Line [tex]$JK$[/tex] passes through points [tex]$J(-3, 11)$[/tex] and [tex]$K(1, -3)$[/tex]. What is the equation of line [tex]$JK$[/tex] in standard form?

A. [tex]$7x + 2y = -1$[/tex]
B. [tex]$7x + 2y = 1$[/tex]
C. [tex]$14x + 4y = -1$[/tex]
D. [tex]$14x + 4y = 1$[/tex]



Answer :

To determine the equation of the line passing through the points [tex]\( J(-3, 11) \)[/tex] and [tex]\( K(1, -3) \)[/tex] in standard form, we will go through the following steps:

### Step 1: Calculate the Slope of the Line

The formula for the slope ([tex]\(m\)[/tex]) between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is given by:
[tex]\[ m = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]

Substitute the coordinates of points [tex]\( J \)[/tex] and [tex]\( K \)[/tex]:
[tex]\[ m = \frac{-3 - 11}{1 - (-3)} = \frac{-14}{4} = -3.5 \][/tex]

So, the slope of the line is [tex]\( m = -3.5 \)[/tex].

### Step 2: Calculate the Y-Intercept

The slope-intercept form of a line is [tex]\( y = mx + b \)[/tex], where [tex]\( m \)[/tex] is the slope and [tex]\( b \)[/tex] is the y-intercept. We can use the coordinates of point [tex]\( J(-3, 11) \)[/tex] to find [tex]\( b \)[/tex]:
[tex]\[ 11 = -3.5(-3) + b \][/tex]
[tex]\[ 11 = 10.5 + b \][/tex]
[tex]\[ b = 11 - 10.5 = 0.5 \][/tex]

So, the y-intercept is [tex]\( b = 0.5 \)[/tex], and the equation of the line in slope-intercept form is:
[tex]\[ y = -3.5x + 0.5 \][/tex]

### Step 3: Convert to Standard Form

The standard form of a line is [tex]\( Ax + By = C \)[/tex]. To convert from the slope-intercept form, we rearrange the equation:
[tex]\[ y = -3.5x + 0.5 \][/tex]
[tex]\[ 3.5x + y = 0.5 \][/tex]

To eliminate the decimal, we multiply every term by 2:
[tex]\[ 2(3.5x) + 2(y) = 2(0.5) \][/tex]
[tex]\[ 7x + 2y = 1 \][/tex]

### Step 4: Compare with Given Choices

Now we compare this standard form equation with the given choices:
1. [tex]\( 7x + 2y = -1 \)[/tex]
2. [tex]\( 7x + 2y = 1 \)[/tex]
3. [tex]\( 14x + 4y = -1 \)[/tex]
4. [tex]\( 14x + 4y = 1 \)[/tex]

We see that the equation [tex]\( 7x + 2y = 1 \)[/tex] matches our derived equation.

### Conclusion

The equation of line [tex]\( J K \)[/tex] in standard form is:
[tex]\[ \boxed{7x + 2y = 1} \][/tex]