Solve the problem of the distance between a straight line and a point in high school mathematics

1. The straight line passing through the point (-5, -4) and the intersection point of the two coordinate axes are divided into three situations: first, it intersects with the negative half axis of the X axis and the positive half axis of the Y axis; second, it intersects with the positive half axis of the x-axis and the negative half axis of the y-axis; Thirdly, it intersects with the negative half axis of the x-axis and intersects with the negative half axis of the y-axis. Clause.

In the second case, the slope is positive, but the signs of the intercepts of the two axes are opposite; In the third case, the slope is negative, and the area of the triangle enclosed by the coordinate axis is greater than 4 5 20, so the third case is not considered.

According to the above analysis, the linear equation can be written as: y k(x+5)-4, where k>0

When x 0, the y-axis intercept is 5k 4

When y 0, the x-axis intercept is 4 k-5

Therefore, the area of the triangle enclosed by the coordinate axis is: -(1 2)(5k-4)(4 k-5)=5

Simplification: 25k 2-50k+16=0

Solution: k1 = 8 5, k2 = 2 5

So the two linear equations are: y=8x 5+4, y=2x 5-2

2. L1 and L2 are parallel, so the X and Y coefficients correspond to equal proportions (in fact, the slopes are required to be equal):

i.e.: a (a-1) = (-b) 1

The distance between the coordinate origin and the two lines is equal, so the y-intercept sign of the two lines is opposite (the above parallel condition determines that the x-intercept sign is also opposite, and vice versa, so one intercept condition is sufficient).

l1: when x 0, y 4 b;

l2: when x 0, y b;

So 4 b (b).

That is: b 2 = 4, so there are two values: b 1 = 2 and b = -2, substituting a = b (1 + b) to get :

b1=2，a1=2/3；b2=-2，a2=2

1 Let y=kx+5k-4(k is not equal to 0).

Let y=0,x=(4-5k) k

Let x=s=1 2** be an absolute value. ）

Let k>4 5, then (5k-4) 2 k=10, and the solution is k=8 5, k=2 5 (round).

At this point y=8 5x+4

When 0 is < 0, k is not solved.

So y=8 5x+4or y=2 5x-2

2 If l1 l2, then there is: a b = (a-1) (-1) a b = 1-a

a/(1-a)=b

If the distance from the coordinate origin to the two lines is equal, then there is:

4/√(a^2+b^2)=|b|/√[(a-1)^2+1]16/(a^2+b^2)=b^2/[(a-1)^2+1]16[(a-1)^2+1)]=a^2/(a-1)^2*[a^2+a^2/(a-1)^2]

16[(a-1)^2+1]=a^2/(a-1)^2*a^2/(a-1)^2*[(a-1)^2+1)]

16=a^4/(a-1)^4

a/(a-1)=(+/-)2

1)a=2)a=2/3,b=2/3/(1-2/3)=-2

1. Let the straight line y+4=k(x+5), which intersects with the two axes (0,5k-4),(4 k-5,0), so 1 2*|5k-4|*|4/k-5|=5, K1=2 5, K2=8 5, so the straight line is Y+4=2 5(X+5) or Y+4=8 5(X+5).

2. Because L1 is parallel to L2, -a B=A-1, because the distance from the coordinate origin to L1 and L2 is equal, the intercept of the two straight lines can be obtained from the image is equal, so B=4 B, so B=, A=2 3;When b = -2, a = 2.

The formula for the distance from a point to a straight line in high school math isd=│axo＋byo＋c│/√a²＋b²）。

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

d=│axo+byo+c│ /a²+b²）。

The point-to-straight distance is the length of the perpendicular segment that connects a point outside the line with points on the line, the shortest, and the perpendicular segment.

Formula Description:

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

Of all the line segments connecting a point outside a line to a point on a line, the perpendicular segment is the shortest, and the length of this perpendicular segment is called the distance from the point to the line.

The distance from the point to the line is formulated as follows:

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

Definitional method proof:

By definition, the distance from the point p(x,y) to the line l:ax+by+c=0 is the length of the perpendicular segment from the point p to the line l.

Let the perpendicular line from point p to the line be 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).

Put 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 the two points:

pq^2=[（b^2x_-aby_-ac）/（a^2+b^2）-x0]^2+[（a^2y_-abx_-bc）/（a^2+b^2）-y0]^2=[（a^2x_-aby_-ac）/（a^2+b^2）]^2

The formula for the distance from the point to the straight dan rental line:

1. If the point p(x0,y0) is on the straight line ax+by+c=0 (a, b are not 0 at the same time), then the stove is late ax0+by0+c=0.

2. If the point p(x0,y0) is not on the straight line ax+by+c=0 (a, b is not 0 at the same time), then ax0+by0+c≠0, at this time the point p(x0,y0) straight line ax+by+c=0 (a, b is not hidden and 0 at the same time) distance d = the distance formula from the point to the straight line.

It can be solved by using the two-point bridge lifting and eliminating the suspicion of the straight line equation, and the answer is "three times the number three fifths of the correct knowledge".

Distance from point to line formula:

1. If the point p(x0,y0) is on the straight line ax+by+c=0 (a, b are not 0 at the same time), then ax0+by0+c=0.

2. If the rent point p(x0,y0) is not on the straight line ax+by+c=0 (a, b is not 0 at the same time), then ax0+by0+c ≠ hidden 0, at this point p(x0,y0) straight line ax+by+c=0 (a, b is 0 when different stoves are late) distance d = point to the straight line distance formula.

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

1. If the point p(x0,y0) is on the straight line ax+by+c=0 (a, b are 0 when they are different), then ax0+by0+c=0.

2. If the point p(x0,y0) is not on the straight line ax+by+c=0 (a, b is not 0 at the same time), then ax0+by0+c≠0, at this point p(x0,y0) straight line d=by+c=0 (a, b is not 0 at the same time) distance d = the distance formula from the point to the straight line.

1) The length of the perpendicular segment from the point to the line.

2) Because the distance between the perpendicular foot and p is the minimum of the distance between the point on the line and p, the distance from the point p to the straight line l can be determined or defined as the minimum value of the distance between the point on l and p.

3) i) Seek the foot q. firstThe equation for a straight line perpendicular to p(1,2) and l:x+y-5=0 is x-y+1=0, and the two lines intersect q(2,3),pq|=√2.

ii) Let the point m(x,5-x) on l, then Wu Wu.

pm|=√x-1)^2+(5...2. Mathematics is solved in a variety of ways, and cannot be directly solved by the point-to-straight distance formula.

Senior 1 Mathematics. It can be solved in a variety of ways, and the non-orange number can be directly solved by using the point-to-straight distance formula.

Find the distance from point a(2,3,1) to the line l: x=t-7,y=2t-2, z=3t-2;

Solution: Change the parametric equation of the line l into a standard equation: (x+7) 1=(y+2) 2=(z+2) 3;

Therefore the direction vector of l n=;

The equation for the plane of the direction vector n of l and the plane of the crossing point (2,3,1) is:

x-2)+2(y-3)+3(z-1)=0, i.e., x+2y+3z-11=0....

Obviously, the plane is straight line l; Then substitute the parametric equation of the line l into the equation to obtain:

t-7)+2(2t-2)+3(3t-2)-11=14t-28=0, so t=2;

Substituting t=2 into the parametric equation, the coordinates of the intersection point b of the line l and the plane are (-5,2,4);

Then ab = [(2+5) +3-2) +1-4) ] = (49+1+9)= 59 is the distance from point A to the line l.

The first step is to find the plane perpendicular to the straight line. The direction vector of this straight line is (u, v, w), then u=1*1-(-1)*(1)=0, v=-1*2-1*1=-3, w=1*(-1)-1*2=-3, [here is the basic knowledge, that is, the direction vector of the cross product of the two plane normal vectors that are joined]. Therefore, the plane is 0(x-3)-3(y+1)-3(z-2)=0, i.e., y+z-1=0

In the second step, find the intersection point of the above plane with a given line. Synthesis, which separates the system of equations and equations, the intersection of the solution p'(1,-1 2,3 2).

The third part, ask for pp'distance is enough. i.e. [(3-1) 2+(-1+1 2) 2+(2-3 2) 2] (1 2)=3 2 2

The direction vector of a straight line is l:i j k

3j-3k means that its direction vector is (0, -3, -3).

Take any point q on a straight line: (1,1,3).

then pq=(2,-2,-1).

d=|pq x l |/ |l|

pq=i j k

3i-6j+6k

Then: d= [(3) 2+(-6) 2+6 2] (-3) 2+(-3) 2

pq x l|: is the area of a parallelogram, |l|for its side. Therefore =|pq x l|/|l|is the height of a parallelogram. This is the distance from a point to a straight line.

The distance from a point to a straight line should belong to high school math.