Question

帮我做2道英文数学题 谢谢

1.Find the equation of the line which passes through the point (6.-4).And is parallel to the line 2x-3y+3=0.

2.Find the equation of the line which passes through the points P(0,A) and Q(A,2A) .

3.Find the equation of the line perpendicular to 5x+2y-1=0 and passing through the point (-2,7).

谢谢了= =..........帮我做下吧.......过程最好也写下吧...
是3道..!- -

Answer 1

1.找一条直线过点(6,-4),并且平行于2x-3y+3=0.
解:因为平行于2x-3y+3=0,所以设直线为2x-3y+b=0
将(6,-4)代入,得12+12+b=0,b=0
所以直线方程为:
2x-3y=0

2.找一条直线过点P和点Q
点斜式学过吗?斜率k=(2A-A)/(A-0)=1
设方程为y-y'=k(x-x')
将P(0,A)代入,y-A=x
所以直线方程为:
y=x+A
PS:如果没学过.就设最简单的ax+by+c=0再代入

3.求一条直线垂直于 5x+2y-1=0 并过点(-2,7)
因为垂直,所以设方程为-2x+5y+b=0
将(-2,7)代入,得:4+35+b=0,b=-39
所以直线方程为:
-2x+5y-39=0

Answer 2

1. assume the equation is y=ax+b. Because 2x-3y+3=0, y=(2/3)x+1, the slope is equal to 2/3. Because y=ax+b is parallel to the line 2x-3y+3=0, a=2/3. The point (6,-4) is on the line y=(2/3)x+b, so -4=(2/3)*6+b, so b=-8. Hence, the equation is y=(2/3)x-8
1.找一条直线过点(6,-4),并且平行于2x-3y+3=0.
解:因为平行于2x-3y+3=0,所以设直线为2x-3y+b=0
将(6,-4)代入,得12+12+b=0,b=0
所以直线方程为:
2x-3y=0

2. Assume that the equation is y=ax+b, the line passes through the point P(0,A), Q(A,2A), so A=b and 2A=Aa+b, 2A=Aa+A, finally, a=1. Therefore, the equation is y=x+A
2.找一条直线过点P和点Q
点斜式学过吗?斜率k=(2A-A)/(A-0)=1
设方程为y-y'=k(x-x')
将P(0,A)代入,y-A=x
所以直线方程为:
y=x+A

3. 5x+2y-1=0, y=(-5/2)x+(1/2), the slope is K=-5/2, because the equation of the line that we want to look for is perpendicular to 5x+2y-1=0, assume the equation is y=ax+b, so a*K=-1, so a=2/5, y=2/5x+b, the point (-2,7) is on the equation of line, so 7=(2/5)*(-2)+b, b=39/5. Hence, the equation is y=(2/5)x+(39/5).
3.求一条直线垂直于 5x+2y-1=0 并过点(-2,7)
因为垂直,所以设方程为-2x+5y+b=0
将(-2,7)代入,得:4+35+b=0,b=-39
所以直线方程为:
-2x+5y-39=0

Answer 3

1. assume the equation is y=ax+b. Because 2x-3y+3=0, y=(2/3)x+1, the slope is equal to 2/3. Because y=ax+b is parallel to the line 2x-3y+3=0, a=2/3. The point (6,-4) is on the line y=(2/3)x+b, so -4=(2/3)*6+b, so b=-8. Hence, the equation is y=(2/3)x-8

2. Assume that the equation is y=ax+b, the line passes through the point P(0,A), Q(A,2A), so A=b and 2A=Aa+b, 2A=Aa+A, finally, a=1. Therefore, the equation is y=x+A

3. 5x+2y-1=0, y=(-5/2)x+(1/2), the slope is K=-5/2, because the equation of the line that we want to look for is perpendicular to 5x+2y-1=0, assume the equation is y=ax+b, so a*K=-1, so a=2/5, y=2/5x+b, the point (-2,7) is on the equation of line, so 7=(2/5)*(-2)+b, b=39/5. Hence, the equation is y=(2/5)x+(39/5).

够详细吧