A line passes through P(-4, 1) and meets the | vedclass

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(A) Let the coordinates of $A$ be $(a, 0)$ and $B$ be $(0, b)$.Since $P(-4, 1)$ divides $AB$ in the ratio $1:2$,using the section formula:$-4 = \frac{

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$x-2y+6=0$

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(A) Let the coordinates of $A$ be $(a, 0)$ and $B$ be $(0, b)$.Since $P(-4, 1)$ divides $AB$ in the ratio $1:2$,using the section formula:$-4 = \frac{1(0) + 2(a)}{1+2} \implies -4 = \frac{2a}{3} \implies a = -6$.$1 = \frac{1(b) + 2(0)}{1+2} \implies 1 = \frac{b}{3} \implies b = 3$.Thus,the intercepts are $a = -6$ and $b = 3$.The equation of the line in intercept form is $\frac{x}{a} + \frac{y}{b} = 1$.$\frac{x}{-6} + \frac{y}{3} = 1$.Multiplying by $-6$,we get $x - 2y = -6$,or $x - 2y + 6 = 0$.

$A$ line passes through $P(-4, 1)$ and meets the coordinate axes at points $A$ and $B$. If $P$ divides the segment $AB$ internally in the ratio $1:2$,then the equation of the line is

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