Question

Please solve this and previous question.......

find the image of the point P(-1,2) in the line mirror 2x - 3y +4=0 ?

Answer 1

Hello mate,

This Question can be solved in 2 methods

▶To verify just find the mId-point of the line.

⭐The image of P(-1,2) about the line

➡2x - 3y +4=0 is

$\frac{x - 1}{2} = \frac{y - 2}{ - 3} = - 2 \frac{(2 \times ( - 1) - 3 \times 2 + 4)}{ {2}^{2} +( { - 3}^{2}) }$
By solving --

$\frac{x + 1}{2} = \frac{y - 2}{ - 3} = \frac{8}{13}$

➡Furthur equating and solving ,

Here we get the Equation

➡13x +13 = 13

➡x = 3/13

➡13y - 26 = -24

➡y = 2/13

•°• Now , these were the cordinates of the mirror image ,

$( \frac{3}{13} , \frac{2}{13} )$

Answer 2

Hello mate, 

This Question can be solved in 2 methods 

▶To verify just find the mId-point of the line.

⭐The image of P(-1,2) about the line 

➡2x - 3y +4=0 is 

$\dfrac{x - 1}{2} = \dfrac{y - 2}{ - 3} = - 2 \dfrac{(2 \times ( - 1) - 3 \times 2 + 4)}{ {2}^{2} +( { - 3}^{2}) }$

By solving --

$\dfrac{x + 1}{2} = \dfrac{y - 2}{ - 3} = \dfrac{8}{13}$

➡Furthur equating and solving ,

Here we get the Equation

➡13x +13 = 13

➡x = 3/13

➡13y - 26 = -24 

➡y = 2/13

•°• Now , these were the cordinates of the mirror image ,