I think c .. I already took a course .. and the test had that question.. I cant remember the answer though
Slope of line = tan(120) = -tan(60) = - √3
Distance from origin = 8
Let equation be Ax+By+C=0
then -A/B=-√3, or
B=A/√3.
Equation becomes
Ax+(A/√3)y+C=0
Knowing that line is 8 units from origin, apply distance formula
8=abs((Ax+(A/√3)y+C)/sqrt(A^2+(A/√3)^2))
Substitute coordinates of origin (x,y)=(0,0) =>
8=abs(C/sqrt(A^2+A^2/3))
Let A=1 (or any other arbitrary finite value)
solve for positive solution of C
8=C/√(4/3) => C=8*2/√3 = (16/3)√3
Therefore one solution is
x+(1/√3)+(16/3)√3=0
or equivalently
√3 x + y + 16 = 0
Check:
slope = -1/√3 .....ok
distance from origin
= (√3 * 0 + 0 + 16)/(sqrt(√3)^2+1^2)
=16/2
=8 ok.
Similarly C=-16 will satisfy the given conditions.
Answer The required equations are
√3 x + y = ± 16
in standard form.
You can conveniently convert to point-slope form if you wish.
Answer:
c²=√181=13.45, or 13.5
Step-by-step explanation:
The pythagorean theorem is a²+b²=c²
Given this, we will plug in our numbers like this: 9²+10²=c²
9²+10²=181, but we have to find the square root of that.
c²=√181=13.45, or 13.5
I hope this helps!! :)
