please help

What is the point-slope form of the equation for the line with a slope of 6/19(6 on the top and 19 on the bottom) that passes through the point (−1,7/5)?(7/5= 7 on the top and 5 on the bottom)

A.y+7/5=6/19(x−1)

B.y−7/5=6/19(x+1)

C.y−1=6/19(x+7/5)

D.y+1=6/19(x−7/5)

[tex]\bf (\stackrel{x_1}{-1}~,~\stackrel{y_1}{\frac{7}{5}})~\hspace{10em} slope = m\implies \cfrac{6}{19} \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\cfrac{7}{5}=\cfrac{6}{19}[x-(-1)]\implies y-\cfrac{7}{5}=\cfrac{6}{19}(x+1)[/tex]

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please helpWhat is the point-slope form of the equation for the line with a slope of 6/19(6 on the top and 19 on the bottom) that passes through the point (−1,7/5)?(7/5= 7 on the top and 5 on the bottom)A.y+7/5=6/19(x−1)B.y−7/5=6/19(x+1)C.y−1=6/19(x+7/5)D.y+1=6/19(x−7/5)​

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please help

What is the point-slope form of the equation for the line with a slope of 6/19(6 on the top and 19 on the bottom) that passes through the point (−1,7/5)?(7/5= 7 on the top and 5 on the bottom)

A.y+7/5=6/19(x−1)

B.y−7/5=6/19(x+1)

C.y−1=6/19(x+7/5)

D.y+1=6/19(x−7/5)