Find the equation for a straight line with points A 1,2 and points B 2,4.

Solution: Let the equation for the straight line be.

y=ax+b

Substituting a(-1,2) and b(2,4) into the equation respectively gives (1)-a+b=2(2)2a+b=4, and solves (1) and (2).

a=2 3, b=8 3, so, the linear equation is.

y=2/3x+8/3

Solve the equation of the straight line as y=kx+b

The point a(-1,2) and the point b(2,4) are crossed by a straight line

i.e. -k+b=2

2k+b=4

The simultaneous solution yields k = 2 3 and b = 8 3

Therefore, the linear equation is y=2x 3+8 3

Find the equation for a straight line with points a(-1,2) and points b(2,4).

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The slope of the straight line is (4-2) (2+1)=2 3

Let the linear equation be y=2x 3+b, and substitute the coordinates of a to obtain b=8 3

So y=2x 3+8 3

Let the equation for the straight line be .

y=kx+b。

Substituting two points gives two equations.

2=-k+b（1）

4=2k+b（2）

Subtract the two formulas to get k = 2 3, and substitute k for 1 to solve the solution with b = 8 3.

So the equation for the straight line is.

y=(2/3)*x+(8/3)

The equation for a straight line that crosses the points a(1,-2) and b(-1,-4) is y=x-3.

Analysis: If a(1,-2),b(-1,-4) is known, then the equation is y=kx+b, and then a,b is substituted into -2=k+b, -4=-k+b; The addition of the two formulas gives -6=2b, b=-3, k=1, then y=x-3.

Limitations of the various forms of linear equations:(1) Neither the point oblique nor the oblique truncation can represent a straight line where the slope does not exist;

2) The two-point formula cannot represent a straight line parallel to the coordinate axis;

3) The intercept type cannot represent a straight line parallel to the coordinate axis or past the origin;

4) In the general equation of a straight line equation, the coefficients a and b cannot be zero at the same time.

Let the equation be y=kx+b

Substitute A and B for it.

2=k+b4=-k+b

The sum of the two formulas.

6=2bb=-3

k=1y=x-3

Since the direction vector of the straight line is v=m1m2=(-4,2,1), the equation for the straight line m1m2 is (x-3) (-4)=(y+2) 2=(z-1) 1.

m1m2=(-3,4,-6),m1m3=(-2,3,-1), therefore, the plane normal vector is n=m1m2 m1m3=(14,9,-1), therefore, the equation for plane m1m2m3 is 14(x-2)+9(y+1)-(z-4)=0, which is simplified to 14x+9y-z-15=0.

The direction vector is n(-1-3,0+2,2-1)=(-4,2,1).

So the equation for the straight line is.

x-3)/(-4)=(y+2)/2=(z-1)/1.

Solve the equation of the straight line as y=kx+b

The point a(-1,2) and the point b(2,4) are crossed by a straight line

i.e. -k+b=2

2k+b=4

The simultaneous solution yields k = 2 3 and b = 8 3

Therefore, the linear equation is y=2x 3+8 3

k=(4-0)/(2-1) =4

ab equation: y-0=4(x-1) ....Point oblique.

That is: 4x-y-4=0......General.

You are welcome to ask.

Solution: Since the straight line passes through the points a(1,0) and b(2,4), the slope formula k=(y2-y1) (x2-x1) yields:

The slope of the straight line ab k=(4-0) (2-1)=4, and the point slope of the straight line y-y0=k(x-x0) obtains:

The equation for the straight line ab is: y-0=4(x-1).

That is: 4x-y-4=0......

The equation for a straight line is y=kx+b

Substituting two points into the equation and solving k and b to obtain the linear equation.

k+b=02k+b=-3

The solution of the two formulas gives k=-1 and b=-1

Then the equation for the straight line is y=-x-1

From the two-point equation of the straight line: (y-y1) (y2-y1) = (x-x1) (x2-x1), the equation of the straight line is :

y+1)/(2+1)=(x-3)/(4-3)(y+1)/3=(x-3)/(7)

Reduced to a general formula, i.e.:

3x+7y-2=0

The equation for the straight line of the crossing points a(3,-1),b(-4,2) is.

y+1)/(x-3)=(2+1)/(4-3)(y+1)/(x-3)=-3/7

3(x-3)+7(y+1)=0

Simplification yields 3x+7y-2=0.

Because the straight line passes through the points a(-1,4), b(0,2), and the balance is empty.

The equation for this straight line that can be obtained from the two simple ignition point equations is:

y-2) (4-2) = (x-0) (1-0) is reduced to a general formula:

2x+y-2=0。