Question

If the two lines of regression are $4x+3y+7=0$ and $3x+4y+8=0$, then the means of $x$ and $y$ are

• A. $-\frac{4}{7},-\frac{11}{7}$
• B. $-\frac{4}{7},\frac{11}{7}$
• C. $\frac{4}{7},-\frac{11}{7}$
• D. $4,7$

Answer 1

The correct option is A

$-\frac{4}{7},-\frac{11}{7}$

Explanation for the correct option.

Step 1. Concept used:

When two lines of regression under the variables $x$ and $y$ intersect they intersect at the point $(\overline{x},\overline{y})$.

So the $\mathrm{x}$ coordinate of the point of intersection gives the mean of $x$ and the $y$ coordinate of the point of intersection gives the mean of $y$.

Step 2. Eliminate $\mathrm{x}$ and find the value of $y$.

Multiply the equation $4x+3y+7=0$ by $3$ on both sides.

$$\begin{gathered} 3\times \left(4x+3y+7\right)=3\times 0 \\ \Rightarrow 12x+9y+21=0...\left(1\right) \end{gathered}$$

Multiply the equation $3x+4y+8=0$ by $4$ on both sides.

$\begin{array}{rcl}4\times \left(3x+4y+8\right)& =& 4\times 0\\ & \Rightarrow & 12x+16y+32=0...\left(2\right)\end{array}$

Now subtract equation $1$ from equation $2$:

$$\begin{gathered} 12x+16y+32-(12x+9y+21)=0 \\ \Rightarrow 12x+16y+32-12x-9y-21=0 \\ \Rightarrow 7y+11=0 \\ \Rightarrow 7y=-11 \\ \Rightarrow y=-\frac{11}{7} \end{gathered}$$

Step 3. Find the value of $\mathrm{x}$

Substitute $y=-\frac{11}{7}$ in the equation $4x+3y+7=0$.

$$\begin{gathered} 4x+3\times \left(-\frac{11}{7}\right)+7=0 \\ \Rightarrow 4x=\frac{33}{7}-7 \\ \Rightarrow 4x=\frac{33-49}{7} \\ \Rightarrow 4x=-\frac{16}{7} \\ \Rightarrow x=-\frac{4}{7} \end{gathered}$$

So, the mean of $x$ is $-\frac{4}{7}$ and the mean of $y$ is $-\frac{11}{7}$.

Hence, the correct option is A.