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write an equation for a line parallel to $y = -4x + 6$ and passes throu…

Question

write an equation for a line parallel to $y = -4x + 6$ and passes through the point $(-13, 63)$.\
step 1:\
identify the slope of the new equation.\
$m = -4$\
part 2 of 3\
step 2:\
substitute $m$, $x_1$, and $y_1$ into point - slope form.\
$y - y1 = m(x - x1)$\
$y - \square = \square (x - \square )$

Explanation:

Step1: Identify point coordinates

The point is \((-13, 63)\), so \(x_1 = -13\), \(y_1 = 63\).

Step2: Substitute into point - slope form

We know \(m=-4\), \(x_1 = - 13\), \(y_1=63\). Substitute into \(y - y_1=m(x - x_1)\).
So \(y - 63=-4(x - (-13))\), which simplifies to \(y - 63=-4(x + 13)\) when filling the blanks: \(y-\boldsymbol{63}=\boldsymbol{-4}(x-\boldsymbol{(-13)})\) (or \(y - 63=-4(x+13)\) and the blanks are 63, - 4, - 13 respectively).

Answer:

The blanks should be filled with 63, - 4, - 13 (in order), so the equation in point - slope form is \(y - 63=-4(x+13)\) (and the filled blanks are \(y - \boldsymbol{63}=\boldsymbol{-4}(x-\boldsymbol{(-13)})\)).