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| To find the length of the perpendicular for the point (x1,y1) on the line ax+by+c=0 |
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| To find the length of the perpendicular from a point (x1,y1) to the line ax+by+c=0 |
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| Let AB represent the line ax + by + c = 0. |
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| The coordinates of A and B respectively are, |
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To find the distance of a line from the point (x1, y1)
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| i) Replace x by x1 and y by y1 in the LHS of the expression. |
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| (RHS being zero). |
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| Aliter: |
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| Let the equation of the line AB be ax + by + c = 0 … (i) |
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| Let P(x1, y1) be the given point. |
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| Draw PN perpendicular to AB. Let the equation of PN be |
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| bx - ay + k = 0 …(ii) |
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| (ii) - (iii) = b(x - x1) - a(y - y1) = 0 …(iv) |
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| Let N be (x2,y2). |
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| Squaring and adding (vi) and (vii), we get |
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| Remark: |
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| If c > 0 in the equation ax + by + c = 0 of the straight line and the given point and the origin are on the same side of the given straight line, then the required length of the perpendicular is positive and if the given point and the origin are on the opposite sides of the given straight line, then the perpendicular distance will be negative. |
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| Corollary: |
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The length of the perpendicular drawn from the origin to the
line ax + by + c = 0 is given by  |
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