Q.If $P(r, c)$ is the mid point of a line segment between the axes, then show that $\dfrac{x}{r} + \dfrac{y}{c} = 2$.

Reflecting $(x,y)$ in the line $y=-x$ gives $(-y,-x)$, so $(2,3)\to(-3,-2)$.

The line $y=-x$ passes through the origin at $135^\circ$ to the x-axis. Reflection in this line swaps the coordinates and negates both: the general rule is $(x,y)\mapsto(-y,-x)$.

(This can be verified using the reflection formula for a line $x+y=0$: reflecting $(x_1,y_1)$ gives $\left(x_1-2\cdot\dfrac{x_1+y_1}{2},\,y_1-2\cdot\dfrac{x_1+y_1}{2}\right)=(-y_1,-x_1)$.)

Applying to $(2,3)$: the image is $(-3,-2)$.

Find the equation of the line through the two given points and read off the condition satisfied by any point on it.

Slope $=\dfrac{4-3}{-1-2}=\dfrac{1}{-3}=-\dfrac13$.

Equation: $y-3=-\dfrac13(x-2)\Rightarrow 3(y-3)=-(x-2)\Rightarrow 3y-9=-x+2\Rightarrow x+3y=11$.

So any point $(\alpha,\beta)$ on this line satisfies $\alpha+3\beta=11$.

Perpendicular slopes are negative reciprocals; use the point-slope form with the given point.

Rewrite $3x+4y=6$ as $y=-\dfrac34x+\dfrac64$, so its slope is $-\dfrac34$.

A line perpendicular to it has slope $\dfrac43$ (negative reciprocal).

Using point-slope form through $(1,-1)$:

$y-(-1)=\dfrac43(x-1)\Rightarrow 3(y+1)=4(x-1)\Rightarrow 3y+3=4x-4\Rightarrow 4x-3y-7=0.$

Find where the given line meets the $y$-axis, then draw the perpendicular there.

The given line $x-y+5=0$, i.e. $y=x+5$, has slope $1$. It meets the $y$-axis where $x=0$: $y=0+5=5$, so the point of intersection is $(0,5)$.

A line perpendicular to a line of slope $1$ has slope $-1$ (product of slopes of perpendicular lines is $-1$). The required line passes through $(0,5)$ with slope $-1$:

$y-5=-1(x-0)\ \Rightarrow\ y=-x+5\ \Rightarrow\ x+y-5=0.$

Writing $\dfrac{x}{a}-\dfrac{y}{b}=0$ as $y=\dfrac{b}{a}x$ shows the slope is $b/a$; setting this equal to 1 gives $a=b$.

$\dfrac{x}{a}-\dfrac{y}{b}=0 \Rightarrow \dfrac{y}{b}=\dfrac{x}{a} \Rightarrow y = \dfrac{b}{a}x$.

This is the equation of a line through the origin with slope $m=\dfrac{b}{a}$.

For the slope to equal 1: $\dfrac{b}{a}=1 \Rightarrow a=b$ (with $a,b\ne 0$).

The slope of the line through $(2,3)$ and $(-1,4)$ is $-1/3$; writing its equation and substituting $(\alpha,\beta)$ gives $\alpha+3\beta=11$.

Slope $m = \dfrac{4-3}{-1-2} = \dfrac{1}{-3} = -\dfrac{1}{3}$.

Using point-slope form through $(2,3)$: $y-3 = -\dfrac{1}{3}(x-2)$.

Multiply by 3: $3y-9 = -(x-2) = -x+2$, so $x+3y = 11$.

Any point $(\alpha,\beta)$ on this line therefore satisfies $\alpha+3\beta=11$.

Computing $y$-intercept $=c/b$ for each line, $x+2y=3$ gives the largest magnitude, $1.5$.

For $ax+by=c$, setting $x=0$ gives the $y$-intercept $y=c/b$.

1. $3x+4y=5 \Rightarrow y\text{-intercept} = 5/4 = 1.25$
2. $2x+3y=4 \Rightarrow y\text{-intercept} = 4/3 \approx 1.33$
3. $4x+5y=6 \Rightarrow y\text{-intercept} = 6/5 = 1.2$
4. $x+2y=3 \Rightarrow y\text{-intercept} = 3/2 = 1.5$ Comparing magnitudes $1.25, 1.33, 1.2, 1.5$, the largest is $1.5$, from option (d).