In Cartesian coordinate system, if the ordinates and abscissa of a point are all integers, then the point is called the integral point. Let K be an integer. When the intersection of the line y = x + 2 and y = KX + 4 is the integral point, how many values of K can be taken at most?

In Cartesian coordinate system, if the ordinates and abscissa of a point are all integers, then the point is called the integral point. Let K be an integer. When the intersection of the line y = x + 2 and y = KX + 4 is the integral point, how many values of K can be taken at most?

For straight lines: y = x + 2 (1) and y = KX + 4 (2)
Substituting (1) into (2) leads to
x + 2 = kx + 4
(1-k)x = 2
The solution is x = 2 / (1-k)
To guarantee that x is an integer, we must make 1-k = ± 2 or 1-k = ± 1
From the solution of 1-k = ± 2, k = - 1 or K = 3 can be obtained
From the solution of 1-k = ± 1, k = 0 or K = 2 can be obtained
That is, when the intersection of the line y = x + 2 and the line y = KX + 4 is the entire point, the value of K can be taken as - 1,0,2,3 at most

As shown in the figure, in the rectangular coordinate system, the quadrilateral oabc is a rectangular trapezoid, the coordinates of point a are (10,0), the coordinates of point B are (6,3), and the moving points P and Q come out from two points c and a at the same time
Let P move from a to o at a speed of 2 units per second. When point Q stops moving, point P also stops moving. Let t (0 ≤ t ≤ 5) be the motion time. (1) when t is what, the quadrilateral pqab is a parallelogram? (2) when t is what, the quadrilateral pqab is an isosceles trapezoid?

(1) Q (10-2t, 0) P (T 3) ∵ BP "AQ ∵ pqab when a parallelogram, BP = AQ, BP = 6T, AQ = 10-2t, BP = AQ introduces the solution of 6T = 10-2t, t = 2 (2) ∵ BP" AQ ∵ pqab isosceles trapezoid, QP = AB, ab = √ (4 ^ 2, 3 ^ 2) = 5pq = √ [(10-3

As shown in the figure, in the plane rectangular coordinate system, the quadrilateral oabc is a right angled trapezoid, CB ‖ OA, ∠ OCB = 90 °, CB = 1, ab = 5, and the straight line y = − 12x + 1 passes through point a and intersects with the Y axis at point D (1) Find the coordinates of points a and B; (2) try to explain: ad ⊥ Bo; (3) if point m is a moving point on the straight line ad, is there another point n on the X axis, so that the quadrilateral with O, B, m and N as the vertex is a parallelogram? If it exists, request the coordinates of point n; if not, please explain the reason

(1) When y = 0, - 12x + 1 = 0, the solution is x = 2, the coordinates of a point are (2,0), and the coordinates of B point are (1,2); (2) when x = 0, y = -

Given a point Q (2,0) and a circle x ^ 2 + y ^ 2 = 1 on the rectangular coordinate plane, the ratio of the tangent length Mn (n is the tangent point) from the moving point m to the circle O to MQ is constant λ (λ〉 0)

Let m (x, y), then MO & # 178; = x & # 178; + Y & # 178;, the length of the tangent Mn from m to ⊙ o | Mn | = √ (X & # 178; + Y & # 178; - 1)
According to the meaning of the question √ (X & # 178; + Y & # 178; - 1) = λ √ [(X-2) &# 178; + Y & # 178;]
The two sides are squared
(1-λ²)x²+4λ²x+(1-λ²)y²=1+4λ² ①
It is known that λ > 0
If λ = 1, then ① becomes x = 5 / 4 and is a vertical line
If λ ≠ 1, then because the coefficients of X &# 178; and Y &# 178; in ① are equal, there is no cross term in ① and the right side > 0, we know that the trajectory of M is a circle