A line passes through the point (3, 4) and cu | vedclass

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(A) Let the intercepts on the $x$-axis and $y$-axis be $a$ and $b$ respectively.Given that $a + b = 14$,so $a = 14 - b$.The equation of the line in in

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$4x + 3y = 24$

Vedclass · Answer

(A) Let the intercepts on the $x$-axis and $y$-axis be $a$ and $b$ respectively.Given that $a + b = 14$,so $a = 14 - b$.The equation of the line in intercept form is $\frac{x}{a} + \frac{y}{b} = 1$.Substituting $a = 14 - b$,we get $\frac{x}{14 - b} + \frac{y}{b} = 1$.Since the line passes through $(3, 4)$,we have $\frac{3}{14 - b} + \frac{4}{b} = 1$.Multiplying by $b(14 - b)$,we get $3b + 4(14 - b) = b(14 - b)$.$3b + 56 - 4b = 14b - b^2$.$b^2 - 15b + 56 = 0$.$(b - 7)(b - 8) = 0$.So,$b = 7$ or $b = 8$.If $b = 7$,then $a = 14 - 7 = 7$. The equation is $\frac{x}{7} + \frac{y}{7} = 1 \Rightarrow x + y = 7$.If $b = 8$,then $a = 14 - 8 = 6$. The equation is $\frac{x}{6} + \frac{y}{8} = 1 \Rightarrow 4x + 3y = 24$.Comparing with the given options,$4x + 3y = 24$ is the correct choice.

$A$ line passes through the point $(3, 4)$ and cuts off intercepts from the coordinate axes such that their sum is $14$. The equation of the line is

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