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) .

Similar Problems and Solutions

ISI MStat 2019 PSA Problem 14 — Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

Related

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) .

Similar Problems and Solutions

ISI MStat 2019 PSA Problem 14 — Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

Related

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