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Question Number 110948 by bemath last updated on 01/Sep/20

(√(bemath)) If each point on the line 3x+4y=2 is transformed by matrix M= (((2 0)),((0 1)) ) , the image is a line ___

$$\:\:\:\:\:\sqrt{\mathrm{bemath}} \\ $$$$\mathrm{If}\:\mathrm{each}\:\mathrm{point}\:\mathrm{on}\:\mathrm{the}\:\mathrm{line}\:\mathrm{3x}+\mathrm{4y}=\mathrm{2} \\ $$$$\mathrm{is}\:\mathrm{transformed}\:\mathrm{by}\:\mathrm{matrix}\:\mathrm{M}=\begin{pmatrix}{\mathrm{2}\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\mathrm{1}}\end{pmatrix}\:,\:\mathrm{the} \\ $$$$\mathrm{image}\:\mathrm{is}\:\mathrm{a}\:\mathrm{line}\:\_\_\_ \\ $$

Commented by bemath last updated on 01/Sep/20

hshahaha....bemath sir. bemath macho

$$\mathrm{hshahaha}....\mathrm{bemath}\:\mathrm{sir}.\:\mathrm{bemath}\:\mathrm{macho} \\ $$

Commented by Rasheed.Sindhi last updated on 01/Sep/20

Thank you my dear!

$$\mathcal{T}{hank}\:{you}\:{my}\:{dear}! \\ $$

Commented by bobhans last updated on 01/Sep/20

standart way (((x′)),((y′)) ) = (((2 0)),((0 1)) ) ((x),(y) ) ⇒ ((x),(y) ) = (1/2) (((1 0)),((0 2)) ) (((x′)),((y′)) ) ((x),(y) ) = (1/2) (((x′)),((2y′)) ) = ((((1/2)x′)),(( y′)) ) hence the image of line is (3/2)x′+4y′=2 multiply by 2 ⇒ 3x+8y−4=0

$$\mathrm{standart}\:\mathrm{way}\: \\ $$$$\begin{pmatrix}{\mathrm{x}'}\\{\mathrm{y}'}\end{pmatrix}\:=\:\begin{pmatrix}{\mathrm{2}\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\mathrm{1}}\end{pmatrix}\:\begin{pmatrix}{\mathrm{x}}\\{\mathrm{y}}\end{pmatrix}\:\Rightarrow\:\begin{pmatrix}{\mathrm{x}}\\{\mathrm{y}}\end{pmatrix}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\begin{pmatrix}{\mathrm{1}\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\mathrm{2}}\end{pmatrix}\:\begin{pmatrix}{\mathrm{x}'}\\{\mathrm{y}'}\end{pmatrix} \\ $$$$\begin{pmatrix}{\mathrm{x}}\\{\mathrm{y}}\end{pmatrix}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\begin{pmatrix}{\mathrm{x}'}\\{\mathrm{2y}'}\end{pmatrix}\:=\:\begin{pmatrix}{\frac{\mathrm{1}}{\mathrm{2}}\mathrm{x}'}\\{\:\:\:\mathrm{y}'}\end{pmatrix} \\ $$$$\mathrm{hence}\:\mathrm{the}\:\mathrm{image}\:\mathrm{of}\:\mathrm{line}\:\mathrm{is}\:\frac{\mathrm{3}}{\mathrm{2}}\mathrm{x}'+\mathrm{4y}'=\mathrm{2} \\ $$$$\mathrm{multiply}\:\mathrm{by}\:\mathrm{2}\:\Rightarrow\:\mathrm{3x}+\mathrm{8y}−\mathrm{4}=\mathrm{0} \\ $$

Commented by Rasheed.Sindhi last updated on 01/Sep/20

(√(bemath))=(√b)(√e)(√m)(√a)(√t)(√h) :)

$$\left.\:\sqrt{\mathrm{bemath}}=\sqrt{\mathrm{b}}\sqrt{\mathrm{e}}\sqrt{\mathrm{m}}\sqrt{\mathrm{a}}\sqrt{\mathrm{t}}\sqrt{\mathrm{h}}\:\:\::\right) \\ $$

Commented by bemath last updated on 01/Sep/20

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Commented by Rasheed.Sindhi last updated on 01/Sep/20

BTW, be-math or bem-ath or something else?

$$\mathrm{BTW},\:{be}-{math}\:{or}\:{bem}-{ath} \\ $$$${or}\:{something}\:{else}? \\ $$

Answered by bobhans last updated on 01/Sep/20

say the image is a line ℓ≡ x determinant (((3 4)),((0 1)))+y determinant (((2 0)),((3 4)))−2. determinant (((2 0)),((0 1)))=0 ℓ≡ 3x +8y−4 = 0

$$\mathrm{say}\:\mathrm{the}\:\mathrm{image}\:\mathrm{is}\:\mathrm{a}\:\mathrm{line}\:\ell\equiv\:\mathrm{x}\begin{vmatrix}{\mathrm{3}\:\:\:\mathrm{4}}\\{\mathrm{0}\:\:\:\mathrm{1}}\end{vmatrix}+\mathrm{y}\begin{vmatrix}{\mathrm{2}\:\:\:\mathrm{0}}\\{\mathrm{3}\:\:\:\mathrm{4}}\end{vmatrix}−\mathrm{2}.\begin{vmatrix}{\mathrm{2}\:\:\:\:\mathrm{0}}\\{\mathrm{0}\:\:\:\:\mathrm{1}}\end{vmatrix}=\mathrm{0} \\ $$$$\ell\equiv\:\mathrm{3x}\:+\mathrm{8y}−\mathrm{4}\:=\:\mathrm{0}\: \\ $$

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