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Question Number 95260 by i jagooll last updated on 24/May/20
if the line 3x+2y−1=0 transformed by matrix A= (((1 a)),((b 2)) ) such that the image is the line 2x+8y+c=0 find the value of a×b×c
$$\mathrm{if}\:\mathrm{the}\:\mathrm{line}\:\mathrm{3x}+\mathrm{2y}−\mathrm{1}=\mathrm{0}\:\mathrm{transformed} \\ $$$$\mathrm{by}\:\mathrm{matrix}\:\mathrm{A}=\begin{pmatrix}{\mathrm{1}\:\:\:\mathrm{a}}\\{\mathrm{b}\:\:\:\mathrm{2}}\end{pmatrix}\:\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{the}\:\mathrm{image}\:\mathrm{is}\:\mathrm{the}\:\mathrm{line}\:\mathrm{2x}+\mathrm{8y}+\mathrm{c}=\mathrm{0} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{a}×\mathrm{b}×\mathrm{c}\: \\ $$
Commented by mr W last updated on 24/May/20
(x,y)→(u,v) with A= (((1 a)),((b 2)) ) u=x+ay v=bx+2y 2(x+ay)+8(bx+2y)+c=0 2(1+4b)x+2(a+8)y+c=0 ≡3x+2y−1=0 2(1+4b)=3 ⇒b=(1/8) 2(a+8)=2 ⇒a=−7 c=−1 ⇒a×b×c=(7/8)
$$\left({x},{y}\right)\rightarrow\left({u},{v}\right)\:{with}\:\:\mathrm{A}=\begin{pmatrix}{\mathrm{1}\:\:\:\mathrm{a}}\\{\mathrm{b}\:\:\:\mathrm{2}}\end{pmatrix}\: \\ $$$${u}={x}+{ay} \\ $$$${v}={bx}+\mathrm{2}{y} \\ $$$$\mathrm{2}\left({x}+{ay}\right)+\mathrm{8}\left({bx}+\mathrm{2}{y}\right)+{c}=\mathrm{0} \\ $$$$\mathrm{2}\left(\mathrm{1}+\mathrm{4}{b}\right){x}+\mathrm{2}\left({a}+\mathrm{8}\right){y}+{c}=\mathrm{0} \\ $$$$\equiv\mathrm{3}{x}+\mathrm{2}{y}−\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{2}\left(\mathrm{1}+\mathrm{4}{b}\right)=\mathrm{3}\:\Rightarrow{b}=\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$\mathrm{2}\left({a}+\mathrm{8}\right)=\mathrm{2}\:\Rightarrow{a}=−\mathrm{7} \\ $$$${c}=−\mathrm{1} \\ $$$$\Rightarrow{a}×{b}×{c}=\frac{\mathrm{7}}{\mathrm{8}} \\ $$
Commented by mr W last updated on 24/May/20
i am not sure if i understand the question correctly. please check!
$${i}\:{am}\:{not}\:{sure}\:{if}\:{i}\:{understand}\:{the} \\ $$$${question}\:{correctly}.\:{please}\:{check}! \\ $$
Commented by i jagooll last updated on 24/May/20
thank you sir. you and sir john got the same result
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}.\: \\ $$$$\mathrm{you}\:\mathrm{and}\:\mathrm{sir}\:\mathrm{john}\:\mathrm{got}\:\mathrm{the}\:\mathrm{same}\:\mathrm{result} \\ $$
Answered by john santu last updated on 24/May/20
(((x′)),(y^′ ) ) = (((1 a)),((b 2)) ) ((x),(y) ) = (((x+ay)),((bx+2y)) ) image of the line 2x′+8y′+c = 0 ⇒ 2(x+ay)+8(bx+2y)+c = 0 it similar to line 3x+2y−1 = 0 ⇒(2+8b)x+(2a+16)y+c = 0 we get 2+8b = 3 ⇒b = (1/8) 2a+16 = 2 ⇒ a = −7 and c = −1 so a×b×c = (7/8) .
$$\begin{pmatrix}{\mathrm{x}'}\\{\mathrm{y}^{'} }\end{pmatrix}\:=\:\begin{pmatrix}{\mathrm{1}\:\:\:\mathrm{a}}\\{\mathrm{b}\:\:\:\mathrm{2}}\end{pmatrix}\:\begin{pmatrix}{\mathrm{x}}\\{\mathrm{y}}\end{pmatrix}\:=\:\begin{pmatrix}{\mathrm{x}+\mathrm{ay}}\\{\mathrm{bx}+\mathrm{2y}}\end{pmatrix} \\ $$$$\mathrm{image}\:\mathrm{of}\:\mathrm{the}\:\mathrm{line}\:\mathrm{2x}'+\mathrm{8y}'+\mathrm{c}\:=\:\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{2}\left(\mathrm{x}+\mathrm{ay}\right)+\mathrm{8}\left(\mathrm{bx}+\mathrm{2y}\right)+\mathrm{c}\:=\:\mathrm{0} \\ $$$$\mathrm{it}\:\mathrm{similar}\:\mathrm{to}\:\mathrm{line}\:\mathrm{3x}+\mathrm{2y}−\mathrm{1}\:=\:\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{2}+\mathrm{8b}\right)\mathrm{x}+\left(\mathrm{2a}+\mathrm{16}\right)\mathrm{y}+\mathrm{c}\:=\:\mathrm{0} \\ $$$$\mathrm{we}\:\mathrm{get}\:\mathrm{2}+\mathrm{8b}\:=\:\mathrm{3}\:\Rightarrow\mathrm{b}\:=\:\frac{\mathrm{1}}{\mathrm{8}} \\ $$$$\mathrm{2a}+\mathrm{16}\:=\:\mathrm{2}\:\Rightarrow\:\mathrm{a}\:=\:−\mathrm{7}\: \\ $$$$\mathrm{and}\:\mathrm{c}\:=\:−\mathrm{1}\:\mathrm{so}\:\mathrm{a}×\mathrm{b}×\mathrm{c}\:=\:\frac{\mathrm{7}}{\mathrm{8}}\:.\: \\ $$

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