length of perpendicular from origin to line 4x+3y-2=0is.....unit
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The answer is given below :
FORMULA :
Let us consider any line be
ax + by + c = 0 .....(i)
Let, any point be (x1, y1).
Then, the required distance of the line (i) from the point (x1, y1) is given by

So, the required distance from the origin O (0,0,0) is

SOLUTION :
The given line is
4x + 3y - 2 = 0 .....(ii)
Thus, the required distance of the line (ii) from the origin O (0,0,0) is

which is the required length of the perpendicular line.
Thank you for your question.
FORMULA :
Let us consider any line be
ax + by + c = 0 .....(i)
Let, any point be (x1, y1).
Then, the required distance of the line (i) from the point (x1, y1) is given by
So, the required distance from the origin O (0,0,0) is
SOLUTION :
The given line is
4x + 3y - 2 = 0 .....(ii)
Thus, the required distance of the line (ii) from the origin O (0,0,0) is
which is the required length of the perpendicular line.
Thank you for your question.
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