Question

Consider the equation of the line, c(c2 + 2)y = c − x where c > 0 is a constant. (a) Find the coordinates of the x-intercept and the y-intercept. x-intercept ( , ) y-intercept ( , ) (b) Find a formula for the area of the triangle enclosed between the line, the x-axis, and the y-axis. A(c) =

Answer 1

The equation of the line indicates that the x and y-intercept and the formula for the area of the triangle are;

x-intercept (c, 0)

y-intercept (0, 1/(c² + 7))

A(c) = c/(2·(c² + 7))

The steps used to find the x-intercept and the y-intercept are presented as follows;

The equation of the line c·(c² + 7)·y = c - x, can be expressed in the slope intercept form to find the coordinates of the x-intercept and the coordinates of the y-intercept as follows;

c·(c² + 7)·y = c - x

y = (c - x)/(c·(c² + 7))

y = c/(c·(c² + 7)) - x/(c·(c² + 7))

y = 1/((c² + 7)) - x/(c·(c² + 7))

The above equation is in the slope-intercept form, y = m·x + c

Where c is the y-coordinate of the y-intercept, and (0, c) ids the coordinate of the y-intercept; Therefore, the coordinates of the y-intercept is; (0, 1/((c² + 7)))

The coordinate of the x-intercept can be found by plugging in y = 0, in the above equation to get;

0 = 1/((c² + 7)) - x/(c·(c² + 7))

x/(c·(c² + 7)) = 1/((c² + 7))

x = (c·(c² + 7))/((c² + 7))

(c·(c² + 7))/((c² + 7)) = c

x = c

Therefore coordinates of the x-intercept is; (c, 0)

The triangle enclosed by the line and the x-axis and y-axis is a right triangle, therefore;

The positive x and y-values of the x-intercept and y-intercept indicates that the area of the triangle is the product half the distance from the origin to the y-intercept and the distance from the origin to the x-intercept

Area = (1/2) × (1/((c² + 7)) - 0) × (c - 0)

(1/2) × (1/((c² + 7))) × (c) = c/(2·(c² + 7)

Area of the triangle, A(c) = c/(2·(c² + 7)

The complete question found through search can be presented as follows;

Consider the equation of the line c·(c² + 7)·y = c - x where c > 0 is a constant

(a) Find the coordinates of the x-intercept and the y-intercept

x-intercept ( , )

y-intercept ( , )

(b) Find a formula for the area of the triangle enclosed between the line, the x-axis and the y-axis

Answer 2

Part a)

The y-intercept is the point
`(0,(1)/(c^(2)+2))`

The x-intercept is the point
`(c,0)`

Part b) The formula of the area of triangle is equal to

`A(c)=((c)/(2c^(2)+4))`

Explanation:

step 1

Find the coordinates of the x-intercept and the y-intercept

we have that

The equation of the line is equal to

`c(c^(2)+2)y=c-x`

Find the y-intercept (value of y when the value of x is equal to zero)

For x=0

`c(c^(2)+2)y=c-0`

`c(c^(2)+2)y=c`

simplify

`(c^(2)+2)y=1`

`y=(1)/(c^(2)+2)`

The y-intercept is the point
`(0,(1)/(c^(2)+2))`

Find the x-intercept (value of x when the value of y is equal to zero)

For y=0

`c(c^(2)+2)(0)=c-x`

`0=c-x`

`x=c`

The x-intercept is the point
`(c,0)`

step 2

Find the formula for the area of triangle enclosed between the line, the x-axis, and the y-axis

The formula for the area of triangle is equal to

`A=(1)/(2)(b)(h)`

In this problem

`b=c` -----> the x-intercept

`h=(1)/(c^(2)+2)` ----> the y-intercept

substitute

`A(c)=(1)/(2)(c)((1)/(c^(2)+2))`

`A(c)=((c)/(2c^(2)+4))`