Given the slope of the straight line k = 2, P1 (3,5) P2 (x2,7) P3 (- 1, Y3) is the three points on the straight line to find x2 and Y3

Given the slope of the straight line k = 2, P1 (3,5) P2 (x2,7) P3 (- 1, Y3) is the three points on the straight line to find x2 and Y3

7-5=2(x2-3),x2=4
y3-5=2(-1-3),y3=-3

We know that the slope of the straight line k = 2. P1 (3,5), P2 (x2,7), P3 (- 1, Y3) are the three points on the straight line
Ask x2 and Y3. Remember the process, thank you

Linear equation: Y-5 = 2 (x-3) y = 2x-1 x = (y + 1) / 2
x2=(7+1)/2=4
y3=2*(-1)-2=-4

Given that a straight line L passes through two points P1 (2,1) P2 (m, 2) (m ∈ R), the slope k is obtained. If the inclination angle of the straight line L is 45 degrees, the value of M is obtained

K=（2-1）/(m-2)；
The inclination angle is 45 degrees, M = 3

The line L passes through the points P1 (1,2), P2 (m, 3). Find the slope and inclination angle of the line L

Slope k = (3-2) / (m-1) = 1 / (m-1)
m> 1, inclination angle = arctan 1 / (m-1)
m

Find the slope of line L1 passing through two points P1 (2,1) and P2 (m, 2) (M belongs to R), and find the inclination angle α and its value range

1. If M = 2, the slope does not exist and the inclination angle is 90 degrees;
2. If M ≠ 2, then the slope k = 1 / (m-2)
① If m2, the inclination angle is arctan [1 / (m-2)]

Find the slope of the straight line L passing through two points P1 (2,1) and P2 (m, 2) (M is not equal to 2), and find out the inclination angle a of L and its value range
I will start school tomorrow, and I haven't finished my winter vacation homework today. Please help me!

Just draw a picture. P2 moves on the line y = 2, except (2,2), so it's very clear

If the point P (2a + B, a-3b) is known to be on the angle dividing line of the second and fourth quadrants, what is 4b-6a-8 equal to

2a+b+a-3b=0
3a-2b=0
4b-6a-8=-2(3a-2b)-8=-8

Given the slope of the straight line k = 1 / 2, P1 (- 2,3), P2 (X2, - 2), P3 (1 / 2, Y3) three points on a straight line, find X2, Y3

Because k = 1 / 2 = 0.5
So the difference of Y / the difference of x = 0.5
And because 0.5 is a positive number, so when y increases, X also increases, and vice versa
P1P2P3 on the same line
P1p2: the difference of y = 3 - (- 2) = 5, so the difference of X is 10
Because 3 to - 2 is smaller, so - 2 - 10 = - 12 = x2
P1p3: the difference of x = 1 / 2 - (- 2) = 2.5, so the difference of Y is equal to 1.25, because x is bigger, so y is bigger 3 + 1.25 = 4.25 = Y3

If two points P1 (4,9) and P2 (6,3) are known, then the equation for a circle with diameter p1p2 is______ ．

Let the midpoint of line p1p2 be m, ∵ P1 (4,9) and P2 (6,3), the center of circle m (5,6), and | p1p2 | = (4 − 6) 2 + (9 − 3) 2 = 210, and the radius of circle 12 | p1p2 | = 10, then the equation of circle is: (X-5) 2 + (y-6) 2 = 10. So the answer is: (X-5) 2 + (y-6) 2 = 10

Given two points P1 (4,1), P2 (- 2, - 3), the equation for finding a circle with the diameter of line p1p2 is obtained

The center coordinates are: x = (4-2) / 2 = 1, y = (1-3) / 2 = - 1
The radius is r, R & sup2; = (- 1-1) & sup2; + (1-4) & sup2; = 13
The equation of circle is: (x-1) & sup2; + (y + 1) & sup2; = 13