You can start by computing the coordinates of the vertices of the triangle ABC
4x -5y = -16 and y = 0 gives A(-4,0)
2x +3y = 14 and y = 0 gives B(7,0)
2x +3y = 14 and 4x -5y = -16 gives C(1,4)
A triangle is formed by the intersection of the lines 2x +3y = 14, 4x -5y = -16 and the x-axis. Find the area of the triangle.
I started by attempting to convert both given equations into slope intercept form, so I could begin to graph the lines
2x -2x +3y = 14 -2x
3y = -2x +14
y = -2x +4.666666667
4x -4x -5y = -16 -4x
-5y = -4x -16
y = -4x + 3.2
The y intercept, 4.666666667 is impossible to graph by hand and this is stopping me from going any further.
What am i doing wrong?
Thanks for your help so far.
I actually managed to go back and work it out.
I now have a triangle plotted with the area; 7cm x 2.5cm /2 = 8.75cm2
The next part of the question asks me to find the vertices of a triangle with half the area of the triangle just found.
Well the area will be 4.375cm2.
How do I get the vertices from that triangle?
Ok, it took me a little while to see how you did that calculation but the formula is exactly the same as:
Ax(By - Cy) + Bx(Cy - Ay) + Cx(Ay - By) /2
By inputting the vertices co-ordinates I get the same answer.
The next part of the problem asks me to find the co-ordinates of the vertices of a triangle with half the area of the one just found.
Obviously the area of this new triangle will be 11, multiplied by 2 giving 22. How do i begin to deconstruct the formula to find the vertices?