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:
Please explain more further. It is not described very well.
Step-by-step explanation:
Answer:
add all the numbers
Step-by-step explanation:
Answer:
see explanation
Step-by-step explanation:
To find the y- intercept let x = 0 in the function
g(0) = 0 - 0 - 84 = - 84 ← y- intercept
To find the x- intercepts let y = 0, that is
x² - 5x - 84 = 0
To factor the quadratic
Consider the factors of the constant term (- 84) which sum to give the coefficient of the x- term
The factors are - 12 and + 7, since
- 12 × 7 = - 84 and - 12 + 7 = - 5, thus
(x - 12)(x + 7) = 0
Equate each factor to zero and solve for x
x - 12 = 0 ⇒ x = 12
x + 7 = 0 ⇒ x = - 7
x- intercepts are x = - 7 and x = 12
