SOLUTION: Consider the line x/3 + y/ 8 = k, where k is a positive number a) Write the coordinates for A and B, the points where the line meets the x and y axes respectively b) The area of

Algebra ->  Linear-equations -> SOLUTION: Consider the line x/3 + y/ 8 = k, where k is a positive number a) Write the coordinates for A and B, the points where the line meets the x and y axes respectively b) The area of       Log On


   



Question 1129132: Consider the line x/3 + y/ 8 = k, where k is a positive number
a) Write the coordinates for A and B, the points where the line meets the x and y axes respectively
b) The area of triangle AOB where O is the origin, is 144 units squared. Find the value for k

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
Consider the line x%2F3+%2B+y%2F+8+=+k, where k is a positive number

a) Write the coordinates for A and B, the points where the line meets the x and y axes respectively

the points where the line meets the x is where y=0
x%2F3+%2B+0%2F+8+=+k
x++=+3k=>+A (3k, 0)

the points where the line meets the y is where x=0
0%2F3+%2B+y%2F+8+=+k
y%2F+8+=+k
y+=+8k=>B(0, 8k)


b) The area of triangle AOB where+O is the origin, is 144 units squared. Find the value for+k+

The area of triangle AOB:

area=%28base%2Ah%29%2F2 where

base is distance from origin to x the point A
OA=3k
height is distance from origin to y the point B
OB=8k

so, we have:
144=%283k%2A8k%29%2F2
144=12k%5E2
144%2F12=k%5E2
12=k%5E2
k=sqrt%2812%29
k=sqrt%284%2A3%29
k=2sqrt%283%29

then,
OA=3k=>OA=3%2A2sqrt%283%29=>OA=6sqrt%283%29
OB=8k=>OB=8%2A2sqrt%283%29>=>OB=16sqrt%283%29


check the area:
area=%28base%2Ah%29%2F2

area=%286sqrt%283%29%2A16sqrt%283%29%29%2F2
area=6sqrt%283%29%2A8sqrt%283%29
area=6%2A8%28sqrt%283%29%29%5E2
area=6%2A8%2A3
area=144