CUET UG 2024 — Mathematics Calculus

Let's find the intercepts of the line:

x-intercept (when $y = 0$y=0):

$$\dfrac{x}{7 \sqrt{3} a}+\dfrac{0}{b}=4$$73​ax​+b0​=4

$$\dfrac{x}{7 \sqrt{3} a}=4$$73​ax​=4

$$x = 28\sqrt{3}a$$x=283​a

y-intercept (when $x = 0$x=0):

$$\dfrac{0}{7 \sqrt{3} a}+\dfrac{y}{b}=4$$73​a0​+by​=4

$$\dfrac{y}{b}=4$$by​=4

$$y = 4b$$y=4b

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The triangle has vertices at $(0,0)$(0,0), $(28\sqrt{3}a, 0)$(283​a,0), and $(0, 4b)$(0,4b).

Area of the triangle $= \dfrac{1}{2} \times \text{base} \times \text{height}$=21​×base×height

$$= \dfrac{1}{2} \times 28\sqrt{3}a \times 4b$$=21​×283​a×4b

$$= \dfrac{1}{2} \times 112\sqrt{3}ab$$=21​×1123​ab

$= 56\sqrt{3}ab$=563​ab square units