line jk passes through points j(-3, 11) and k(1, -3). what is the equation of line jk in standard form?\n7x…

Explanation:

Step1: Calculate the slope

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

Step2: Use the 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=-7x+7). Rearrange to (7x+2y=1).

Answer:

(7x + 2y = 1) (Option 2)