The combined equation of the straight lines p | vedclass

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Question

(A) Let the intercepts of the line on the coordinate axes be $a$ and $b$. Given that $a+b = -1$,so $b = -(1+a)$. The intercept form of the line is $\f

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$(3x-2y-6)(x-2y+2)=0$

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(A) Let the intercepts of the line on the coordinate axes be $a$ and $b$. Given that $a+b = -1$,so $b = -(1+a)$. The intercept form of the line is $\frac{x}{a} + \frac{y}{b} = 1$. Substituting $b$,we get $\frac{x}{a} - \frac{y}{1+a} = 1$. Since the line passes through $(4,3)$,we have $\frac{4}{a} - \frac{3}{1+a} = 1$. Multiplying by $a(1+a)$,we get $4(1+a) - 3a = a(1+a)$. $4 + 4a - 3a = a + a^2$ $\Rightarrow 4 + a = a + a^2$ $\Rightarrow a^2 = 4$. Thus,$a = 2$ or $a = -2$. If $a = 2$,then $b = -(1+2) = -3$. The equation is $\frac{x}{2} + \frac{y}{-3} = 1 \Rightarrow 3x - 2y - 6 = 0$. If $a = -2$,then $b = -(1-2) = 1$. The equation is $\frac{x}{-2} + \frac{y}{1} = 1$ $\Rightarrow -x + 2y = 2$ $\Rightarrow x - 2y + 2 = 0$. The combined equation is $(3x - 2y - 6)(x - 2y + 2) = 0$.

The combined equation of the straight lines passing through the point $(4,3)$ and each line making intercepts on the coordinate axes whose sum is $-1$,is

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