2. According to the meaning of the question, the inclination angle is X: tanx = (0-(-√ 3))/(6-5) = √ 3, so the inclination angle is X = atan (√ 3) = 60.
3. The straight line according to the meaning of the question is: y+2=[0-(-2)/(3-0)]*(x-3), that is, y=(2/3)x-4.
4.ab=√[(- 1-7)^2+(-2-(-5))^2]=√[64+9]=√73
bc=√[(7-4)^2+(-5-3)^2]=√[9+64]=√73
ac=√[(- 1-4)^2+(-2-3)^2]=√[25+25]=5√2
AB linear equation: y+2 = [(-2-(-5))/(-1-7)] * (x+1), that is, y=-3/8x- 19/8.
BC linear equation: y+5=[(3-(-5))/(4-7)]*(x-7), that is, y=-8/3x+4 1/3.
Ac linear equation: y+2 = [(3-(-2))/(4-(-1))] * (x+1), which means y=x- 1.
5. y+1= [(2-(-5))/(1-1)] * (x+1), that is, y=(7/2)x+5/2.
6. l1:y = [(2-0)/(0-(-3))] (x-2), that is, y=2/3x-4/3.
L2: y-2/3 = [(0-2/3)/(1-0)] x, that is, y=-2/3x+2/3.
L3:x=-2
7. Because the slope of the straight line x+3y-2=0 is-1/3, the straight line is: y+5=(- 1/3)*(x- 1), that is, Y =- 1/3x+.
8. The intersection of the straight line x-y- 1=0 and the straight line 2x+y-2=0 is the solution of the equation:
x-y- 1=0
2x+y-2=0
Solution:
x= 1，y=0
That is, the intersection point is: (1, 0)
The slope of the straight line 2x-3y+2=0 is 2/3.
So the linear equation is: y-0=(2/3)*(x- 1), that is, y=2/3x-2/3.
9. The slope of the straight line BC is: (7-3)/(6-0)=2/3.
So the high slope on BC is -3/2.
So the equation of height is: y-0=-3/2*(x-4)=-3/2x+6.
10. The slope of the line 3x- 18y-4=0 is 1/6.
So let the intersection of the straight line L and the Y axis be (0, d), then the straight line is Y =1/6x+d.
So the point where the straight line intersects the X axis is: (-6D, 0).
And 0.5*D*D=3, so d = √ 6.
So the straight line is: y = 1/6x √ 6.
1 1. Solve the equation:
y=x+ 1
(x+2)？ +(y- 1)？ =2
Solution:
x=- 1，y=0
So when a straight line is tangent to a circle, two different roots intersect, only one is tangent, and if there is no solution, it will be separated.
12. Circle (x+3)? +y？ = 16 and circle x? +(y+3)？ The center distance of =36 is √ [(-3-0) 2+(0-(-3)) 2] = 3 √ 2.
The sum of the radii of two circles is 4+6=20.
Because 3 √ 2 < 20, two circles intersect.
Solving the coordinates of common points is solving equations;
(x+3)？ +y？ = 16
x？ +(y+3)？ =36
Solution:
x =- 19/6+ 1/6 *√287； y= 1/6+ 1/6*√287
and
x =- 19/6- 1/6 *√287； y= 1/6- 1/6*√287
That is to say, the intersection point is:
(- 19/6+ 1/6*√287, 1/6+ 1/6*√287)
and
(- 19/6- 1/6*√287, 1/6- 1/6*√287)
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I wrote all the words by hand. It's too hard. I hope the landlord will give me some points:)