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Find the equation of the set of points which are equidistant from the points (1, 2 , 3) and (3, 2, -1)
a) x + 2z = 0
b) y + 2z = 0
c) x – 2z = 0
d) x – 2y = 0
Correct answer is option 'C'. Can you explain this answer?
a) x + 2z = 0
b) y + 2z = 0
c) x – 2z = 0
d) x – 2y = 0
Correct answer is option 'C'. Can you explain this answer?
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Find the equation of the set of points which are equidistant from the ...
Pt. A(1, 2 , 3)
Pt. B(3, 2, -1)
Let P(x,y,z)
So, AP = BP
((x-1)2 + (y-2)2 + (z-3)2)1/2 = ((x-3)2 + (y-2)2 + (z+1)2)1/2
(x-1)2 + (y-2)2 + (z-3)2) = (x-3)2 + (y-2)2 + (z+1)2
x2 +1 -2x + y2 + 4 - 4y + z2 + 9 – 6z = x2 +9 -6x + y2 + 4 - 4y + z2 + 1 + 2z
4x – 8z = 0
x – 2z = 0
Pt. B(3, 2, -1)
Let P(x,y,z)
So, AP = BP
((x-1)2 + (y-2)2 + (z-3)2)1/2 = ((x-3)2 + (y-2)2 + (z+1)2)1/2
(x-1)2 + (y-2)2 + (z-3)2) = (x-3)2 + (y-2)2 + (z+1)2
x2 +1 -2x + y2 + 4 - 4y + z2 + 9 – 6z = x2 +9 -6x + y2 + 4 - 4y + z2 + 1 + 2z
4x – 8z = 0
x – 2z = 0
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Find the equation of the set of points which are equidistant from the ...
Equation of set of points equidistant from two given points is the equation of the perpendicular bisector of the line segment joining these two points. Hence, we need to find the midpoint of the line segment joining (1, 2, 3) and (3, 2, -1) and then find the equation of the plane perpendicular to this line passing through this midpoint.
Midpoint of the line joining (1, 2, 3) and (3, 2, -1) is ((1+3)/2, (2+2)/2, (3-1)/2) = (2, 2, 1).
Let the equation of the required plane be ax + by + cz + d = 0. Since the plane is perpendicular to the line joining the given points, the direction ratios of the normal to the plane are the direction ratios of the line joining the given points, which are (3-1, 2-2, -1-3) = (2, 0, -4).
Hence, the equation of the plane is 2(x-2) + 0(y-2) - 4(z-1) = 0, which simplifies to x - 2z = 0.
Hence, option C is the correct answer.
Midpoint of the line joining (1, 2, 3) and (3, 2, -1) is ((1+3)/2, (2+2)/2, (3-1)/2) = (2, 2, 1).
Let the equation of the required plane be ax + by + cz + d = 0. Since the plane is perpendicular to the line joining the given points, the direction ratios of the normal to the plane are the direction ratios of the line joining the given points, which are (3-1, 2-2, -1-3) = (2, 0, -4).
Hence, the equation of the plane is 2(x-2) + 0(y-2) - 4(z-1) = 0, which simplifies to x - 2z = 0.
Hence, option C is the correct answer.
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Find the equation of the set of points which are equidistant from the points (1, 2 , 3) and (3, 2, -1)a)x + 2z = 0b)y + 2z = 0c)x – 2z = 0d)x – 2y = 0Correct answer is option 'C'. Can you explain this answer?
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