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Question

(A) Let the equation of the line passing through $(4, 3)$ be $y - 3 = m(x - 4)$,where $m < 0$ for the line to cut the first quadrant.$y = mx - 4m + 3$

Vedclass · Accepted Answer

$3x + 4y = 24$

Vedclass · Answer

(A) Let the equation of the line passing through $(4, 3)$ be $y - 3 = m(x - 4)$,where $m $y = mx - 4m + 3$.The $x$-intercept is found by setting $y = 0$: $0 = mx - 4m + 3 \implies x = 4 - \frac{3}{m}$.The $y$-intercept is found by setting $x = 0$: $y = 3 - 4m$.The area $A$ of the triangle formed in the first quadrant is $A = \frac{1}{2} 	imes (x	ext{-intercept}) 	imes (y	ext{-intercept}) = \frac{1}{2} (4 - \frac{3}{m})(3 - 4m) = \frac{1}{2} (12 - 16m - \frac{9}{m} + 12) = 12 - 8m - \frac{9}{2m}$.To minimize the area,differentiate $A$ with respect to $m$: $\frac{dA}{dm} = -8 + \frac{9}{2m^2} = 0$.$8 = \frac{9}{2m^2} \implies m^2 = \frac{9}{16} \implies m = -\frac{3}{4}$ (since $m Substituting $m = -\frac{3}{4}$ into the line equation: $y - 3 = -\frac{3}{4}(x - 4)$.$4y - 12 = -3x + 12 \implies 3x + 4y = 24$.

Find the equation of a line passing through the point $(4, 3)$,which cuts a triangle of minimum area from the first quadrant.

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