Q. 53.7( 7 Votes )

Find the image of

Answer :

Let line AB be x – 3y + 4 = 0 and point P be (1, 2)

Let the image of the point P(1, 2) in the line mirror AB be Q(h, k).



Since line AB is a mirror. Then PQ is perpendicularly bisected at R.


Since R is the midpoint of PQ.


We know that,


Midpoint of a line joining (x1, y1) & (x2, y2)


So, Midpoint of the line joining (1, 2) & (h, k)


Since point R lies on the line AB. So, it will satisfy the equation of line AB x – 3y + 4 = 0


Substituting the in abthe ove equation, we get




3 + h – 3k = 0


h – 3k = -3 …(i)


Also, PQ is perpendicular to AB


We know that, if two lines are perpendicular then the product of their slope is equal to -1


Slope of AB × Slope of PQ = -1



Now, we find the slope of line AB i.e. x – 3y + 4 = 0


We know that, the slope of an equation is



and here, a = 1 & b = -3




= - 3


Now, Equation of line PQ formed by joining the points P(1, 2) and Q(h, k) and having the slope – 3 is


y2 – y1 = m(x2 – x1)


k – 2 = (-3)(h – 1)


k – 2 = -3h + 3


3h + k = 5 …(ii)


Now, we will solve the eq. (i) and (ii) to find the value of h and k


h – 3k = -3 …(i)


and 3h + k = 5 …(ii)


From eq. (i), we get


h = -3 + 3k


Putting the value of h in eq. (ii), we get


3(-3 + 3k) + k = 5


-9 + 9k + k = 5


-9 + 10k = 5


10k = 5 + 9


10k = 14



Putting the value of k in eq. (i), we get



5h – 21 = -3 × 5


5h – 21 = -15


5h = -15 + 21


5h = 6



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