SOLUTION: Please help me with the following question: A Straight line has the equation {{{x-2y=12}}} and it meets the axes at A and B. Find the distance of the mid-point of AB from the orig

Question 223775: Please help me with the following question:
A Straight line has the equation and it meets the axes at A and B. Find the distance of the mid-point of AB from the origin, Giving your answer in the form
Found 2 solutions by Alan3354, edjones:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
A Straight line has the equation and it meets the axes at A and B. Find the distance of the mid-point of AB from the origin,
-------------------
x-2y = 12
x-intercept is at y = 0
x = 12
y-intercept is at x = =
y-intercept = -6
---------------
A(12,0)
B(0,-6)
Midpoint is (6,-3) ( =1/2 of x and of y)
------------------
Distance from Origin =
=
=

Answer by edjones(8007) (Show Source): You can put this solution on YOUR website!
x-2y=12
-2y=12-x
y=x/2 -6
the y intercept where x=0 is y=-6
x/2 -6=0
x/2=6
x=12
The x intercept where y=0 is x=12
.
a^2+b^2=c^2
6^2+12^2=c^2
36+144=180
sqrt(180)
=sqrt(36*5)
=6sqrt(5) length of AB
3sqrt(5) midpoint of AB
.
tan=opp/adj=12/6=2 angle opposite line from midpoint to origin.
tan^-1(2)=63.42... deg
.
Law of cosines:
a^2=b^2+c^2-2bc cos(a)
a^2=6^2+(3sqrt(5))^2-2*6*3sqrt(5)*cos(63.42...)
a^2=36+45-36sqrt(5)*cos A
=45
sqrt(45)
=sqrt(9*5)
=3sqrt(5)
.
Ed
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