The equation of a line passing through two points (2, - 3) and parallel to two points (1,2), (- 4,5)

The equation of a line passing through two points (2, - 3) and parallel to two points (1,2), (- 4,5)

Let y = ax + B bring in (1,2) (- 4,5)
The equation y = - 3 / 5 x + 13 / 5 is obtained
Because the two lines are parallel, the slope is the same
Let y = - 3 / 5, x + C bring in (2, - 3)
The equation y = - 3 / 5 X-9 / 5 is obtained

The equation (1) of a straight line satisfying the following conditions passes through point a (3,2) and is parallel to the line 4x + Y-2 = 0
(2) The line (3) passing through point B (2, - 3) and parallel to points m (1,2) and n (- 1, - 5) passes through point C (3,0) and is perpendicular to the line 2x + Y-5 = 0

(1) It is parallel to the line 4x + Y-2 = 0, k = - 4, substituting a to get Y-2 = - 4 (x-3), that is 4x + y-14 = 0 (2), Mn slope = (2 + 5) / (1 + 1) = 7 / 2, k = 7 / 2, substituting B to get y + 3 = 7 / 2 (X-2), that is 7x-2y-20 = 0 (3), perpendicular to the line 2x + Y-5 = 0, K ′ = - 2, k = 1 / 2, substituting C to get y = 1 / 2 (x-3), that is x-2y-3 = 0

The equation of a straight line passing through point a (2,3) is obtained under the following conditions: (1) parallel to the straight line 2x + Y-5 = 0, (2) perpendicular to the straight line x-y-2 = 0
This is a separate question. Please explain it in detail

1) Parallel to the straight line 2x + Y-5 = 0,
Let the equation be: 2x + 5Y + C = 0
The straight line passes through point a (2,3), that is, point a (2,3) is on the straight line,
So: (the coordinates of point a fit the equation)
2*2+5*3+c=0
c=-19
The equation of the line is 2x + y-19 = 0
(2) Perpendicular to the line x-y-2 = 0
Let the equation be: x + y + D = 0
The straight line passes through point a (2,3), that is, point a (2,3) is on the straight line,
So: (the coordinates of point a fit the equation)
1*2+1*3+d=0
c=-5
The equation of the line is x + Y-5 = 0
The line parallel to AX + by + C1 = 0 is usually set as ax + by + C = 0;
The line perpendicular to AX + by + C1 = 0 is usually set as BX ay + C = 0 or - BX + ay + C = 0

A light rod with length of L = 0.5m has a small ball with mass of M = 2kg at one end. The light rod rotates uniformly in the vertical plane with the other end o as the center, taking g = 10m / ^ 2
(1) If the ball passes through the lowest point of the circular track at the speed of 3m / s, calculate the force of the light bar on the ball
(2) If the force of the light bar on the ball is 16N when the ball passes through the highest point, the velocity of the ball is calculated
(3) In order to make the ball pass through the highest point and the lowest point, the force direction of the light bar on the ball is the same, what conditions should the speed of the light bar meet?
As the title!

(1) Let f be the tension of light bar
F-mg=mV^2/L
F=56N
(2) In two cases, the pulling force is 16N and the supporting force is 16N
Tension time
F+mg=mV^2/L
V=3m/s
Support time
mg-F=mV^2/L
V=1m/s
(3) When the tension is zero at the highest point
mg=mV^2/L
V=√5m/s
V=ωL
ω = 2 √ 5 radians / S
The direction of the force is the same, it can only be upward
be
Rotational speed ω '

How many kilogram, kilogram, gram is one ton

1 ton = 1000 kg = 1000 kg = 1000000 G

An object is weighed in the air with a spring scale, and the reading is 1.6 n. if the object is completely immersed in water, the reading of the spring scale is 0.6 n
Calculation: 1. Buoyancy of the object; 2. Volume of the object; 3. Density of the object

F floating = f air-f water = 1.6-0.6 = 1 (n)
F float = P water GV row, V row = f float / P water g = 1 / 1000 x 10 = 0.0001 (M3)
P = f / V row, g = 1.6 / 0.0001x10 = 1.6X10 & # 179; (kg / M & # 179;)

The height of the bottom edge of an isosceles triangle is 5cm. If the height of this triangle is increased by 1cm, the area will be increased by 3cm square

The height of the bottom edge of an isosceles triangle is 5cm. If the height of this triangle is increased by 1cm, the area will be increased by 3cm square
Original triangle bottom length = 6cm
Original triangle area = 15cm & # 178;

A certain mass object a, which was pulled by an extension spring in the original uniform motion elevator, is still on the floor, as shown in the figure. Now it is found that a is suddenly pulled to the right by the spring. It can be judged that the motion of the elevator at this time may be ()
A. Accelerate rise B. decelerate rise C. accelerate fall D. decelerate fall

When the elevator is stationary, the static friction of the floor to the object is balanced with the elastic force of the spring, and the static friction may be less than or equal to the maximum static friction. When the elevator has downward acceleration, it will inevitably reduce the positive pressure of the object on the floor, which also reduces the maximum static friction force. At this time, the maximum static friction is less than the electric force

What is the order of mixed decimal operations

1. First multiply, then divide, and finally add and subtract; 2. Operate from left to right at the same level in order; 3. If there are brackets, first small, then middle, and finally large, and then calculate in order

A horizontal disk with radius R is used to fix an object with mass M. when the disk moves uniformly around the O-axis with angular velocity ω, and the linear velocity of the object is V, the centripetal force on the object is ()
A. mv2RB. mω2RC. mg-mv2RD. mωv

Several formulas of centripetal force: FN = mv2r = m ω 2R = m ω v. no matter which formula is chosen, the first thing to know is which quantity. In this question, we know that angular velocity ω and linear velocity are V, so the correct choice should be d