Question

line $jk$ passes through points $j(-3, 11)$ and $k(1, -3)$. what is the equation of line $jk$ in standard form?
$\bigcirc$ $7x + 2y = -1$
$\bigcirc$ $7x + 2y = 1$
$\bigcirc$ $14x + 4y = -1$
$\bigcirc$ $14x + 4y = 1$

Explanation:

Step1: Calculate slope of line JK

The slope formula is $m=\frac{y_2-y_1}{x_2-x_1}$. Using $J(-3,11)$ and $K(1,-3)$:
$m=\frac{-3-11}{1-(-3)}=\frac{-14}{4}=-\frac{7}{2}$

Step2: Use point-slope form

Point-slope form: $y-y_1=m(x-x_1)$. Use point $J(-3,11)$:
$y-11=-\frac{7}{2}(x+3)$

Step3: Convert to standard form

Multiply all terms by 2 to eliminate fraction:
$2(y-11)=-7(x+3)$
Expand both sides:
$2y-22=-7x-21$
Rearrange to $Ax+By=C$:
$7x+2y=1$

Step4: Verify with the other point

Substitute $K(1,-3)$ into $7x+2y$:
$7(1)+2(-3)=7-6=1$, which matches the right-hand side.

Answer:

B. $7x + 2y = 1$