For a triangle whose sides are formed by the x−axis, the y−axis , we can consider the distance between origin (0,0) & the triangle's vertex of the y-axis to be height & the distance between origin (0,0) & the triangle's vertex of the x-axis vertex to be base of the triangle. These two are y-intercept & x intercept of the given line.
y intercept : when x=0, \(y = −\frac{3}{4}*0 + 9\) ==>\(y=9\)
x intercept : when y=0, \(0 = −\frac{3}{4} + 9\) ==> \(-9 = −\frac{3}{4}x\) ==>\(3 = \frac{1}{4}x\) ==>\(12 = x\)
Hence the area =\(\frac{1}{2} *12*9\) =\(54\)

Hence I would go for option D.
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we just need to calclate the X & Y intercepts. This can be calculated by putting X=0 & Y=0, one by one in the equation of the third side of the triangle.
1. when y=0 then -3/4X + 9=0 i.e. 3/4x=9. i.e x=12
2. when x=0 then -3/4*0 + 9=Y i.e. Y=9.
Now are of the triangle is 1/2bh => 1/2 * 12 * 9 = 54.

Answer is D.

This is the equation of line. Let's change it to intercept form so that x and y intercepts can be calculated easily.
\(y = −\frac{3}{4}x + 9\)

\(y + \frac{3}{4}x = 9\)

\(\frac{y}{9} + \frac{3}{4.9}x = 1\).

\(\frac{y}{9} + \frac{x}{12} = 1\).

Now Line intersects the x axis at (12,0) and Y axis at (0,9).

If you could imagine a line passing thru these points (12,0) ,(0,9) ,x axis and y-axis.
Height & base of Triangle is 9 & 12. You can also consider vice-versa.

Area of Triangle = \(\frac{1}{2} * 12* 9 = \frac{108}{2}\)= 54.

Ans - D

Base of triangle on x-axis= x intercept= 12

Height of triangle on y-axis= y intercept= 9

Area of triangle= (9*12)/2= 54

Ans D

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