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(A) The equation of a line in the intercept form is $\frac{x}{a} + \frac{y}{b} = 1$ ..... $(i)$Here,$a$ and $b$ are the intercepts on $x$ and $y$ axes

Vedclass · Accepted Answer

$x+2y-6=0$ and $2x+y-6=0$

Vedclass · Answer

(A) The equation of a line in the intercept form is $\frac{x}{a} + \frac{y}{b} = 1$ ..... $(i)$Here,$a$ and $b$ are the intercepts on $x$ and $y$ axes respectively.It is given that $a + b = 9 \Rightarrow b = 9 - a$ ..... $(ii)$From equation $(i)$ and $(ii)$,we obtain $\frac{x}{a} + \frac{y}{9 - a} = 1$ ..... $(iii)$It is given that the line passes through point $(2,2)$. Therefore,equation $(iii)$ becomes $\frac{2}{a} + \frac{2}{9 - a} = 1$$\Rightarrow 2 \left( \frac{1}{a} + \frac{1}{9 - a} \right) = 1$$\Rightarrow 2 \left( \frac{9 - a + a}{a(9 - a)} \right) = 1$$\Rightarrow \frac{18}{9a - a^2} = 1$$\Rightarrow 18 = 9a - a^2$$\Rightarrow a^2 - 9a + 18 = 0$$\Rightarrow a^2 - 6a - 3a + 18 = 0$$\Rightarrow a(a - 6) - 3(a - 6) = 0$$\Rightarrow (a - 6)(a - 3) = 0$$\Rightarrow a = 6$ or $a = 3$If $a = 6$,then $b = 9 - 6 = 3$. The equation is $\frac{x}{6} + \frac{y}{3} = 1 \Rightarrow x + 2y - 6 = 0$.If $a = 3$,then $b = 9 - 3 = 6$. The equation is $\frac{x}{3} + \frac{y}{6} = 1 \Rightarrow 2x + y - 6 = 0$.

Find the equation of the line passing through the point $(2,2)$ and cutting off intercepts on the axes whose sum is $9$.

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