Question:
find the distance of the point (-1,1) from the line 3(x 5)=4(y 3).
Answer:
Given, equation of line is:
3(x + 5) = 4(y + 3)
=> 3x + 15 = 4y + 12
=> 3x + 15 - 4y - 12 = 0
=> 3x - 4y + 3 = 0
Given, point is (-1, 1)
Now, the distance d = {3*(-1) - 4*1 + 3}/√(32 + 42 )
= {-3 - 4 + 3}/√(9 + 16)
= -4//√(25)
= -4/5
Since, distance can not be negative,
So, the distance d = 4/5
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