Linear slope range If two points a (2, - 3) B (- 3, - 2) are known, a straight line passes through P (1,1) and intersects with line AB, the slope range of the straight line can be calculated 2 let the range of inclination angle a of line l be 60 degrees

Linear slope range If two points a (2, - 3) B (- 3, - 2) are known, a straight line passes through P (1,1) and intersects with line AB, the slope range of the straight line can be calculated 2 let the range of inclination angle a of line l be 60 degrees

(2) when the slope does not exist, the line is perpendicular to the X axis. To sum up, K belongs to {K | K greater than 4 / 3 or K less than - 4} 2, because the slope is Tana (a is the inclination angle), so tan60 * = root

It is known that the linear L1: Χ + μ YD + 6 = 0 and L2: (μ - 2) Χ + 3 μ YD + 2 = 0
Q: when μ =? Two straight lines are vertical? (please be careful)

The slope is multiplied by - 1
Slope K1 = - 1 / M K2 = (2-m) / 3M
(m-2)/3m^2=-1
m-2=-3m^2
3m^2+m-2=0
(3m-2)(m+1)=0
m=-1 m=2/3

The equation for finding a circle whose center is on the straight line 3x + 2Y = 0 and whose intersection points with X axis are (- 2,0), (6,0) respectively

Let the coordinates of the center of a circle be (x, y)
Then find the distance from (- 2,0) and (6,0) to the center of the circle
Using the distance between the two points and the center of the circle, we can think of the equation
（-2-x）²+（0-y）²=（6-x）²+（0-y)²
X = 2 of the solution
Then substitute x = 2 into 3x + 2Y = 0
The solution is y = - 3
The center coordinates of the circle are (2, - 3)
Substituting the coordinates of the center of the circle and a point on the circle into (x1-x2) square + (y1-y2) under the root sign
The radius of the solution is 5
Finally, it is substituted into the standard equation of the circle
（x-a）²+（y-b）²=r²
We obtain (X-2) & sup2; + (y + 3) & sup2; = 25

Circle square x ^ 2 + y ^ 2 = 25, through M (- 4,3), make straight line Ma, MB and circle intersection a, B, and Ma, MB are symmetric with respect to straight line y = 3, find the slope of ab

Using graphic method, first of all, point m is on the circle, do x = - 4, the line intersects the circle at point Z, and the line y = 3 intersects the circle at point H
It can be seen from the figure that the line x = - 4 intersects the line y = 3 at point m, then the line Ma and the line MB are symmetrical about y = 3. It can be seen that the point a (or b) can be found on the arc zmh, and the point B (or a) can be found on the arc zh
It can be seen that the slope of AB is a range, so it is necessary to find the limit state. From the figure, it can be seen that the limit state is when point a or B coincides with point m and ab coincides with point Z and h. The slope of AB is - 1 and infinite respectively. Is it positive or negative infinity? Take point a (0,5) (- 5,0) for verification, and the negative value less than - 1 is obtained
Therefore, the slope range of AB is K

Solving the equation of known circle x2 + y2 = 4 with respect to symmetric circle y = x + 3
"That is | X-Y + 3 | / √ 2 = 3 √ 2 / 2
The solution is x = - 3, y = 3“
Isn't X & Y two variables?
thanks

To make a circle symmetrical to a circle about a straight line
That is to find the symmetry of the center of the circle about the straight line
That is to find the symmetric point of (0,0) with respect to y = x + 3
Let the symmetric point be (x, y)
y/x=-1
The distance from the point (x, y) to the line y = x + 3 is the distance from the origin to the line 3 √ 2 / 2
That is | X-Y + 3 | / √ 2 = 3 √ 2 / 2
The solution is x = - 3, y = 3
So the symmetric circle equation is (x + 3) & sup2; + (Y-3) & sup2; = 4

Given that the coordinates of two points AB are a (0, - 4) B (0,4), and the product of the slopes of the line Ma and MB is - 1, the trajectory equation of point m is obtained

Let m (x, y), then: [(y + 4) / (x-0)] × [(y-4) / (x-0)] = - 1
So, y ^ 2-16 = - x ^ 2
That is: x ^ 2 + y ^ 2 = 16, where x ≠ 0

In the triangle ABC, if the angle B = 60 degrees and the square of side B = AC, then the triangle ABC must be? Why? Prove it

COSb=(a²+c²-b²)/2ac=(a²+c²-ac)/2ac=1/2
A = C isosceles triangle is obtained
B=60°
A triangle is an equilateral triangle

In the triangle ABC, we know that the three sides of a, B and C are a, B and C, and the square of B = AC
Find the function y = 1 + sin2b / SINB + CoSb

The formula of senior two: sin2b = 2sinbconb, CoSb = (a ^ 2 + C ^ 2-B ^ 2) / 2Ac
So y = 1 + 3cosb = (a ^ 2 + C ^ 2 + AC) / 2Ac

In △ ABC, point D is the midpoint of AC, 3aE = AB, BD and CE intersect at point P, and vector AB and vector AC are used to represent vector AP

Let BP = λ BD = λ (AC / 2-AB) EP = μ EC = μ (ac-ab / 3) BP = ep-2ab / 3 = μ (ac-ab / 3) - 2Ab / 3 = μ AC - (μ / 3 + 2 / 3) ab ∵ BP = BP ∫ μ AC - (μ / 3 + 2 / 3) AB = λ AC / 2 - λ AB, μ = λ / 2, μ / 3 + 2 / 3 = λ, λ = 4 / 5, μ = 2

In the triangle ABC, D and E are on AB and AC respectively, and BD = CE, M is the midpoint of be and N is the midpoint of CD. Through M and N, make a straight line intersection AB to P and AC to Q. find: AP = a

Take the midpoint f on BC and connect MF and NF
MF and NF were the median lines of triangular BEC and CDB
MF=1/2EC
NF=1/2BD
MF=NF
Angle FMN = angle FNM
Because AB parallel FM AC parallel MF
Angle AQP = angle FMN
Angle Apq = angle FNM
So angle AQP = angle Apq
AP=AQ