题目内容：

一条直线3x+y-2=0与x+5y+10=0间的线段被点（2,-3）平分,求这条直线的方程
2．球的表面积等于圆柱的侧面积

优质解答

设这条直线和3x + y - 2 = 0的交点为(a,b).
则,
3a + b - 2 = 0,b = 2 - 3a ...(1)
设这条直线和x + 5y + 10 = 0的交点为(c,d)
则,因为点(2,-3) 为(a,b)和（c,d）连线的中点.
a + c = 2*2 = 4,c = 4 - a ...(2)
b + d = 2*(-3) = -6,
d = -6 - b = -6 -(2-3a) = -8 + 3a ...(3)
又,
（c,d）在直线x + 5y + 10 = 0上.
所以,
c + 5d + 10 = 0 ...(4)
将（2）,（3）带入（4）,
4 - a + 5[-8 + 3a] + 10 = 0,
14a = 26,a = 26/14 = 13/7.
由（1）,
b = 2 - 3a = 2 - 3*(13/7) = -25/7.
所以,所求直线的方程为,
(x - 2)/(13/7 - 2) = (y + 3)/(-25/7 + 3),
(x-2)/(-1) = (y+3)/(-4),
4(x - 2) = (y+3),
y = 4x - 11