Question

Find the area of a triangle bounded by the y axis, the line f (x) = 3-1/7x, and the line perpendicular to f (x) that passes through the origin

Answer 1

Let us the line with the equation:

`f(x)=y=\text{ 3-}(1)/(7)x`

According to this equation, the slope of this line is:

`m\text{ =-}(1)/(7)`

now, the slope of the line perpendicular to the line f(x) would be:

`m_p=7\text{ }`

Since the perpendicular line to f(x), passes through the origin, we have that the y-intercept of this line is 0, so the equation of the perpendicular line passing through origin is:

`y\text{ = 7x}`

now, to find the intersecting point between the lines, we must equalize both equations of the lines:

`7x\text{ = 3-}(1)/(7)x`

then, solve for x:

`x\text{ = }(21)/(50)`

the y-coordinate corresponding to this x, is:

`y\text{ = 7(}(21)/(50)\text{)=}(147)/(50)`

thus, we obtain the point:

`(x,y)=((21)/(50),(147)/(50))`

now, the y-intercept of the given line f(x) is when x=0, then:

`f(0)=y=\text{ 3-}(1)/(7)(0)=3`

thus, we get the point:

`(x,y)=(0,3)`

Therefore, we already have the 3 points that define the triangle

so the triangle is bounded by the points