<u>ANSWER: </u>
The area of the triangle bounded by the y-axis is 
<u>SOLUTION:</u>
Given, 
Consider f(x) = y. Hence we get
--- eqn 1

On rewriting the terms we get
4x + 7y – 63 = 0
As the triangle is bounded by two perpendicular lines, it is an right angle triangle with y-axis as hypotenuse.
Area of right angle triangle =
where a, b are lengths of sides other than hypotenuse.
So, we need find length of f(x) and its perpendicular line.
First let us find perpendicular line equation.
Slope of f(x) = 
So, slope of perpendicular line = 
Perpendicular line is passing through origin(0,0).So by using point slope formula,

Where m is the slope and 

--- eqn 2
4y = 7x
7x – 4y = 0
now, let us find the vertices of triangle, one of them is origin,
second one is point of intersection of y-axis and f(x)
for points on y-axis x will be zero, to get y value, put x =0 int f(x)
0 + 7y – 63 = 0
7y = 63
y = 9
Hence, the point of intersection is (0, 9)
Third vertex is point of intersection of f(x) and its perpendicular line.
So, solve (1) and (2)

Put x value in (2)

So, the point of intersection is 
Length of f(x) is distance between
and (0,9)

Now, length of perpendicular of f(x) is distance between 

Now, area of right angle triangle = 

Hence, the area of the triangle is 