Q54. - Algebra 2

Question

54. Choose the equation of the line through $\left(\frac{\mathbf{1}}{\mathbf{2}},-\frac{\mathbf{3}}{\mathbf{2}}\right)$ and $\left(-\frac{\mathbf{1}}{\mathbf{2}},\frac{\mathbf{1}}{\mathbf{2}}\right)$

1. $\mathbf{y}=-\mathbf{2}\mathbf{x}-\frac{\mathbf{1}}{\mathbf{2}}$        (B) $\mathbf{y}=-\mathbf{3}\mathbf{x}$ (C) $\mathbf{y}=\mathbf{2}\mathbf{x}-\frac{\mathbf{5}}{\mathbf{2}}$                (D) $\mathbf{y}=\frac{\mathbf{1}}{\mathbf{2}}\mathbf{x}+\mathbf{1}$

Step-by-Step Solution

Answer

The equation of the required straight line is $y=-2x-\frac{1}{2}$. So, (A) is the correct option.

Step 1 – State the concept

The slope intercept form of a straight-line equation is $y=mx+c$ where $m$ is the slope and $c$ is the y-intercept.

The equation of a straight-line passing through the points $\left(h,k\right)$ and $\left(a,b\right)$ is given as $\frac{y-k}{x-h}=\frac{b-k}{a-h}$.

Step 2 – List the given data

It is given that the line passes through $\left(\frac{1}{2},-\frac{3}{2}\right)$ and $\left(-\frac{1}{2},\frac{1}{2}\right)$.

Then, $\left(h,k\right)=\left(\frac{1}{2},-\frac{3}{2}\right)$ and $\left(a,b\right)=\left(-\frac{1}{2},\frac{1}{2}\right)$.

Step 3 – Write the equation

Put $\left(h,k\right)=\left(\frac{1}{2},-\frac{3}{2}\right)$ and $\left(a,b\right)=\left(-\frac{1}{2},\frac{1}{2}\right)$ in $\frac{y-k}{x-h}=\frac{b-k}{a-h}$ to get,

 

$\frac{y-\left(-\frac{3}{2}\right)}{x-\frac{1}{2}}=\frac{\frac{1}{2}-\left(-\frac{3}{2}\right)}{-\frac{1}{2}-\frac{1}{2}}$

 

$\frac{y+\frac{3}{2}}{x-\frac{1}{2}}=\frac{\frac{1}{2}+\frac{3}{2}}{-1}$  (Simplify)

 

$\frac{y+\frac{3}{2}}{x-\frac{1}{2}}=\frac{2}{-1}$  (Simplify)

 

$\frac{y+\frac{3}{2}}{x-\frac{1}{2}}=-2$  (Simplify)

 

$\frac{y+\frac{3}{2}}{x-\frac{1}{2}}\left(x-\frac{1}{2}\right)=-2\left(x-\frac{1}{2}\right)$  (Multiply both sides by $x-\frac{1}{2}$)

 

$y+\frac{3}{2}=-2\left(x-\frac{1}{2}\right)$  (Simplify)

 

$y+\frac{3}{2}=-2x+1$  (Distributive property)

 

$y+\frac{3}{2}-\frac{3}{2}=-2x+1-\frac{3}{2}$  (Subtract $\frac{3}{2}$ from both sides)

 

$y=-2x-\frac{1}{2}$  (Simplify)

 

So, the required equation of the straight line is $y=-2x-\frac{1}{2}$.