Which equation in point -slope form describes the line that passes through the point (4,-5) and is perpendicular to the line represented by -7x+2y=14.

Linear equations are typically organized in slope-intercept form: [tex]y=mx+b[/tex] where m is the slope and b is the y-intercept (the value of y when the line crosses the y-axis)
Point-slope form: [tex]y-y_1=m(x-x_1)[/tex] where the given point is [tex](x_1,y_1)[/tex] and m is the slope
Perpendicular lines always have slopes that are negative reciprocals (ex. 2 and -1/2, 3/4 and -4/3)

1) Determine the slope (m)

[tex]-7x+2y=14[/tex]

First, rearrange this given equation into slope-intercept form ([tex]y=mx+b[/tex]) so we can easily find the slope.

Add 7x to both sides to isolate 2y

[tex]-7x+2y+7x=7x+14\\2y=7x+14[/tex]

Divide both sides by 2 to isolate y

[tex]y=\frac{7}{2} x+7[/tex]

Now, we can identify clearly that the slope of this line is [tex]\frac{7}{2}[/tex]. Because perpendicular lines have slopes that are negative reciprocals, we know that the slope of the line we're currently solving for will have a slope of [tex]-\frac{2}{7}[/tex].

[tex]m=-\frac{2}{7}[/tex]

2) Plug all necessary values into [tex]y-y_1=m(x-x_1)[/tex]

[tex]y-y_1=m(x-x_1)[/tex]

We know that [tex]m=-\frac{2}{7}[/tex] and that the given point is (4,-5). Plug the slope into the equation

[tex]y-y_1=-\frac{2}{7} (x-x_1)[/tex]

Plug the point into the equation

[tex]y-(-5)=-\frac{2}{7} (x-4)\\y+5=-\frac{2}{7} (x-4)[/tex]

I hope this helps!