5.Find the equation of the straight line joining the points (a,b) and {(a+b), (a-b)}

Dear Student,

Please find below the solution to the asked query :

$$\begin{gathered} \mathrm{Let} \mathrm{the} \mathrm{given} \mathrm{points} \mathrm{are} : \\ \mathrm{A}\left(\mathrm{a},\mathrm{b}\right) \mathrm{and} \mathrm{B}\left(\mathrm{a}+\mathrm{b}, \mathrm{a}-\mathrm{b}\right) \\ \mathrm{Now}, \\ \frac{\mathrm{y}-\mathrm{b}}{\mathrm{x}-\mathrm{a}} = \frac{\mathrm{a}-\mathrm{b}-\mathrm{b}}{\mathrm{a}+\mathrm{b}-\mathrm{a}} \\ \Rightarrow \frac{\mathrm{y}-\mathrm{b}}{\mathrm{x}-\mathrm{a}} = \frac{\mathrm{a}-2\mathrm{b}}{\mathrm{b}} \\ \Rightarrow \mathrm{by}-{\mathrm{b}}^2 = \left(\mathrm{x}-\mathrm{a}\right)\left(\mathrm{a}-2\mathrm{b}\right) \\ \Rightarrow \mathrm{by}-{\mathrm{b}}^2 = \mathrm{ax}-2\mathrm{bx}-{\mathrm{a}}^2+2\mathrm{ab} \\ \Rightarrow \mathrm{by}-\mathrm{ax}+{\mathrm{a}}^2-{\mathrm{b}}^2-2\mathrm{bx}-2\mathrm{ab} = 0 \\ \Rightarrow \mathrm{by}-\left(\mathrm{a}+2\mathrm{b}\right)\mathrm{x} + \left({\mathrm{a}}^2-{\mathrm{b}}^2-2\mathrm{ab}\right) = 0 \end{gathered}$$

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