The perpendicular distance from a point (x1, y1) to a line Ax + By + C = 0 is given by the formula:
d = |Ax1 + By1 + C| / sqrt(A^2 + B^2)
Given that the point (-4,2) lies on the line and the perpendicular distance from the origin is 2 units, we have:
d = 2
(-4A + 2B + C) / sqrt(A^2 + B^2) = 2
To find the equations of the lines passing through (-4,2) and having a perpendicular distance of 2 units from the origin, we can substitute (-4,2) into the equation of the line equation.
1) Let C = 0
(-4A + 2B)/sqrt(A^2 + B^2) = 2
Square both sides:
(-4A + 2B)^2 = 4(A^2 + B^2)
16A^2 - 16AB + 4B^2 = 4A^2 + 4B^2
12A^2 - 16AB = 0
A(3A - 4B) = 0
A = 0 or 3A - 4B = 0
If A = 0, then the line is of the form Bx + C = 0
Assuming B = 1, C = -5
So the line equation is x - 5 = 0
2) If 3A - 4B = 0
B = 3/4A
Substitute this into the distance formula relation:
(-4A + 2(3/4)A) / sqrt(A^2 + (3/4A)^2) = 2
(-4A + 3A) / sqrt(A^2 + 9/16 A^2) = 2
(15A)/sqrt((16 + 9)A^2 / 16) = 2
(15A)(4 / 5A) = 2
12 = 2
Which is not possible.
Therefore, the only line passing through (-4,2) and 2 units from the origin is x - 5 = 0.
Find the equations of two lines passing through the point (-4,2) whose perpendicular distance from the origin is 2 units
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