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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-




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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