Find the point to straight distance formula tangent equation formula for a circle parabolic formula

y=ax²+bx+c（a≠0）

When y=0, i.e., ax +bx+c=0(a≠0) is the parabolic equation. Knowing the three conditions, you can determine the three coefficients of a, b, and c.

Three conditions: 1. It can be three known points. 2. Two points and axis of symmetry x=-b (2a).

3. A vertex of a point and a parabola [-b (2a), (4ac-b) (4a)]. 4. The other three conditions.

Determination of vertices: 1. Matching method. y=ax²+bx+c=a（x-b/2a）²+4ac-b²）/(4a）。

2. Calculated with vertex formula. x=-b/（2a），y=（4ac-b²）/(4a）。

Opening direction: Determined only by the positive or negative of a. a>0, opening up: a<0, opening down.

m=|ax+by+c root number [a2+b2]|This is the point-to-straight distance formula.

x,y] denotes the dot |. denotes the absolute value a2 denotes the square of a.

Point p(x0,y0), the equation for the straight line ax+by+c=0 The formula for the distance from the point to the straight line.

d=|ax0+by0+c|[ (a 2+b 2)](a 2+b 2) indicates a squared plus b squared under the root number.

The tangent distance from a point to a circle is formulated as (x -a) (x-a) + (y -b) (y-b) = r and (a, b) is a point on the circle.

The theorem of the tangent is that a straight line that passes through the outer end of the radius and is perpendicular to this radius is a tangent of a circle, and the tangent of a circle is perpendicular to the radius of this circle.

The tangent of a circle is perpendicular to the radius of its tangent point; A straight line that passes through the non-centered end of the radius and is perpendicular to that radius is a tangent of the circle. The two tangents that lead the circle from a point outside the circle are equal in length, and the line connecting the center of the circle and this point bisects the angle between the two tangents.

1. In the plane Cartesian coordinate system xoy, there are two different points a(x1,y1) and b(x2,y2), then the distance between the two points ab is: |ab|=[x2-x1) 2+(y2-y1) 2]. Straight line ax+by+c=0 coordinates (xo,yo) then the distance from this point to this straight line is:

The equation for the straight line in the formula is ax+by+c=0, and the coordinates of the point p are (x0, y0).

2. The horizontal and vertical coordinates of the points on the bisector of the two quadrant angles are opposite to each other.

3. One point is translated up and down, and the abscissa is unchanged, that is, the abscissa of the point on the straight line parallel to the y-axis is the same. The points on the y-axis, the abscissa are all points on the axis, and the ordinate is 0. Points on the axes do not belong to any quadrants.

A point with respect to the x-axis symmetry does not change the abscissa and the ordinate becomes the opposite of the original coordinate. The reverse is also true. The horizontal and vertical coordinates of a point symmetrical with respect to the origin are all the opposite of the original coordinates.

When the x-axis is axially symmetrically transformed, x does not change, and y becomes the opposite number. When the axis is symmetrically transformed with the y axis, y does not change, and x becomes the opposite number.

1. The distance from the point to the straight line.

Formula: d=|(x1-x0,y1-y0,z1-z0)×(l,m,n)|/l2+m2+n2)。

2. The distance from the point to the straight line, that is, the perpendicular line of the target straight line through this point.

From this point to the distance of the foot of the land.

3. Functional forensics: The minimum value of the distance from the point p to any point on the line is the distance from the point p to the friend chain of the line. In order to take advantage of the condition above the above equation to deform it, the matching coefficient is processed:

It is said that only the equal sign should be taken at that time, so the minimum value is the distance from the point to the straight line.

The formula for the distance from the center of the circle to the straight line: for p(x0,y0), it is to the distance of the straight line ax+by+c=0, and the formula is used to envy d=|ax0+by0+c|a + b) indicates that the distance from the center of the circle to the chord is called the chord centroid distance.

The distance formula from the point to the straight line is often used by students when doing problems, so what are the distance formulas from the point to the straight line, and how is the cover width derived? The following noise is the "point to line distance formula, point to line distance formula derivation method" I have compiled for you, for reference only, welcome to read this article.

The formula for the distance from a point to a straight lineDistance formula: d= (axo+byo+c) a +b ) Formula description: The linear equation in the formula is ax+by+c=0, and the coordinates of the point p are (x0, y0).

The distance from the point to the straight line, that is, the perpendicular line from this point to the vertical foot of the target straight line.

The perpendicular line of the straight line is made by crossing the point, and the perpendicular segment obtained is the distance from the point to the straight line.

For example, if the equation for a straight line is: ax+by+c=0, and the point coordinates are: (x,y), then there is the distance formula |ax+by+c|A 2+b 2) point-to-straight distance refers to the length of the perpendicular segment.

Find the equation of the straight line that is illuminated by the object cover and perpendicular to the known straight line ax+by+c=0 (a and b are not zero), and then the system of simultaneous equations is used to find the coordinates of the perpendicular n point, and then the distance from the point to the straight line is calculated by using the distance formula between the two points.

Method for deriving the formula for the distance from a point to a line

The formula for the distance from a point to a straight line is:

Let the equation for the line l be ax+by+c=0 and the coordinates of the point p be (x0,y0), then the distance from the point p to the line l is:

In the same way, when p(x0,y0), the analytic formula of the straight line l is y=kx+b, then the distance from the point p to the straight line l is: only guess Zen.

Consider the point (x0, y0, z0) and the spatial line x-x1 l=y-y1 m=z-z1 n, with d=|(x1-x0,y1-y0,z1-z0)×(l,m,n)|l + m trillion sell + n ).

Proof Method:

Definition forensics: According to the definition, the distance from the point p (x refers to dust, y) to the line l: ax+by+c=0 is the length of the perpendicular segment from the point p to the line l, and the perpendicular line from the point p to the line is l', the vertical foot is q, then l'The slope of is b a then l'The analytic formula is y-y = (b a) (x-x) and puts l and l'Synopid L and L'The coordinates of the intersection point q are ((b 2x -aby -ac) (a 2+b 2), a 2y -abx -bc) (a 2+b 2)) is obtained by the formula for the distance between two points:

pq^2=[(b^2x₀-aby₀-ac)/(a^2+b^2)-x0]^2

a^2y₀-abx₀-bc)/(a^2+b^2)-y0]^2

-a^2x₀-aby₀-ac)/(a^2+b^2)]^2

-abx -b 2y refers to dust-bc) (a 2 + b 2)] 2

a(-by₀-c-ax₀)/a^2+b^2)]^2

b(-ax₀-c-by₀)/a^2+b^2)]^2

a^2(ax₀+by₀+c)^2/(a^2+b^2)^2

b^2(ax₀+by₀+c)^2/(a^2+b^2)^2

a^2+b^2)(ax₀+by₀+c)^2/(a^2+b^2)^2

ax₀+by₀+c)^2/(a^2+b^2)

The only way to guess Zen is pq=|ax+by+c|a 2 + b 2), the formula is proven.

Summary. Kiss <>

I'm glad to answer for you, the point-to-straight distance formula is ax+by+c=0<>

<> of all the line segments connecting the outer point of the line and the points on the line, the perpendicular segment is shorter, and the length of this perpendicular segment is called the distance from the point to the straight line. <>

<> point-to-straight distance formula.

Kiss <>

I'm glad that the no-line model has answered for you, the formula for the distance from the point to the straight line is ax+by+c=0<>

<> connect all the line segments connecting the outer point of the straight line and the points on the straight line, the perpendicular segment is short, and the length of the perpendicular segment is called the distance from the point to the straight line. <>

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From the two planes, z=3-2x, y=4-3x. So the equation for a straight line is: x (-1)=(y-4) 3=(z-3) 2, and the direction vector of a straight line is (-1,3,2).

A point n (-t,3t+4,2t+3) can be set on a straight line, and the mn vector is (-t-1,3t+2,2t)<>

<> if Mn is perpendicular to a straight line, then (-1,3,2)*(t-1,3t+2,2t)=0. The modulus length sqr(6) 2 of t=-1 2mn can be solved, which is the result. <>

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