# Find n if the area of the triangle determined by the graph of the function f=nx+n-4 and the x,y axis where is A=n/2.

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The function f=nx+n-4 is a straight line graph that:

1. intercept the x-axis at x=(4-n)/n {obtained by equating the expression to 0 and making x the subject};

2. intercept the y-axis at y=n-4 {obtained by f(0), ie substituting x=0 into the expression}

Hence the triangle is a right-angled triangle with the 3 points being the origin, the x-intercept and the y-intercept.

Classically,

Area of triangle

= 1/2 . Base . Height

= 1/2 . (n-4) . (4-n)/n

As we are only concerned with the absolute distance between the origin and the intercepts,

The area formula can be re-written as

A

= 1/2 . (n-4) . Abs( (n-4)/n )

= 1/2n . (n-4)^2

Given also that A=n/2 .

Equating these 2 expressions:

1/2n . (n-4)^2 = n/2

Multiplying both sides by 2n,

(n-4)^2 = n^2

Expand and manipulate:

n^2 -8n +16 = n^2

8n=16

n = 16/8 = 2

The area of the triangle whose sides are the graph of f(x) and x, y axis is the area of a right angle triangle: half-product of cathetus.

The cathetus of the triangle are x and y axis, where the values of x and y are the intercepts of the graph with x and y axis.

A = x*y/2

We'll calculate x intercept of f(x). We'll put y = 0.

f(x) = 0

nx+n-4 = 0

We'll isolate x to the left side:

nx = 4 - n

x = (4-n)/n

We'll calculate y intercept of f(x). We'll put x = 0.

f(0) = n - 4

y = n - 4

Now, we'll calculate the area:

A = (4-n)*(n-4)/2n

We know, from enunciation, that A=n/2.

n/2 = -(n-4)^2/2n

We'll cross multiply:

2n^2 = -2(n-4)^2

We'll divide by 2:

n^2 = (n-4)^2

We'll subtract (n-4)^2 both sides:

n^2 - (n-4)^2 = 0

We'll expand the square:

n^2 - n^2 + 8n - 16 = 0

We'll eliminate like terms:

8n - 16 = 0

We'll add 16

8n = 16

**n = 2**

f(x) = nx+n-4 is the equation of a straight line.

The intercept of the line on x axis (X) is obtained by putting f(x) = 0 in the equation f(x) = nx+(n-4) and solving for x:

0 = nx+n-4. So x = X = (4-n)/n.

Put x= 0 in the equation and solve for f(x) in f(x) = nx+n-4 to get the y intercept Y .

f(0) = n*0 +n-4. Therefore f(0) = Y = n-4

Therefore the area of the triangle XOY formed by f(x) with x and y axis is given by:

Area XOY = (1/2)X*Y = (1/2){(4-n)/n}{n-4} which is given to be n/2.

Therefore (1/2) {(n-4)/n} (n-4) = n/2.

(n-4)^2 /n = n.

(n-4)^2 = n^2 .

n^2-8n+16 = n^2.

-8n+16 = 0.

-8n = - 16.

v = -16/-8 = 2.

So **n = 2.**