Problem: equations of the lines that contain the three altitudes of triangle (0,0), (12,6), and (18,0)
Need help on special segments
Apparently you mean the three corners of the triangle.
Let's start with the easy one. We have two points that lie
on the x-axis: (0, 0) and (18, 0). The slope of that line
is zero, and we can see that the y-intercept is zero.
Therefore, the equation is y = 0.
The next one to consider is the line with points at (0, 0)
and (12, 6). Obviously, the y-intercept is 0.The slope is the
change in y divided by the change in x. The slope has to be 6/12,
which is 1/2. Why? As the line moves 12 spaces to the right, it
moves 6 spaces up. So, the equation is y = 1/2 x.
Likewise, we can calculate the slope of the third line. As it moves
backwards from (18, 0) to (12, 6), the x value changes by -6, while
the y value changes by 6. The slope is 6/-6, which is -1.
All we need now is the y-intercept of this last line. From the equation
in slope-intercept form, y = mx + b, we can get b = y - mx. We get
x and y from the point that was given to us in the problem statement.
b = 6 - (-1)12
b = 6 + 12
b = 18
The equation for this line is y = -x + 18.
Our three equations are:
y = 0
y = 1/2 x
y = -x + 18