Concurrency of three lines Questions in English

(B) Step $1$: Find the point of intersection of $x - y + 1 = 0$ $(i)$ and $2x + 3y - 8 = 0$ (ii).
From $(i)$, $y = x + 1$. Substitute this into (ii): $2x + 3(x + 1) - 8 = 0 \implies 2x + 3x + 3 - 8 = 0 \implies 5x = 5 \implies x = 1$.
Then $y = 1 + 1 = 2$. The point of intersection is $(1, 2)$.
Step $2$: The line has an $x$-intercept of $3$, so it passes through $(3, 0)$.
Step $3$: Find the equation of the line passing through $(1, 2)$ and $(3, 0)$.
The slope $m = \frac{0 - 2}{3 - 1} = \frac{-2}{2} = -1$.
Using point-slope form $y - y_1 = m(x - x_1)$: $y - 0 = -1(x - 3) \implies y = -x + 3 \implies x + y - 3 = 0$.

(D) The given equation is $(2 + k)x + (1 + k)y = 5 + 7k$.
Rearrange the equation to group terms with $k$:
$2x + kx + y + ky = 5 + 7k$
$(2x + y - 5) + k(x + y - 7) = 0$
For this line to pass through a fixed point for all values of $k$, both expressions in the parentheses must be zero:
$2x + y - 5 = 0$ ---$(1)$
$x + y - 7 = 0$ ---$(2)$
Subtracting $(2)$ from $(1)$:
$(2x + y - 5) - (x + y - 7) = 0$
$x + 2 = 0 \implies x = -2$
Substitute $x = -2$ into $(2)$:
$-2 + y - 7 = 0 \implies y = 9$
The fixed point is $(-2, 9)$.
The distance $d$ from the origin $(0, 0)$ is $\sqrt{(-2)^2 + 9^2} = \sqrt{4 + 81} = \sqrt{85}$.
Therefore, $d^2 = 85$.

(C) Given the equation of the family of lines: $4ax + 3by + c = 0$ $(1)$.
Given the condition: $a + b + c = 0$, which implies $c = -a - b$ $(2)$.
Substitute $(2)$ into $(1)$:
$4ax + 3by - a - b = 0$
Rearrange the terms to group $a$ and $b$:
$a(4x - 1) + b(3y - 1) = 0$
For this equation to hold for all real constants $a$ and $b$, the coefficients of $a$ and $b$ must be zero independently:
$4x - 1 = 0 \implies x = \frac{1}{4}$
$3y - 1 = 0 \implies y = \frac{1}{3}$
Thus, the lines are concurrent at the point $(\frac{1}{4}, \frac{1}{3})$.