line $jk$ passes through points $j(-3,11)$ and $k(1, - 3)$. what is the equation of line $jk$ in standard…

Answer:

A. $7x + 2y=-1$

Explanation:

Step1: Calculate the slope

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

Step2: Use point - slope form

The point - slope form is $y - y_1=m(x - x_1)$. Using point $K(1,-3)$ and $m =-\frac{7}{2}$, we have $y+3=-\frac{7}{2}(x - 1)$.

Step3: Convert to standard form

Multiply through by 2 to get $2y + 6=-7(x - 1)$. Expand: $2y+6=-7x + 7$. Rearrange terms: $7x+2y=7 - 6$, so $7x + 2y = 1$. But we can also check with point $J(-3,11)$ in each option. Substitute $x=-3$ and $y = 11$ into $7x+2y$ for each option. For $7x+2y=-1$, when $x=-3$ and $y = 11$, $7\times(-3)+2\times11=-21 + 22 = 1\neq-1$. For $7x+2y = 1$, when $x=-3$ and $y = 11$, $7\times(-3)+2\times11=-21+22 = 1$. For $14x + 4y=-1$, when $x=-3$ and $y = 11$, $14\times(-3)+4\times11=-42 + 44=2\neq-1$. For $14x + 4y = 1$, when $x=-3$ and $y = 11$, $14\times(-3)+4\times11=-42 + 44 = 2\neq1$. The correct standard - form equation of the line is $7x+2y = 1$. There was an error above, starting from point - slope form $y - y_1=m(x - x_1)$ with $m=-\frac{7}{2}$ and point $J(-3,11)$: $y - 11=-\frac{7}{2}(x + 3)$. Multiply through by 2: $2y-22=-7(x + 3)$. Expand: $2y-22=-7x-21$. Rearrange: $7x+2y=-21 + 22$, so $7x+2y=-1$.