2.
Let (x, y) be the solution to the system. We know it is true that
y = (3x + 15) / 4
We also know it is true that:
x2 + y2 = 36
Since y = 3x+154 we can substitute 3x+154 in for y
x2 + (3x+154)2 = 36
x2 + (3x+154)(3x+154) = 36
x2 + 9x2+90x+22516 = 36
Multiply through by 16 to eliminate the denominator
16x2 + 9x2 + 90x + 225 = 576
Combine like terms
25x2 + 90x + 225 = 576
Subtract 576 from both sides of the equation
25x2 + 90x - 351 = 0
We can use the quadratic formula to solve for x
x = −9±12√35
We can use the equation y = (3x + 15) / 4 to find the y-coordinates of the intersection points.
When x = −9 + 12√35 , y = 14⋅( 3( −9 + 12√35 )+15 ) = 12 + 9√35
When x = −9 − 12√35 , y = 14⋅( 3( −9 − 12√35 )+15 ) = 12 − 9√35
So the intersection points are:
(−9 + 12√35, 12 + 9√35) and (−9 − 12√35, 12 − 9√35)
Check: https://www.desmos.com/calculator/jfndjchrhd
We can find the distance between those two points using the distance formula.
distance = √(−9 + 12√35 − −9 − 12√35)2 + (12 + 9√35 − 12 − 9√35)2
distance = √(24√35)2 + (18√35)2
distance = √172825 + 97225
distance = √270025
distance = 6√3
.1.
P is a point on the line y = x and Q is a point on the line y = 7x.
So we can say...
Let P be the point (a, a) where a > 0
Let Q be the point (b, 7b) where b > 0
And O is the point (0, 0)
Now we can make this equation:
the distance between O and P = the distance between O and Q
√[ a2 + a2 ] = √[ b2 + (7b)2 ]
Square both sides of the equation
a2 + a2 = b2 + (7b)2
a2 + a2 = b2 + 49b2
Combine like terms
2a2 = 50b2
Divide both sides by 2
a2 = 25b2
Take the positive square root of both sides
a = 5b
the slope of PQ = the slope between the points (a, a) and (b, 7b)
the slope of PQ = a−7ba−b
Substitute 5b in for a
the slope of PQ = 5b−7b5b−b
Combine like terms
the slope of PQ = −2b4b
Reduce the fraction by 2b
the slope of PQ = −12
Here is a graph for this problem: https://www.desmos.com/calculator/c9sjxvalqw
Using trig we can make this triangle:
z = -3 + 3√3 i
Substitute 6 cos( 2π/3 ) in for -3 and 6 sin( 2π/3 ) in for 3√3
z = 6 cos( 2π/3 ) + 6 sin( 2π/3 ) i
Move the i to the front of the term
z = 6 cos( 2π/3 ) + 6 i sin( 2π/3 )
Factor 6 out of both terms
z = 6 [ cos( 2π/3 ) + i sin( 2π/3 ) ]
cis( x ) = cos( x ) + i sin( x ) so cos( 2π/3 ) + i sin( 2π/3 ) = cis( 2π/3 )
z = 6 cis( 2π/3 )