Let L denote the line in the xy -plane with x and y intercepts as 3...

Equation of line is

$\dfrac{x}{3} + \dfrac{y}{1} = 1 ⇒ x + 3y - 3 = 0$

if image is\left({x}_{1,}{y}_{1}\right)$</p><p>Then,</p><p>$\dfrac{{x}_{1} + 1}{1} = \dfrac{{y}_{1} + 4}{3} = - 2\left(\dfrac{- 1 - 12 - 3}{10}\right)$</p><p>${x}_{1} + 1 = \dfrac{{y}_{1} + 4}{3} = \dfrac{16}{5}$</p><p>$⇒ {x}_{1} = \dfrac{11}{5}, {y}_{1} = \dfrac{28}{5}