$$x+3y-5=0$$

$\therefore 3t=5-x\\ \therefore y=\dfrac {-1}3x+\dfrac 53 ..... (1)\\ And \space \space , 4x+3y-8=0\\ \therefore 3y=-4x+8\\ \therefore y=\dfrac {-4x+8}3\\ \therefore y=\dfrac {-4x}3+\dfrac 83...... (2)\\ Let \space \space b_1=\dfrac {-1}3 \space \space and \space \space b_3=\dfrac {-4}3 \\ Since \space \space |b_1| \lt |b_2|, b_{yx}=b_1=\dfrac {-1}3 \space \&\space b_{xy}=\dfrac 1{b_2}=\dfrac {-3}4........(3)$

Hence, equation (1) is regression equation of Y on X type & equation (2) is regression equation of X on Y type.

$$\therefore r=\pm \sqrt{b_{yx}.b_{xy}}$$

$=\pm\sqrt{\dfrac {-1}3\times \dfrac {-3}4}\space \space \space from (3)\\ =\pm \dfrac {-1}2$

Since, $b_{yx}$ and $b_{xy}$ are both negative, ‘r’ is negative.

$$\therefore r=\dfrac {-1}2$$

$\therefore $ Given Statement is false.