Respuesta :
ANSWER:
The area of the triangle bounded by the y-axis is [tex]\frac{7938}{4225} \sqrt{65} \text { unit }^{2}[/tex]
SOLUTION:
Given, [tex]f(x)=9-\frac{-4}{7} x[/tex]
Consider f(x) = y. Hence we get
[tex]f(x)=9-\frac{-4}{7} x[/tex] --- eqn 1
[tex]y=9-\frac{4}{7} x[/tex]
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 = [tex]\frac{1}{ab}[/tex] 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) = [tex]\frac{-x \text { coefficient }}{y \text { coefficient }}=\frac{-4}{7}[/tex]
So, slope of perpendicular line = [tex]\frac{-1}{\text {slope of } f(x)}=\frac{7}{4}[/tex]
Perpendicular line is passing through origin(0,0).So by using point slope formula,
[tex]y-y_{1}=m\left(x-x_{1}\right)[/tex]
Where m is the slope and [tex]\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)[/tex]
[tex]y-0=\frac{7}{4}(x-0)[/tex]
[tex]y=\frac{7}{4} x[/tex] --- 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)
[tex]\begin{array}{l}{9-\frac{4}{7} x=\frac{7}{4} x} \\\\ {9 \times 4-\frac{4 \times 4}{7} x=7 x} \\\\ {36 \times 7-16 x=7 \times 7 x} \\\\ {252-16 x=49 x} \\\\ {49 x+16 x=252} \\\\ {65 x=252} \\\\ {x=\frac{252}{65}}\end{array}[/tex]
Put x value in (2)
[tex]\begin{array}{l}{y=\frac{7}{4} \times \frac{252}{65}} \\\\ {y=\frac{441}{65}}\end{array}[/tex]
So, the point of intersection is [tex]\left(\frac{252}{65}, \frac{441}{65}\right)[/tex]
Length of f(x) is distance between [tex]\left(\frac{252}{65}, \frac{441}{65}\right)[/tex] and (0,9)
[tex]\begin{aligned} \text { Length } &=\sqrt{\left(0-\frac{252}{65}\right)^{2}+\left(9-\frac{441}{65}\right)^{2}} \\ &=\sqrt{\left(\frac{252}{65}\right)^{2}+0} \\ &=\frac{252}{65} \end{aligned}[/tex]
Now, length of perpendicular of f(x) is distance between [tex]\left(\frac{252}{65}, \frac{441}{65}\right) \text { and }(0,0)[/tex]
[tex]\begin{aligned} \text { Length } &=\sqrt{\left(0-\frac{252}{65}\right)^{2}+\left(0-\frac{441}{65}\right)^{2}} \\ &=\sqrt{\left(\frac{252}{65}\right)^{2}+\left(\frac{441}{65}\right)^{2}} \\ &=\frac{\sqrt{(12 \times 21)^{2}+(21 \times 21)^{2}}}{65} \\ &=\frac{63}{65} \sqrt{65} \end{aligned}[/tex]
Now, area of right angle triangle = [tex]\frac{1}{2} \times \frac{252}{65} \times \frac{63}{65} \sqrt{65}[/tex]
[tex]=\frac{7938}{4225} \sqrt{65} \text { unit }^{2}[/tex]
Hence, the area of the triangle is [tex]\frac{7938}{4225} \sqrt{65} \text { unit }^{2}[/tex]
