Respuesta :
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
[tex]\frac{(y-3)}{(8-3)} =\frac{(x-2)}{(4-2)}[/tex]
⇒(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 ([tex]\frac{2+4}{2},\frac{3+8}{2}[/tex])≡(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)
