Mathematics textbook compulsory two answers

1. Because the straight line ax+3y-5=0 passes through point A (2 1).

So ax2+3x 1-5=0.

So a= 1

So the real number a= 1.

2. From the known (3+a)/(-a-5)= 1, a=-4, and the distance between these two points is d= radical number 2.

3.B

4. Obviously, n is n< equals 0, so there is y=-m/nx+ 1/n, so there is n < 0, m>0.

5. Let the linear equation be x/a+y/b= 1.

Then there are 1/2|ab|=5 and -5? The solution of a-4/b= 1 is a=5, b=-2 or A =-5/2, and B = 4.

Therefore, the linear equation is 2x-5y- 10=0 or 8x-5y+20=0.

6. Let a straight line be y-6=k(x-5).

So 5-k/6=2(6-5k).

K=- 1/2 or k=6/5.

Therefore, the linear equation is x+2y- 17=0 or 6x-5y=0.

7.a= 1 cong 1/a=a/ 1 is not equal to (2a+2)/(a+ 1).

8. According to the meaning of the question, (x-1) 2+(y-3) 2 = (x-5) 2+(y-1) 2.

That is 3x+y+4=0.

9. Let A(4, t) or (-4, t).

T=-5 is obtained by (4- 1) 2+(T+ 1) 2 = 5 under the radical sign.

T=- 1 cong (-4- 1) 2+(t+ 1) 2 = 5 under the radical sign.

The coordinates of point A are (4,3) or (4,5) or (-4, 1).

10. Use (|a+6|)/ (2 2+3 2 under radical number) = 2, a=2 radical number 13-6 or a=-2 radical number 13-6.

The equation of 1 1. circle can be changed to (x-2) 2+(y+2) 2 = 4.

The distance from the center of the circle (2, -2) to the straight line x-y-5=0 d= root number 2/2.

Therefore, the truncated chord length is the root sign 14.

12. The point equidistant from the three points is the center of the circumscribed circle of △ABC, which is the intersection of three vertical lines. The equation of the BC vertical line is x=2 1.

The vertical line on the AC side is 16x+5y-28 1=0.

It can be concluded that the coordinates of points with the same distance are (2 1,-1 1).

13. There are two types of straight lines with equal intercepts on two coordinate axes: (1) with a slope of-1(2) passing through the origin.

(1) When the slope is-1, let the tangent be x+y+t=0.

Then the distance from the center of the circle (0, -5) to the straight line d=(|0-5+t|)/ (root number 2)= root number 3.

The square root of t = 5+6 or t = 5-6.

So the tangent equation is the square root of x+y+5+6 =0 or the square root of x+y+5-6 =0.

(2) When the tangent crosses the origin, let the tangent be kx-y=0.

Then (|kx0+5|)/ (under the root number (1+k 2)) = root number 3, so k= root number 66/3 or-root number 66/3.

So the tangent equation at this time is y= root number 66/3x or-root number 66/3.

So there are four cases of tangent equation, namely, the square root of x+y+5+6 =0 or the square root of x+y+5-6 =0 or y= root number 66/3x or-root number 66/3.

14. The equation of the first circle can be changed to (x+1/2) 2+(y-1) 2 = 85/4.

The length of the straight line connecting two circles is d= 5/2 of the radical sign.

So the five-root symbol 85/2

The midpoint q of 15.Mn (3,-1)

When l crosses the q point, the linear equation is 3x+2y-7=0.

When l is parallel to MN, the linear equation is 4x+y-6=0.

16. Omit

17. Establish a rectangular coordinate system with the midpoint of the bridge as the origin.

Then a (-18.7,0) b (18.7,0 0) c (0 0,7.2)

Let the equation of a circle be x 2+(y-b) 2 = r 2.

Then there are (7.2-b) 2 = r 2 and 18.7 2+b 2 = r 2.

The solution is b =-20.7 and r 2 = 778.2.

So the equation for finding a circle is x 2+(y+20.7) 2 = 778.2.

18. the straight line y=kx+2 runs through C (0 0,2), the slope of AC is (2- 1)/(0+4) = 1/4, and the slope of BC is (2+1)/(0

Therefore, the range of real number K is (-infinity,-1) and [1/4, positive infinity].

19. Prove that the original equation is transformed into

x-2y+2+(4x+3y- 14)k=0

X = 2 and Y = 2 are obtained from the solutions of x-2y+2=0 and 4x+3y- 14=0.

So the coordinate (2,2) always satisfies the linear equation.

Therefore, the straight line must pass through the fixed points (2, 2).

20. the set m represents the circle x 2+y 2 = 4 and its interior point set, and the set n represents the circle (x-1) 2+(y-1) 2 = r 2 and its interior point set.

And because m crosses N=N, n is included in m, so the root number is 2+R.

Therefore, the value range of R is 0.

2 1. Let n ′ (5,2), | PM-PN | = | PM-PN ` < = Mn` intersect with X axis and P 1

Therefore, when p moves to P 1, |PM-PN| has the maximum value MN'.

At this point, the coordinate of point P is (13,0).

22. It is proved that a rectangular coordinate system is established with A as the origin and AB as the positive direction of X axis. Let A (0 0,0 0) E (a a,0 0) F (2a 2a,0 0) D (0 0,a a) C (3a 3a,a), then the linear equation of AC is y= 1/3x.

The linear equation of DF is y =-1/2x+a.

Add them together to get G(6/5a, 2/5a).

So the slope of GE is 2, and the slope of DF is-1/2.

So the slope of GE times the slope of DF is-1, so EG⊥DF.

23. The symmetry point of a (1,2) about the straight line 2x+y- 1=0 is a ` (-7/5,4/5).

So the linear equation of BC is (y+1)/(4/5+1) = (x+1)/(-7/5+1).

That is, the linear equation of BC is 9x+2y+ 1 1=0.

The equation is simultaneous with C (-13/5,31/5).

24. The straight line y=x+b is a parallel moving straight line with a slope of 1. Under the root sign x = (1-y 2), it represents the right half of the circle x 2+y 2 =1.

-1= 1 or b=- root number 2

25.( 1) Let A(x 1, x 1), and its symmetry point B(2-x 1, -x 1) about P (1) is at the radical sign 3x+.

So the root number 3(2-x 1)+3(-x 1)=0 gives the root number 3 of x1=-/kloc-0.

So the equation of straight line AB is (-2+ root number 3)y+(- 1+ root number 3)(x- 1).

(2) Let A(x 1, x 1)B(- radical 3y2, y2).

Then (x 1+y2)/2=(x 1- radical 3y2)/4 and y2/(- radical 3y2-1) = x1(x1-kloc-)

So B(2 1- 12 root number 3, 12-7 root number 3)

So the equation of AB is y=( 12-7 radical number 3)/(20- 12 radical number 3)(x- 1).

26. Point P is a circle with O as the center and 3 as the radius on the plane xOy.

exist

Because the circle with the diameter of chord AB passes through the origin, the equation of this circle can be set to c': x 2+y 2+dx+ey = 0.

The slope of l is 1.