Question

for 3y - 6x = 2:
① find slope of a parallel line.
② give the slope of a ⊥ line.
③ give the slope of a line containing a(-3,4), b(1,-6)
④

x	f(x)
0	3
1	7
2	11
3	15

find.
f(x)=
give any g(x) || f(x).
g(x)=
⑤ write the proof of the triangle angle sum t.
⑥ find an equation || to y = \frac{3}{5}x + 4, through (-5,2)
⑦
⑧ the product of the slopes of 2 ⊥ non - vertical lines is
⑨ give the equation of:
⑩ x axis
⑪ a line || to x axis through (4,3)
⑫ y axis
⑬ a line || to y axis through (4,-3)

Explanation:

Step1: Rewrite $3y - 6x=2$ in slope - intercept form

First, solve $3y - 6x = 2$ for $y$. Add $6x$ to both sides: $3y=6x + 2$. Then divide by 3: $y = 2x+\frac{2}{3}$. The slope of this line is $m = 2$.

Step2: Find slope of parallel line

Parallel lines have the same slope. So the slope of a line parallel to $3y - 6x = 2$ is $m_{parallel}=2$.

Step3: Find slope of perpendicular line

The product of the slopes of two perpendicular non - vertical lines is $- 1$. Let the slope of the perpendicular line be $m_{perp}$. Since $m = 2$, then $2m_{perp}=-1$, so $m_{perp}=-\frac{1}{2}$.

Step4: Find slope of line through $A(-3,4)$ and $B(1,-6)$

Use the slope formula $m=\frac{y_2 - y_1}{x_2 - x_1}$. Here, $x_1=-3,y_1 = 4,x_2 = 1,y_2=-6$. Then $m=\frac{-6 - 4}{1-(-3)}=\frac{-10}{4}=-\frac{5}{2}$.

Step5: Find the linear function for the table

The table shows a linear relationship. The slope $m=\frac{f(1)-f(0)}{1 - 0}=\frac{7 - 3}{1}=4$. Using the point - slope form with the point $(0,3)$ (since when $x = 0,f(0)=3$), the function is $f(x)=4x + 3$. A parallel function $g(x)$ has the same slope, so $g(x)=4x + k$ (where $k
eq3$), for example $g(x)=4x+5$.

Step6: Proof of Triangle Angle Sum Theorem

Draw a triangle $\triangle ABC$. Through point $A$, draw a line $l$ parallel to $BC$. By the alternate interior angles theorem, $\angle B=\angle BAX$ and $\angle C=\angle CAY$ (where $X$ and $Y$ are points on $l$). Since $\angle BAX+\angle BAC+\angle CAY = 180^{\circ}$, then $\angle A+\angle B+\angle C=180^{\circ}$.

Step7: Find parallel and perpendicular lines to $y=\frac{3}{5}x + 4$ through $(-5,2)$

For a parallel line, it has the same slope $m=\frac{3}{5}$. Using the point - slope form $y - y_1=m(x - x_1)$, we have $y - 2=\frac{3}{5}(x + 5)$, which simplifies to $y=\frac{3}{5}x+5$. For a perpendicular line, the slope $m_{perp}=-\frac{5}{3}$. Using the point - slope form, $y - 2=-\frac{5}{3}(x + 5)$, which simplifies to $y=-\frac{5}{3}x-\frac{19}{3}$.

Step8: Properties of axes and parallel lines

The equation of the $x$ - axis is $y = 0$. A line parallel to the $x$ - axis through $(4,3)$ has the equation $y = 3$. The equation of the $y$ - axis is $x = 0$. A line parallel to the $y$ - axis through $(4,-3)$ has the equation $x = 4$.

Answer:

1. Slope of parallel line to $3y - 6x = 2$: $2$
2. Slope of perpendicular line to $3y - 6x = 2$: $-\frac{1}{2}$
3. Slope of line through $A(-3,4)$ and $B(1,-6)$: $-\frac{5}{2}$
4. $f(x)=4x + 3$, $g(x)=4x + 5$ (for example)
5. Proof of Triangle Angle Sum Theorem: See above explanation
6. Parallel line to $y=\frac{3}{5}x + 4$ through $(-5,2)$: $y=\frac{3}{5}x+5$
7. Perpendicular line to $y=\frac{3}{5}x + 4$ through $(-5,2)$: $y=-\frac{5}{3}x-\frac{19}{3}$
8. Equation of $x$ - axis: $y = 0$
9. Line parallel to $x$ - axis through $(4,3)$: $y = 3$
10. Equation of $y$ - axis: $x = 0$
11. Line parallel to $y$ - axis through $(4,-3)$: $x = 4$