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Question Number 29276 by gyugfeet last updated on 06/Feb/18

if the (x/a)+(y/b)=1 passes through the point lf intersection of the lines x+y=3 and 2x−3y=1 and is parallel to the line y=x−6 then find the value of a and b.

$${if}\:{the}\:\frac{{x}}{{a}}+\frac{{y}}{{b}}=\mathrm{1}\:{passes}\:{through}\:{the}\:{point}\:{lf}\:{intersection}\:{of}\:{the}\:{lines}\:{x}+{y}=\mathrm{3}\:{and}\:\mathrm{2}{x}−\mathrm{3}{y}=\mathrm{1}\:{and}\:{is}\:{parallel}\:{to}\:{the}\:{line}\:{y}={x}−\mathrm{6}\:{then}\:{find}\:\:{the}\:{value}\:{of}\:{a}\:\:{and}\:{b}.\: \\ $$

Answered by Rasheed.Sindhi last updated on 06/Feb/18

Intersection point of x+y=3 and 2x−3y=1 y=3−x ⇒2x−3(3−x)=1 x=2 , y=1 ∩tersection point (2,1) The line (x/a)+(y/b)=1 passes through (2,1) ∴ (2/a)+(1/b)=1......................(A) This line is parallel to y=x−6 , so they have same slope (x/a)+(y/b)=1⇒bx+ay=ab y=−(b/a)x+b −(b/a)=1 a=−b (A)⇒ (2/(−b))+(1/b)=1⇒b=−1 ,a=−(−1)=1 a=1,b=−1

$$\mathrm{Intersection}\:\mathrm{point}\:\mathrm{of}\:{x}+{y}=\mathrm{3}\:{and}\:\mathrm{2}{x}−\mathrm{3}{y}=\mathrm{1} \\ $$$${y}=\mathrm{3}−{x}\:\Rightarrow\mathrm{2}{x}−\mathrm{3}\left(\mathrm{3}−{x}\right)=\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}=\mathrm{2}\:,\:{y}=\mathrm{1} \\ $$$$\:\:\:\cap\mathrm{tersection}\:\mathrm{point}\:\left(\mathrm{2},\mathrm{1}\right) \\ $$$$\mathrm{The}\:\mathrm{line}\:\frac{{x}}{{a}}+\frac{{y}}{{b}}=\mathrm{1}\:\mathrm{passes}\:\mathrm{through}\:\left(\mathrm{2},\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\therefore\:\:\frac{\mathrm{2}}{{a}}+\frac{\mathrm{1}}{{b}}=\mathrm{1}......................\left({A}\right) \\ $$$$\mathrm{This}\:\mathrm{line}\:\mathrm{is}\:\mathrm{parallel}\:\mathrm{to}\:{y}={x}−\mathrm{6}\:,\:\mathrm{so}\:\mathrm{they} \\ $$$$\mathrm{have}\:\mathrm{same}\:\mathrm{slope} \\ $$$$\:\:\:\frac{{x}}{{a}}+\frac{{y}}{{b}}=\mathrm{1}\Rightarrow{bx}+{ay}={ab} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{y}=−\frac{{b}}{{a}}{x}+{b} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\frac{{b}}{{a}}=\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{a}=−{b} \\ $$$$\left({A}\right)\Rightarrow\:\frac{\mathrm{2}}{−{b}}+\frac{\mathrm{1}}{{b}}=\mathrm{1}\Rightarrow{b}=−\mathrm{1}\:,{a}=−\left(−\mathrm{1}\right)=\mathrm{1} \\ $$$${a}=\mathrm{1},{b}=−\mathrm{1} \\ $$

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