How to Find Image of a Point About the Line :
Here we are going to see how to find image of a point about the line.
Question 1 :
Find the image of the point (−2, 3) about the line x + 2y − 9 = 0.
Solution :
Let us draw a perpendicular line from the point (-2, 3).
Now we have to find the equation of the line perpendicular to the line x + 2y - 9 = 0.
Let us find the line which is perpendicular to the line x + 2y - 9 = 0.
2x - y + k = 0
The perpendicular line 2x - y + k = 0 is passing through the point (-2, 3)
2(-2) - 3 + k = 0
-4 - 3 + k = 0
k = 7
2x - y + 7 = 0
By solving the above two equations, we get the point B.
x + 2y - 9 = 0 -----(1)
2x - y + 7 = 0 -----(2)
Multiply the first equation by 2, we get
2x + 4y - 18 = 0
(1) - (2)
(2x + 4y - 18) - (2x - y + 7) = 0
2x + 4y - 18 - 2x + y - 7 = 0
5y - 25 = 0
y = 5
By applying the value of y in the first equation, we get
x + 2(5) - 9 = 0
x + 10 - 9 = 0
x + 1 = 0
x = -1
B(-1, 5)
Mid point of AA' = (x_{1} + x_{2})/2, (y_{1} + y_{2})/2
(-2 + a) /2 , (3 + b)/2 = (-1, 5)
(-2 + a)/2 = -1 -2 + a = -2 a = -2 + 2 a = 0 |
(3 + b)/2 = 5 3 + b = 10 b = 10 - 3 b = 7 |
Hence the required point is (0, 7)
Question 2 :
A line is drawn perpendicular to 5x = y + 7. Find the equation of the line if the area of the triangle formed by this line with co-ordinate axes is 10 sq. units.
Solution :
5x - y - 7 = 0
The line which is perpendicular to the above line is 1x + 5y + k = 0
By applying A(x, 0), we get the value of x in terms of k.
1x + 5(0) + k = 0
x + k = 0
x = -k
By applying B(0, y), we get the value of y in terms of k.
1(0) + 5y + k = 0
5y + k = 0
y = -k/5
here, x = base of triangle, y = height of triangle
Area of triangle = 10 square units
(1/2) ⋅ Base ⋅ Height = 10
(1/2) ⋅ (-k) ⋅ (-k/5) = 10
k^{2} = 100 ==> k = ± 10
x + 5y = ± 10
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