If two circles intersect a (1,3). B (m, - 1), and the centers of the two circles are on the straight line X-Y + C = 0, then the value of M C is?

If two circles intersect a (1,3). B (m, - 1), and the centers of the two circles are on the straight line X-Y + C = 0, then the value of M C is?

Geometric theorem: the common chord of two intersecting circles is perpendicular to the connecting line AB and the straight line X-Y + C = 0, so (- 1-3) / (m-1) = - 1, M = 5, and the midpoint (3,1) of two points AB is on the straight line X-Y + C = 0, C = - 2

If two circles intersect at two points (1, 3) and (m, 1), and the centers of the two circles are on the straight line x − y + C2 = 0, then M + C = ()
A. -1B. 2C. 3D. 0

It is known that two circles intersect at two points (1, 3) and (m, 1), and the centers of the two circles are on the straight line X-Y + C2 = 0, the slope of the common chord is: - 1, the common chord passing through point (1, 3) is Y-3 = - 1 (x-1), so x + y-4 = 0, and because (m, 1) is on the common chord, so m + 1-4 = 0, the solution is m = 3

If M ∈ R, L1: (2m-1) x + (M + 1) Y-3 = 0, L2: MX + 2y-2 = 0
A. When m = 2, L1 ‖ L2b. When m ≠ 2, L1 intersects L2. When C. M = 2, L1 ⊥ l2d. For any m ∈ R, L1 is not perpendicular to L2

When m = 2, l1:3x + 3y-3 = 0, l2:2x + 2y-2 = 0, then the two lines coincide. Therefore, option a and C are wrong. When m ≠ 2, the two lines do not coincide, but do not necessarily intersect. For example, when m = 1, the two lines are parallel. Therefore, option D is selected

Given two straight lines L1: MX + Y - (M + 1) = 0 and L2: x + my-2m = 0, the intersection point is in the first quadrant when finding the value of real number M

From MX + Y - (M + 1) = 0, y = (M + 1) - MX - (* *). Substituting it into the second equation, y = (M + 1) - MX - (* *)
x＋m[(m＋1)－mx]－2m=0
(1-m & # 178;) x = - M & # 178; + m [when m = 1, the two straight lines are parallel; when m = - 1, L1: X-Y = 0, L2: X-Y + 2 = 0, the intersection point is not in the first quadrant, then m ≠± 1], then:
X = m / (1 + m), substituting into (* *), y = (2m + 1) / (1 + m)
Then: the intersection is (M / (M + 1), (2m + 1) / (1 + m))
Therefore, M / (M + 1) > 0 and (2m + 1) / (1 + m) > 0
[M > 0 or M-1 / 2 or M0 or M

If the direction vector of L1 is the normal vector of L2, then the value of M is?

The direction vector of L1 is just the normal vector of L2, that is, two straight lines are perpendicular. The slope product = - 1
(-m/(m-2))*(-(2m+1)/m)=-1
m=1/3

Given two straight lines l1:2x + y-6 = 0 L2: x + 2Y + 6 = 0, the equation for finding the angular bisector of two straight lines is obtained
Give me a result

X-y-12 = 0 this is the result, the process is to use the trajectory equation to solve, let a point on the line be x1, Y1, use the distance formula from the point to the line, the principle is that the distance from the bisector of the angle to both sides of the angle is equal

The equation of L2 is ()
A. y=x-1B. y＝13x+53C. 3x+y-7=0D. y=3x+7

Let the inclination angle of line L1 be θ L1 to L2 be 45 °, the inclination angle of line L2 be θ + 45 ° ∵ L1: y = 12x + 2, the slope of line L2 be k = Tan (θ + 45 °) = 3, so the equation of line L2 is Y-1 = 3 (x + 2), that is, y = 3x + 7, so D is selected

Find the line equation which passes through the intersection of the line l1:2x + 3y-5 = 0, l2:3x-2y-3 = 0 and is parallel to the line 2x + Y-3 = 0

2x+3y-5=0 (1)
3x-2y-3=0 (2)
(1)×2+(2)×3
13x-19=0
x=19/13
(1)×3-(2)×2
13y-9=0
y=9/13
Let the linear equation: 2x + y-m = 0
X = 19 / 13, y = 9 / 13,
m=47/13
The linear equation is 2x + y-47 / 13 = 0

It is known that the line L1: x + M2y + 6 = 0, L2: (m-2) x + 3My + 2m = 0. When m is the value, L1 and L2 (1) intersect, (2) parallel, (3) coincide

The simultaneous equations of L1 and L2 are obtained as x + M2y + 6 = 0 (m − 2) x + 3My + 2m = 0, and m (M + 1) (3-m) y = 4 (M-3) can be simplified ① (1) when m ≠ - 1, m ≠ 3, m ≠ 0, equation 1 has a unique solution, line L1 intersects line L2. (2) when m = - 1, M = 0, equation 1 has no real solution, line L1 is parallel to line L2. (3) when m = 3, equation 1 has countless real solutions, line L1 coincides with line L2

Line L1: (2m ^ 2 + M-3) x + (m ^ 2-m) y = 2m, line L2: X-Y = 1, when the real number m takes what value, (1) L1 is perpendicular to L2 (2) L1 / / L2

The slope of line L2: X-Y = 1 K2 = 1
L1 ⊥ L2, then: the slope of line L1 K1 = - 1 / K2 = - 1
So: - (2m ^ 2 + M-3) / (m ^ 2-m) = - 1 (Note: m is not equal to 0 or 1)
The results are as follows
m^2+2m-3=0
(m+3)(m-1)=0
M = - 3 or 1
But M is not equal to 0 or 1, so only m = - 3