Rapid Senior 2 Liberal Arts Mathematics Straight Lines and Equations 10

The straight line i is perpendicular to the straight line x+y 7 0, so let the equation x-y+a=0 then |a|(2+√2)=2

a|=2/(2+√2)=2-√2

a=±(2-√2)

The equation is x-y (2- 2)=0

Because it is perpendicular to the straight line x+y 7 0, the slope of the straight line l is 1, let the straight line l be y=x+a, and because the perimeter of the triangle enclosed by the straight line l and the two coordinate axes is 2, so a=, so the straight line is y=x+

Let the equation for the line l be y=ax+b, and since the line i is perpendicular to the line x+y 7 0, then a*(-1)=1 i.e., a=1

So the l equation is y=x+b, and when x=0, y=b; When y=0, x=-b means that the length of the two right-angled sides of the triangle surrounded by the line l and the two coordinate axes is:|b|

Since the circumference of the triangle enclosed by the line l and the two coordinate axes is 2, then 1 2*|b|The square of = 2, i.e. the square of b = 4

b = 2 or -2

The equation for the straight line l is: y=x+2 or y=x-2

Some math symbols can't be typed, so you'll just look at it.

The equation for l from the vertical relation is: x-y+a=0 a is an arbitrary number.

The circumference of the triangle enclosed by the coordinate axis is 2

The absolute value of a plus the absolute value of a plus the root of a*a=2 is the equation for the perimeter, and the solution of a is 2 3 or -2 3

Let the equation be x a+y b=1

Intercepts, respectively).

Bring in (got.) 2a+b=ab

And because ab=+-8

So 2 formulas are obtained.

a=2a=-2+2√2

a=-2-2√2

Then find B.

y=kx+b, bring 2 and 1 in, that is, 2k+b=1, and because the area is 4, the product of the intercept of the x-axis and the intercept of the y-axis is 4, that is, b multiplied by negative b, k is 4, plus 2k+b=1, and solve this binary system of equations.

Since it is impossible for a straight line to be perpendicular to the x-axis since it is to form a triangle, the equation can be (y-1) (x-2)=k

1) Straight line through 124 quadrants.

When x=0, y=1-2k

When y=0, x=2-1 k

So (1-2k)*(2-1 k)*1 2=4(2) straight through 123 quadrants.

then (2-1 k)*|1-2k|=(2-1 k)*(2k-1)=8(3) straight line through 134 quadrants.

then |2-1/k|*(1-2k)=(1 k-2)*(1-2k)=8 Solving three systems of equations gives you three different ks

Let the equation be y=kx+b, let's (2,1) and substitute it for 1=2k+b, when x=0y=b, y=0

x = b of minus k

Because the area is 4, the formula of bringing in the area can derive negative b square = plus or minus 8k, and then according to 1 = 2k + b, b is pushed out to get 2 or minus 2 plus or minus 2 times the root number 2, and k is obtained by bringing in 1 = 2k + b

You can refer to what I have made now.

Let a(x,0) b(0,y).

Because vector ap=1 2 vector pb

So (-5-x,4-0)=1 2(0-(-5),y-4) i.e. -5-x=(1 2)*5

4=y-4 gives x=-15 2 y=8

So a(-15 2,0) b(0,8).

Know the points a and p, and use the two-point formula to find the equation of the straight line.

Or know the points p and b, and use the two-point formula to find the equation of the straight line.

Let the straight line: y=kx+b

Because p(-5,4) brings in the equation: 4=-5k+bb=4+5k

Straight line: y=kx+4+5k

Since the x and y axes are intersected at ab respectively, when y=0 is calculated as x=-(4+5k) k, i.e., a(-(4+5k) k,0).

When x=0 is calculated as y=4+5k

b（0，4+5k）

ap=(-5+(4+5k)/k)i+4j

pb=5i+5kj

AP vector = 1 2PB vector.

5+(4+5k)/k=5/2(1)

4=5k/2(2)

From (1) and (2), any one of the equations is solved, k=8, 5, so y=8x, 5+12

Because p is the 3rd alipart of AB.

So ao=15 bo=6 equation is 2x-5y+30=0

Let l:y=ax+b, because the point a(-2,2) is on a straight line, we get -2a+b=2 , the intersection of l and the coordinate axis.

The points are (0,b), (b a,0).

Because the area of the triangle enclosed by l and the two coordinate axes is 1, we can get |-b/a*b*(1/2)|=1, tidying up.

Gotta |b^2/2a|=1②。

Arrange b=2+2a

Bring in de, |2*(a+1)^2/a|=1, so 2(a+1) 2=a(a>0) or 2(a+1) 2=-a(a<0).

Sorting out is that 2a 2 + 3a + 2 = 0, the equation = -7, and the equation has no solution.

Arrange the result, 2a 2 + 5a + 2 = 0, (2a + 1) (a + 2) = 0, solve a1 = -1 2, a2 = -2

So b1 = 1 and b2 = -2

The equation for the straight line is: l1 y=-(1 2)x+1, l2 y=-2x-2

Let y=ax+b, point A is brought in to obtain the relationship between a and b, and then find the intersection point of the line and the x and y axes, and use the area of the triangle to obtain the second equation of a and b, and the result is obtained.

Solution: (1).Set point a'The coordinates of are (x',y')

The straight line aa is known from the meaning of the title'Perpendicular to the straight line l.

then k(aa') k(l)=-1

The slope of the straight line l, 3x+y-2 0, is known to be k(l)=-3

So straight aa'The slope of is k(aa')=-1 k(l)=1 3 (you calculated that the answer is correct, the slope is 3 when it is a floating cloud).

then the straight line aa'The equation can be written as:

y-4=1/3 *(x+4)

Solve the system of equations x-3y+16=0, 3x+y-2=0 to find the straight line aa'Coordinates of the intersection point with l.

x=-1,y=5, i.e., the intersection coordinates are (-1,5).

It is easy to know that the focus of this is the line segment AA'The midpoint.

It can be obtained from the midpoint coordinate formula.

x'+(-4)=2*(-1),y’+4＝2*5

Solve x'=2,y'=6

So the coordinates of point a' are (2,6).

2).From the meaning of the title, set any point p(x,y) on the line l, and the symmetry point p on the line l' with respect to the point a'(x',y')

Then it is easy to know that point a is the midpoint of the line segment pp'.

It can be obtained from the midpoint formula.

x+x'=-8,y+y'=8

i.e. x=-x'-8,y=-y'+8 (*

Since the point p(x,y) is on the line l, the coordinates of the point p, i.e., the equation (*), are substituted into the equation of the straight line: 3x+y-2=0

Available: 3*(-x.)'-8)+(y'+8)-2=0

i.e. 3x'+y'+18=0

So the line l is about the symmetrical line l of the point a'The equation can be written as: 3x+y+18=0

(1).

Set a'The coordinates of are (x,y).

Because point a is symmetrical point a with respect to the line l'coordinates.

So the straight line L is the line segment AA'of the perpendicular bisector, so aa'm is on a straight line; The slope product is 1,kaa'*kl=0

aa'The midpoint m is [(x-4) 2,(y+4) 2], kaa'=(y-4)/(x+4) ,kl=-3

The simultaneous equation yields: 3[(x-4) 2]+(y+4) 2-2=0 and the solution yields x=2

[y-4)/(x+4)]*3)=0 y=6

So the point a is symmetrical with respect to the line l'The coordinates are (2,6).

If any point n(0,2) is taken on the line l, then the symmetry point n with respect to the point a is n'（－8，6）

Set the straight line L'The equation of the equation is: 3x+y+c=0

then the symmetry point n'(8,6) must be in a straight line L'on, place n'Substituting 3x+y+c=0 obtains.

c=18 so straight line l'The equation of the equation is: 3x+y+18=0

The aa of the first question you said'The slope is indeed 1 3, yes, this is what I calculated, do you understand?

Let the line l:y=k(x-3).

l1：2x-y-2=0 ②

l2：x+y+3=0 ③

Coordinates of point A (3K-2) (K-2), 4K (K-2).

Coordinates of point b (3k-3) (k+1), 6k (k+1)).

Only find one of x and y).

Then according to p as the midpoint of ab, 4k (k-2)+6k (k+1)=0 is obtained to solve k

(Due to the complexity of input, I will briefly tell you the idea, please try to do it first, if you don't understand, then contact me) Because the fixed point (3,0) can be set to the straight line y=k(x-3), and then the straight line is connected to the line 1 and the line 2 respectively, and the two intersection coordinates (which contain a parameter k) are obtained, and because the p point bisects ab, the k value can be solved with the midpoint coordinate formula, and then the line comes out to pull. It's here.

The equation for the direct call line l is (m 2-2m = 3) x + (2m 2 + m -1) y - 2m + 6 = 0

The straight line l passes through the fixed point p(-1,-1).

Bring in to find, m=5 3 or m=-2

It can be seen that there are two straight lines, just as the hole is a straight line, which is uncertain.

However, the slag chain is dry, no matter which straight line it is, it is constant through the point p(-1,-1), so it is said that the straight line l passes through the fixed point p(-1,-1).

Can't you pass a fixed point if you set a straight line?? Is it possible to set a fixed point on a fixed line??

ab straight line slope: k1 = [1-(-1) (-1-3) = -1 2

AC straight line slope: K2 = [3-(-1) (1-3) = -2

BC straight line slope: K3 = (3-1) [1-(-1)] = 1

Then the equation of the ab line: y=-1 2(x-3)-1=-1 2x+1 2

AC linear equation: y=-2(x-3)-1=-2x+5

BC linear equation: y=(x+1)+1=x+2

Then the region of the triangle abc where a(3,-1) b( -1,1) c(1,3) is the vertex is:

y>=-1/2x+1/2

y<=x+2

y<=-2x+5

The function Z=3X-2Y can be understood as a straight line with a slope of 3 2 and a longitudinal intercept of -Z2, which must have an intersection point with the area of the above triangle ABC, and the number is combined, and it can be seen that when Z=3X-2Y passes through the point B, the minimum value is taken, and Z=(-1)*3-2*1=-5

When z=3x-2y passes through point a, take the maximum value, and z=3*3-2*(-1)=11