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

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
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..!- -

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
1.找一条直线过点(6,-4),并且平行于2x-3y+3=0.
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

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=(...

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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).
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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=(...

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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

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