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Can somebody please explain me example 2 ?

Hi!
 
Let us imagine that wall AB acts as a mirror. Then P’ is the position of the image of point P. The object and its mirror image are at the same distance from the mirror. Therefore, points P and P′ are at the same distance from the wall.
 
 
 
The shortest distance between two points is the straight line joining the two points.
Therefore, the shortest distance between P’ and Q is P’Q. Let us join the points P’ and Q by a straight line which passes through the wall at point R.
Now, P’Q = P’R + RQ
But PR is the mirror reflection of P’R.
Or we can say that PR = P’R. This can be proved using the concept of congruence of two triangles. You will study this concept in the next class.
Therefore, P’Q = PR + RQ
Thus, the path from P to R and then from R to Q is the shortest path which should be followed by Sanjana. Here R is the point of intersection of P’ (the image of point P considering the mirror line as the surface of the wall) and the surface of the wall.
 
Hope! This will help you.
Cheers!

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 Let us imagine that wall AB acts as a mirror. Then P’ is the position of the image of point P. The object and its mirror image are at the same distance from the mirror. Therefore, points P and P′ are at the same distance from the wall.

 
 
 
The shortest distance between two points is the straight line joining the two points.
Therefore, the shortest distance between P’ and Q is P’Q. Let us join the points P’ and Q by a straight line which passes through the wall at point R.
Now, P’Q = P’R + RQ
But PR is the mirror reflection of P’R.
Or we can say that PR = P’R. This can be proved using the concept of congruence of two triangles. You will study this concept in the next class.
Therefore, P’Q = PR + RQ
Thus, the path from P to R and then from R to Q is the shortest path which should be followed by Sanjana. Here R is the point of intersection of P’ (the image of point P considering the mirror line as the surface of the wall) and the surface of the wall.
 
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 Hi!

 
Let us imagine that wall AB acts as a mirror. Then P’ is the position of the image of point P. The object and its mirror image are at the same distance from the mirror. Therefore, points P and P′ are at the same distance from the wall.
 
 
 
The shortest distance between two points is the straight line joining the two points.
Therefore, the shortest distance between P’ and Q is P’Q. Let us join the points P’ and Q by a straight line which passes through the wall at point R.
Now, P’Q = P’R + RQ
But PR is the mirror reflection of P’R.
Or we can say that PR = P’R. This can be proved using the concept of congruence of two triangles. You will study this concept in the next class.
Therefore, P’Q = PR + RQ
Thus, the path from P to R and then from R to Q is the shortest path which should be followed by Sanjana. Here R is the point of intersection of P’ (the image of point P considering the mirror line as the surface of the wall) and the surface of the wall.
 
Hope! This will help you.
Cheers!
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Example 2:

The given figure shows a wall with end points A and B. Sanjana is standing at

position P. She has to come to position Q after touching the wall. Find the

shortest path for Sanjana to come from P to Q.

Solution:

Let us imagine that wall AB acts as a mirror. Then P′ is the position of the image of point P. The object and its mirror image are at the same distance from the mirror. Therefore, points P and P′ are at the same distance from the wall.

The shortest distance between two points is the straight line joining the two points.

Therefore, the shortest distance between P′ and Q is P′Q. Let us join the points P′ and Q by a straight line which passes through the wall at point R.

Now, P′Q = P′R + RQ

But PR is the mirror reflection of P`R.

Or we can say that PR = P′R

Therefore, P′Q = PR + RQ

The path from P to R and then from R to Q is the shortest path which should be followed by Sanjana.

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