Assuming that a straight line works as a plane mirror for a point. Find the image of the point (1,2) in the line x-3y+4=0.
Answers
Given : A straight line works as a plane mirror for a point.
To Find : the image of the point (1,2) in the line x-3y+4=0.
Solution:
Let say point P ( a , b) is the image of point ( 1, 2) in the line x-3y+4=0.
Then mid point of ( a, b) and ( 1, 2) will lie on x-3y+4=0.
=> (a + 1)/2 , (b + 2)/2 will lie on x-3y+4=0.
=> (a + 1)/2 - 3 (b + 2)/2 + 4 = 0
=> a + 1 - 3b - 6 + 8 = 0
=> a - 3b = -3
Also line between ( a, b) and ( 1, 2) will be perpendicular to x-3y+4=0
Slope between ( a, b) and ( 1, 2) = (b - 2)/(a-1)
Slope of x-3y+4=0 => 3y = x + 4 => y = (1/3)x + 4/3 slope = 1/3
(b - 2)/(a-1) * (1/3) = - 1
=> b - 2 = -3(a - 1)
=> b - 2 = -3a + 3
=> 3a + b = 5
a - 3b = -3 Eq1
3a + b = 5 Eq2
Eq2 - 3 *Eq1
10b = 14 => b = 1.4
=> 3a + 1.4 = 5
=> 3a = 3.6
=> a = 1.2
Point is ( 1.2 , 1.4)
image of the point (1,2) in the line x-3y+4=0. is ( 1.2 , 1.4)
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Answer:
what is the equation of the line passing through p and parallel to x-3y+4=0