A line cuts the X and Y axes at the points A | vedclass

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(B) Let $A \equiv (a, 0)$ and $B \equiv (0, b)$.Let $P \equiv (5, 6)$ be the point that divides the line segment $AB$ internally in the ratio $3: 1$.U

Vedclass · Accepted Answer

$2x + 5y = 40$

Vedclass · Answer

(B) Let $A \equiv (a, 0)$ and $B \equiv (0, b)$.Let $P \equiv (5, 6)$ be the point that divides the line segment $AB$ internally in the ratio $3: 1$.Using the section formula,the coordinates of $P$ are given by:$5 = \frac{3 	imes 0 + 1 	imes a}{3 + 1}$ $\Rightarrow 5 = \frac{a}{4}$ $\Rightarrow a = 20$.$6 = \frac{3 	imes b + 1 	imes 0}{3 + 1}$ $\Rightarrow 6 = \frac{3b}{4}$ $\Rightarrow 3b = 24$ $\Rightarrow b = 8$.Thus,the $X$-intercept is $20$ and the $Y$-intercept is $8$.The equation of the line in intercept form is $\frac{x}{a} + \frac{y}{b} = 1$.Substituting the values,we get $\frac{x}{20} + \frac{y}{8} = 1$.Multiplying by $40$,we get $2x + 5y = 40$.

$A$ line cuts the $X$ and $Y$ axes at the points $A$ and $B$ respectively. If the point $(5, 6)$ divides the line segment $AB$ internally in the ratio $3: 1$,then the equation of the line is:

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