This is a problem from ISI MStat 2019 PSA Problem 14. First, try the problem yourself, then go through the sequential hints we provide.
The reflection of the point (1,2) with respect to the line \(x+2 y=15\) is
Straight line
But try the problem first...
Answer: is (5,10)
ISI MStat 2019 PSA Problem 14
Precollege Mathematics
First hint
Find an algorithm to find the reflection,
Find the line perpendicular to \( x+2 y=15\) through (1,2).
Find the point of intersection.
Use Midpoint Segment Result.
Second Hint
The line perpendicular to \( x+2 y=15\) is of the form \(-2x+y=k \) .Now it passes through (1,2) . So, \( -2+2=k \Rightarrow k=0 \)
Hence the line perpendicular to \( x+2 y=15\) through (1,2) is y=2x.
Now we will find point of intersection (Foot of Perpendicular )
(3,6) is the point of intersection i.e the foot of perpendicular.
Final Step
Use Mid-Point Formula (special case of Section formula) to get required point (Foot of perpendicular is mid-point of reflection and original point)
\( (3,6)=( \frac{x+1}{2} , \frac{y+2}{2} ) \) \( \Rightarrow x=5 , y=10 \)
Therefore the reflection point is (5,10) .
This is a problem from ISI MStat 2019 PSA Problem 14. First, try the problem yourself, then go through the sequential hints we provide.
The reflection of the point (1,2) with respect to the line \(x+2 y=15\) is
Straight line
But try the problem first...
Answer: is (5,10)
ISI MStat 2019 PSA Problem 14
Precollege Mathematics
First hint
Find an algorithm to find the reflection,
Find the line perpendicular to \( x+2 y=15\) through (1,2).
Find the point of intersection.
Use Midpoint Segment Result.
Second Hint
The line perpendicular to \( x+2 y=15\) is of the form \(-2x+y=k \) .Now it passes through (1,2) . So, \( -2+2=k \Rightarrow k=0 \)
Hence the line perpendicular to \( x+2 y=15\) through (1,2) is y=2x.
Now we will find point of intersection (Foot of Perpendicular )
(3,6) is the point of intersection i.e the foot of perpendicular.
Final Step
Use Mid-Point Formula (special case of Section formula) to get required point (Foot of perpendicular is mid-point of reflection and original point)
\( (3,6)=( \frac{x+1}{2} , \frac{y+2}{2} ) \) \( \Rightarrow x=5 , y=10 \)
Therefore the reflection point is (5,10) .
