Problem: If the area of triangle with vertices (X, 3) (4,4) and (3,5) is 4 square units find X?

Solution:
To solve the problem, we can use the formula for the area of a triangle which is:

A = 1/2 * base * height

We can find the height of the triangle by finding the perpendicular distance from the vertex (4,4) to the line passing through the other two vertices.

Finding the equation of the line:
Let's first find the equation of the line passing through the vertices (X, 3) and (3,5). We can use the slope-intercept form of the equation of a line which is:

y = mx + b

where m is the slope and b is the y-intercept.

The slope of the line passing through the two vertices is:

m = (5 - 3) / (3 - X) = 2 / (X - 3)

We can substitute one of the vertices in the equation to find the y-intercept. Let's use the vertex (X, 3):

3 = (2 / (X - 3)) * X + b

3 = 2 - 2b / (X - 3)

6 - 3X = 2 - 2b

4 = 2b - 3X

b = (3X + 4) / 2

So the equation of the line passing through the vertices (X, 3) and (3,5) is:

y = (2 / (X - 3)) * x + (3X + 4) / 2

Finding the height:
Now let's find the perpendicular distance from the vertex (4,4) to the line passing through the other two vertices. We can use the point-to-line distance formula which is:

distance = |ax + by + c| / sqrt(a^2 + b^2)

where a, b, and c are the coefficients of the equation of the line, and x and y are the coordinates of the point.

The coefficients of the equation of the line are:

a = 2 / (X - 3)
b = -1
c = (3X + 4) / 2

The coordinates of the point are:

x = 4
y = 4

Substituting these values in the formula, we get:

distance = |(2 / (X - 3)) * 4 - 1 * 4 + (3X + 4) / 2| / sqrt((2 / (X - 3))^2 + (-1)^2)

Simplifying the expression, we get:

distance = |8 / (X - 3) - 4 + (3X + 4) / 2| / sqrt(4 / (X - 3)^2 + 1)

distance = |16 - 8X + (3X + 4)(X - 3)| / sqrt(4 + (X - 3)^2)

distance = |16 - 5X - 9| / sqrt(4 + (X - 3)^2)

distance = |7 - 5X| / sqrt((X - 1)^2 + 4)