Find the linear equation of point P (2,0) and perpendicular to the line 1: x-3y-4 = 0

Find the linear equation of point P (2,0) and perpendicular to the line 1: x-3y-4 = 0

It is known that the slope of the straight line is - 3, because the straight line and the known straight line are perpendicular to each other. The product of the slopes is - 1, so the slope of the straight line is 1 / 3. It is also known that the equation of the straight line is y = 3x-6 by substituting (2,0) into 3x + C + y = o
Hope to adopt

Write an equation of degree 1 with 2 variables so that x = 1, y = 2, x = - 7, y = - 3

Let the binary linear equation be ax + by + C = 0, and substitute x = 1, y = 2, x = - 7, y = - 3 into the equation
a+2b+c=0
-7a-3b+c=0
Combining the two equations, 8A + 5B = 0
So the equation has infinitely many solutions, then any one is good, a = 5, B = - 8, C = 11,
So one of the binary linear equations is 5x-8y + 11 = 0, and so on

The bus goes from place a to place B at the speed of 80 km / h. one hour later, the car starts from place a to catch up with the bus at the speed of 120 km / h
Solving linear equation with one variable

It takes x hours to catch up with the bus
80+80x=120x
80x-120x=-80
-40x=-80
x=2

The degree of ∠ 1 + 2 + 3 + 4 = 24 ° where ∠ 1 = 2, ∠ 3 = 4, and the degree of ∠ 3 is 3 times of ∠ 1. Calculate the degree of four angles

∠ 1 + ∠ 2 + ∠ 3 + ∠ 4 = 24 ° where ∠ 1 = ∠ 2, ∠ 3 = ∠ 4
SO 2 ∠ 1 + 2 ∠ 3 = 24 degree
∠1+∠3=12°
Because the degree of ∠ 3 is three times that of ∠ 1
Therefore, 1 + 3 1 = 12 degree
∠1=3°,∠3=3∠1=9°
∠1=∠2=3°,∠3=∠4=9°

After driving from Shanghai to Nanjing, the original plan was to arrive at 11:30 noon. But after starting, the speed increased by 17 and arrived at 11:00. The next day, I returned from Nanjing at the same time. After driving 120 kilometers according to the original plan, I increased the speed by 16 and arrived in Shanghai at 11:10. How many kilometers is the distance between the two places?

Suppose the original planned speed of the car is XKM, 11:30-11:00 = 30 minutes = 12 hours, 12 ^ [1 ^ (1 + 7)], = 12 ^ [1 ^ [8], = 12 ^ 18, = 4 hours; 11:30-11:10 = 20 minutes = 13 hours & nbsp; 4x-120 = (4-120 ^ X-13) × (1 + 16) x, & nbsp; 4x-120 = (4-120x-13) × 76x, & nbsp; 4x-120 = 1118x-20, & nbsp; & nbsp; 6118x = 1006118x ^ 6118 = 100 ^ 6118, & nbsp; & nbsp; &A: the distance between the two places is 118261 km

If the value of 2A + | 4-5a | + | 1-3a | is a fixed value, the value range of a is obtained

When a ＜ 13, 4-5a ＞ 0, 1-3a ＞ 0, the original formula = 2A + 4-5a + 1-3a = - 6A + 5, not suitable for the problem; when 13 ≤ a ≤ 45, 4-5a ≥ 0, 1-3a ＜ 0, the original formula = 2A + 4-5a + 3a-1 = 3, suitable for the problem; when a ＞ 45, 4-5a ＜ 0, 1-3a ＜ 0, the original formula = 2A + 5a-4 + 3a-1 = 10A-5, not suitable for the problem

A train runs 60 kilometers in 3 / 5 hours. At this speed, it takes 3 hours for the train to go from place a to place B. how long is the railway between place a and place B?

Theoretically, it's 300km, but in fact, the train accelerates and decelerates, and there are temporary speed limits in some sections. It's really hard to calculate. Let's analyze the specific situation

The equation of known ellipse is x ^ 2 / 25 + y ^ 2 / 16 = 1. F1 and F2 are the left and right focus of ellipse respectively
It is known that the equation of the ellipse is x ^ 2 / 25 + y ^ 2 / 16 = 1, F1 and F2 are the left and right focal points of the ellipse respectively, the coordinates of point a are (2,1), and P is a point on the ellipse, then the maximum and minimum values of | PA | + | PF | are?
Sorry, I didn't write it clearly. Finally, I want to find the maximum and minimum of | PA | + | PF2 |

According to the first definition of ellipse: | Pf1 | + | PF2 | = 2A = 10 | PF2 | = 10 - | Pf1|
That is, | PA | + | PF2 | = | PA | + 10 - | Pf1 | = 10 + | PA | - | Pf1|
∫ (pa-pf1) max = | PA | - | Pf1 | = | AF1 | = √ (2 + 3) ^ 2 + 1 = √ 26, that is, the maximum value is 10 + √ 26
(pa-pf1) min = - (| PA | - | Pf1 |) = - √ 26, i.e. the minimum value is 10 - √ 26

There is a pile of sand on the construction site. 40% of the sand is used in the first day, and 10.8 tons are used in the second day. 62.5% of the sand is used in two times. How many tons of sand is used in the first day?

8 (62.5% - 40%) = 10.8 △ 22.5% = 48 tons a: the original sand is 48 tons

Hurry! Hurry! Hurry! We know that OA vector = a, OB vector = B, and point G is the center of gravity of △ ABC
It is known that OA (vector) = a (vector), ob (vector) = B (vector), point G is the center of gravity of △ ABC, and the line PQ passing through point G intersects OA and ob at P and Q respectively
If OP vector = m * (a vector), OQ = n * (b vector), prove: 1 / M + 1 / N = 3

Obviously, og = (a + b) / 3, pgq is collinear. Op-og = K (oq-og) [K is a real number]
ma-（a+b）/3＝k[nb-（a+b）/3].
M - (1 / 3) + (K / 3) = 0, (1 / 3) + K (n-1 / 3) = 0
The results show that (3m-1) (3n-1) = 1, i.e. 1 / M + 1 / N = 3