If we know that the line l passes through the point 3, 2 and the intercepts on the two coordinate

The intercepts on the two coordinate axes are equal, indicating that the angle between l and the x-axis is 45 degrees or 135 degrees, and at 45 degrees: let the l equation be y=x+a, and bring in (3,-2), a=-5, so l:y=x-5;

At 135 degrees: let the l equation be y=-x+a, bring in (3,-2), and get a=1, so l:y=1-x;

If the line L crosses the point a(-2,-3) and the intercepts on the two coordinate axes are equal, let the x-axis intercept point (m,0) and the y-axis intercept point (0,m), then the straight line:

k=(y2-y1)/(x2-x1)=(m-0)/(0-m)=-1；

Let the linear equation be y=kx+b, and substitute k=-1 and the point a(-2,-3) into the equation:

3)=(-1)*(2)+b, so b=-5;

Finally, the linear l equation is:

y=-x-5

The answer is complete.

Assuming that the intercept is a, then the linear equation is x+y=a

substitution points (3, -2).

3 - 2 = a = 1

So the equation for a straight line is: x+y=1

Trap, trap. There is also 2x+3y=0

1, the intercept is 0, y=-2 3x

2. The intercept is not 0, x+y=k

If k=1, then x+y=1

Note: The value on the intercept coordinates, not the distance!

2x 3y 0 or x y 5 0Solution 1: (Solved with the help of point slope) Since the straight line l has intercepts on both axes, the straight line is not perpendicular to the x and y axes, the slope exists, and k ≠ 0

Let the linear equation be y 2 k(x 3), let x 0, then y 3k 2;Let y 0, then x 3Set by the question.

If the line L crosses the point a(-2,-3) and the intercepts on the two coordinate axes are equal, let the x-axis intercept point (m,0) and the y-axis intercept point (0,m), then the straight line:

k=(y2-y1) (x2-x1)=(m-0) Answer(0-m)=-1;

Let the equation of the straight circle and blue line be y=kx+b, and substitute k=-1 and the point a(-2,-3) into the equation:

3)=(-1)*(2)+b, so b=-5;

Finally, the linear l equation is:

y=-x-5

1) Let the intercepts of the straight line l on the x,y axis are a, if a=0, that is, l passes through the points (0,0) and (3,2), the equation for l is y=2 3 x, that is, 2x-3y=0 If a≠0, then let the equation for l be x a +y b =1, l passes through the point (3,2), 3 a +2 a =1, Lingmu regrets a=5, and the cry equation of l is x+y-5=0 In summary, the equation for the straight line l is.

The intercepts are equal and permeable can be zero, or may not be zero. So there are two situations in high school.

1.If it is zero, we can let the equation of the straight line be y=kx, because after (2,2), it is obvious that k=1, i.e., y=x;

2.If the mountain is zero, set it to x a+y a=12 a+2 a=1, a=4

The slope is 1 or -1

That is, y-2=x-2 or y-2=-(x-2) point oblique town pin repentance royal zheng.

So the equation for l is y=x or y=-x+4

Let the equation of the straight line be a(x-1)+b(y-2)=0, so that x=0 obtains y=2+a b, that is, the intercept of the straight line on the y-axis is b=2+a b, so that y=0 obtains x=1+2b a, that is, the intercept of the straight line on the x-axis is a=1+2b a, and a+b=6 is known from the open stool, so the arguing 1+2b a+2+a b=6, Simplifying (b-a)(2b-a)=0 , taking a=b=1 or a=2, b=1, the square journey of the straight line can be obtained as x+y-3=0 or 2x+y-4=0

If the line L is too short of the high spike point a(-2,-3) and the intercepts on the two coordinate axes are equal, and the x-axis intercept point (m,0) and the y-axis intercept point (0,m) are set to the straight line:

k=(y2-y1)/(x2-x1)=(m-0)/(0-m)=-1；

Let the linear equation be y=kx+b, and substitute k=-1 and call a(-2,-3) into the equation:

3)=(-1)*(2)+b, so b=-5;

Finally, the linear l equation is:

y=-x-5

The answer is complete.

Cross-section. is the abscissa of the intersection of the line and the x-axis, and the longitudinal intercept is the ordinate of the intersection of the y-axis.

Let the equation of the straight line.

is y=kx+b, then the longitudinal intercept is b

Let y=0, and get the state or intercept of the x-axis is.

x=-b/k

b then k = -1

Namely. y=-x

b substitutes a(-2,-1) into Fang Tuan Qingcheng, and gets -1

b，b=-3

i.e. y=-x-3

1) If the intercept of the straight line on the two coordinate axes is 0, then the straight line passes through the early limb point of the original orange, so let the linear line equation be y=kx, and substitute (3,-2) to get k=-2 3, so the linear line equation is y=-2x 3;

2) If the intercept of the straight circle line on the two coordinate axes is not 0, then let the linear equation be (x a)+(y a)=1 (intercept formula), and substitute the point (3,-2) to obtain a=1, so the linear equation is x+y-1=0;

In summary, the equation for a straight line is y=-2x 3 or x+y-1=0.

Equal intercepts are divided into two cases.

The first demolition of the old one. Past the origin, set a straight line; l:

y=kx==>2=3k

k=2/3y=2/3x

Second: k = -1

Let the straight line l:x+y= =3+2=5

l:x+y-5=0

Let the linear line system equation for the parallel rise of the straight line brigade is:

2x-3y+2)+λ3x-4y-2)=0

2+3 )x-(6-4)y+(2-2)=0 according to the title:

1=4 7

It is also possible to find the intersection of two lines and solve it with k=1 and k=-1.

1) If the intercept of the straight line on the two coordinate axes is 0, then the straight line passes through the early limb point of the original orange, so let the linear line equation be y=kx, and substitute (3,-2) to get k=-2 3, so the linear line equation is y=-2x 3;

2) If the intercept of the straight circle line on the two coordinate axes is not 0, then let the linear equation be (x a)+(y a)=1 (intercept formula), and substitute the point (3,-2) to obtain a=1, so the linear equation is x+y-1=0;

In summary, the equation for a straight line is y=-2x 3 or x+y-1=0.

Let the distance of the line l on the x and y axes be a, if a=0, that is, l passes through the points (0,0) and (3,2), the equation for the hand calendar l is y=2

x, i.e., 2x-3y=0

If a≠0, then let the equation for l be xay

B1, L (3,2), 3a

The equation for a1, a=5, l is x+y-5=0 In summary, the equation for the line l is .

2x 3y 0 or x y 5 0