Question 6 i) The point $$ R $$, with coordinates $$ \left(-\frac{1}{3}, k\right) $$ is on the line with equation $$ 5 x+2 y=6 $$. Find the value of $$ k $$. ii) The point $$ S $$, with coordinates $$ \left(w,-\frac{3}{5}\right) $$ is on the line with equation $$ 5 x+2 y=6 $$. Find the value of $$ w $$. iii) Show that the lines $$ 5 x+2 y=6 $$ and $$ 10 x+4 y-7=0 $$ are parallel. iv) Show that the lines $$ 5 x+2 y=6 $$ and $$ 8 x-20 y+11=0 $$ are perpendicular.

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UpStudy AI · Accepted Answer

**i)**  
The point $$ R\left(-\frac{1}{3}, k\right) $$ satisfies the equation  
$$
5\left(-\frac{1}{3}\right) + 2k = 6.
$$  
Simplify:  
$$
-\frac{5}{3} + 2k = 6.
$$  
Add $$\frac{5}{3}$$ to both sides:  
$$
2k = 6 + \frac{5}{3} = \frac{18}{3}+\frac{5}{3} = \frac{23}{3}.
$$  
Divide both sides by 2:  
$$
k = \frac{23}{6}.
$$

---

**ii)**  
The point $$ S\left(w, -\frac{3}{5}\right) $$ lies on the line, so substitute into the equation  
$$
5w + 2\left(-\frac{3}{5}\right) = 6.
$$  
This becomes:  
$$
5w - \frac{6}{5} = 6.
$$  
Multiply every term by 5 to clear the fraction:  
$$
25w - 6 = 30.
$$  
Add 6 to both sides:  
$$
25w = 36,
$$  
so  
$$
w = \frac{36}{25}.
$$

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**iii)**  
Consider the two lines:  
$$
5x + 2y = 6 \quad \text{and} \quad 10x + 4y - 7 = 0.
$$  
Rewrite the first equation in slope-intercept form:  
$$
2y = -5x + 6 \quad \Longrightarrow \quad y = -\frac{5}{2}x + 3.
$$  
The slope is $$ m_1 = -\frac{5}{2} $$.

For the second equation, rewrite as:  
$$
4y = -10x + 7 \quad \Longrightarrow \quad y = -\frac{10}{4}x + \frac{7}{4} = -\frac{5}{2}x + \frac{7}{4}.
$$  
The slope is $$ m_2 = -\frac{5}{2} $$.

Since $$ m_1 = m_2 $$, the lines are parallel.

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**iv)**  
Examine the lines:  
$$
5x + 2y = 6 \quad \text{and} \quad 8x - 20y + 11 = 0.
$$  
For the first line, as before, the slope is  
$$
m_1 = -\frac{5}{2}.
$$

For the second line, solve for $$ y $$:  
$$
8x - 20y + 11 = 0 \quad \Longrightarrow \quad -20y = -8x - 11,
$$  
$$
y = \frac{8}{20}x + \frac{11}{20} = \frac{2}{5}x + \frac{11}{20}.
$$  
Thus, the slope is $$ m_2 = \frac{2}{5} $$.

Check for perpendicularity by multiplying the slopes:  
$$
m_1 \cdot m_2 = \left(-\frac{5}{2}\right) \cdot \frac{2}{5} = -1.
$$  
Since the product is $$-1$$, the lines are perpendicular.
**i)** $$ k = \frac{23}{6} $$ **ii)** $$ w = \frac{36}{25} $$ **iii)** The lines are parallel because they have the same slope. **iv)** The lines are perpendicular because the product of their slopes is $$-1$$.

Question 6
i) The point , with coordinates is on the line with equation . Find
the value of .
ii) The point , with coordinates is on the line with equation . Find
the value of .
iii) Show that the lines and are parallel.
iv) Show that the lines and are perpendicular.

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Question 6 i) The point , with coordinates is on the line with equation . Find the value of . ii) The point , with coordinates is on the line with equation . Find the value of . iii) Show that the lines and are parallel. iv) Show that the lines and are perpendicular.