Question

急.在直线l：x＋3y＝0 上求一点p使点p到原点与直线m:x＋3y＋2＝0的距离相等,

Answer 1

设A(-3a,a)

OA = √10|a|

A到直线x+3y+2=0的距离d = |-3a + 3a + 2|/√10 = 2/√10

√10|a| = 2/√10

|a| = 1/5

a = ±1/5

A(-3/5,1/5)或(3/5,-1/5)