get out the graph paper and sketch a picture ... then do the same on a larger poster board for your presentation.
Okay, so I got this assignment at school for Alg 2 trig. And in this, the teacher gave us each a question and we need to answer it with a whole explanation. We will be presenting this to the class. Here is the problem:
A line with a slope of -5 passes through the point (3,6). Find the area of the triangle in the first quadrant formed by this line and the coordinate axis.
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Please include step by step explanation of your work and the actual work in your answer.
Thank you very much!
y = mx + C
m is the gradient, which is -5.
y = -5x+C
You know it passes through the point (3,6)
6 = -5(3) + C
C = 6+15 = 21
therefore y = -5x+21. This is the equation of the line.
A straight line and the coordinate axis in the first quadrant forms a triangle. You need the area of this triangle... So you need to know the length of the base, and the height, since area = 0.5(base)(height).
The base goes from x= 0, to the point where the line cuts the x axis, which happens when y = 0.
y =-5x+21 = 0
Solve for x. And then x is your base.
The height is from y = 0, to where it cuts the y axis, which is when x =0
y = -5(0)+21 = 21
So the height is 21.
You should now be able to find the area.
You know the gradient which is and you know a co-ordinate hence you can find the equation of the line.
Now, when you draw the graph you will find that it will pass though first, second and forth quadrant. The line though second and forth quadrant will go on 'forever'.
Calculate the point just when it is about to go though to the second/forth quadrant from the first quadrant (Co-ordinates: ). Then consider the triangle formed in the first quadrant, through which you will calculate the area.