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June 8, 2020

ISI MStat 2019 PSA Problem 14 | Reflection of a point

This is a problem from ISI MStat 2019 PSA Problem 14. First, try the problem yourself, then go through the sequential hints we provide.

Reflection of a point - ISI MStat Year 2019 PSA Question 14


The reflection of the point (1,2) with respect to the line \(x+2 y=15\) is

  • (3,6)
  • (6,3)
  • (10,5)
  • (5,10)

Key Concepts


Straight line

Check the Answer


Answer: is (5,10)

ISI MStat 2019 PSA Problem 14

Precollege Mathematics

Try with Hints


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.

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.

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) .

ISI MStat 2019 PSA Problem 14
Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

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