**Answer:**

**(1.931, 5.172)**

**Step-by-step explanation:**

**Given that the image of A(2,3) is A'(4,8).**

**Now, the equation of straight line passing through A and A' is **

⇒(y-3) =2/5(x-2)

⇒y =2x/5 +(3-4/5)

⇒ y**= 2x/5 +11/5 ......(1)**

**And the mid point of A and A' is **()≡**(3,5.5)**

Now, **Equation of perpendicular straight line to (1), is given by **

**y= -5x/2 +c ....... (2) {Since, the product of slopes of two mutually perpendicular straight line is always -1}**

Now, equation (2) passes through (3, 5.5) point.

Hence, 5.5 =-5×3/2 +c, ⇒ **c=13.**

So, **the mirror straight line is y=-5x/2 +13 ......(3)**

Now, **parallel straight line to (1), is given by**

**y=2x/5 +c' and the point (4,6) satisfies it.**

So, 6=2×4/5 +c', ⇒ **c' =22/5**.

So, the equation becomes** y=2x/5 +22/5 ........ (4) **

Now, **solving equations (3) and (4) we get **

-5x/2 +13=2x/5 +22/5

⇒ 29x/10 =43/5

⇒ x=86/29=2.965 and from (4), we get y=2/5(86/29) +22/5 =172/145+22/5 =5.586

**Hence, the point becomes (2.965, 5.586)**

**If (h,k) be the required image, then (2.965, 5.586) point must be the mid point of (h,k) and (4,6).**

**Therefore, (h+4)/2 =2.965 ⇒ h= 1.931 and (k+6)/2 =5.586, ⇒k=5.172.**

**So, the image point is (1.931, 5.172). (Answer)**