​)+(4−5​)

    1.

    Find the coordinates of the image of the point (6,-2).

    [2]
    Verified
    Solution

    (7−102−3)(6−2)+(−54)\left(\begin{array}{cc}7 & -10 \\ 2 & -3\end{array}\right)\binom{6}{-2}+\binom{-5}{4}(72​−10−3​)(−26​)+(4−5​) (M1) =(5722)=\binom{57}{22}=(2257​) OR (57,22)(57,22)(57,22) A1 [2 marks]

    2.

    Given that T:(pq)↦2(pq)T:\binom{p}{q} \mapsto 2\binom{p}{q}T:(qp​)↦2(qp​), find the value of p and the value of q.

    [3]
    Verified
    Solution

    (2p2q)=(7−102−3)(pq)+(−54)\binom{2p}{2q}=\left(\begin{array}{cc}7 & -10 \\ 2 & -3\end{array}\right)\binom{p}{q}+\binom{-5}{4}(2q2p​)=(72​−10−3​)(qp​)+(4−5​) 7p−10q−5=2p7p-10q-5=2p7p−10q−5=2p 2p−3q+4=2q2p-3q+4=2q2p−3q+4=2q solve simultaneously: p=13,q=6p=13, q=6p=13,q=6 A1 Note: Award A0\boldsymbol{A0}A0 if 13 and 6 are not labelled or are labelled the other way around.

    3.

    A triangle L with vertices lying on the xy plane is transformed by T. Explain why both L and its image will have exactly the same area.

    [2]
    Verified
    Solution

    det⁡(7−102−3)=−1(OR∣det⁡(7−102−3)∣=1)\operatorname{det}\left(\begin{array}{cc}7 & -10 \\ 2 & -3\end{array}\right)=-1\left(\mathrm{OR}\left|\operatorname{det}\left(\begin{array}{ll}7 & -10 \\ 2 & -3\end{array}\right)\right|=1\right)det(72​−10−3​)=−1(OR​det(72​−10−3​)​=1) scale factor of image area is therefore (∣−1∣=)1(|-1|=) 1(∣−1∣=)1 (and the translation does not affect the area)

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