1. We are asked to determine if the given statements are TRUE or FALSE and justify the answers. (i) Distance between points $(3x, -3x)$ and $(-x, -6x)$ is $-5x$ if $x<0$. Distance formula: $$d = \sqrt{(3x - (-x))^2 + (-3x - (-6x))^2} = \sqrt{(4x)^2 + (3x)^2} = \sqrt{16x^2 + 9x^2} = \sqrt{25x^2} = 5|x|$$ If $x<0$, then $|x| = -x$. So, $$d = 5(-x) = -5x$$ **Answer: TRUE** because the distance simplifies to $-5x$ for $x<0$. (ii) Graph of $x = \sqrt{y - 1}$: Check intercepts. -x intercept: Set $y=0$, then $x = \sqrt{0 - 1} = \sqrt{-1}$ which is not real. -y intercept: Set $x=0$, then $0 = \sqrt{y - 1} \implies y - 1 = 0 \implies y = 1$ So $y$ intercept is at $(0, 1)$. **Answer: FALSE** because there is no $x$-intercept but there is a $y$-intercept. (iii) Midpoint $M(x, y)$ between $(3, 4)$ and $(k, 6)$: $$x = \frac{3 + k}{2}, \quad y = \frac{4 + 6}{2} = 5$$ Given $x + y = 1$, substitute: $$\frac{3 + k}{2} + 5 = 1 \implies \frac{3 + k}{2} = 1 - 5 = -4 \implies 3 + k = -8 \implies k = -11$$ Given statement says $k=11$. **Answer: FALSE** because $k$ equals $-11$, not $11$. (iv) Graph $y = -\sqrt{x + 1}$. Find intercepts: -x intercept set $y=0$: $$0 = -\sqrt{x + 1} \implies \sqrt{x + 1} = 0 \implies x + 1 = 0 \implies x = -1$$ So $x$ intercept is $(-1,0)$. -y intercept set $x=0$: $$y = -\sqrt{0 + 1} = -1$$ So $y$ intercept is $(0,-1)$. Midpoint $(m,n)$ of these: $$m = \frac{-1 + 0}{2} = -\frac{1}{2}, \quad n = \frac{0 + (-1)}{2} = -\frac{1}{2}$$ Sum: $$m + n = -\frac{1}{2} - \frac{1}{2} = -1 \neq 0$$ **Answer: FALSE** because $m+n = -1$, not $0$. Final answers: (i) TRUE (ii) FALSE (iii) FALSE (iv) FALSE