Sure, let's break this down step by step.
Given the equation in point-slope form [tex]\( y - y_1 = m(x - x_1) \)[/tex], Harold must have used some point [tex]\((x_1, y_1)\)[/tex] and a slope [tex]\(m\)[/tex].
The equation Harold wrote is [tex]\( y = 3 - 7(x - ?)\)[/tex]. Let's identify the parts of the equation:
1. The slope [tex]\(m\)[/tex] is clearly [tex]\(-7\)[/tex], as this is the multiplier of the [tex]\((x - ?)\)[/tex] term.
2. The term [tex]\(3\)[/tex] in the equation [tex]\( y = 3 - 7(x - ?)\)[/tex] suggests that it corresponds to [tex]\( y_1 \)[/tex].
Thus, the full point-slope form [tex]\( y - y_1 = m(x - x_1) \)[/tex] of Harold's equation should be:
[tex]\[ y - 3 = -7(x - x_1) \][/tex]
To make this match Harold's format, let's place the point [tex]\((x_1, y_1)\)[/tex] in the equation:
[tex]\[ y = 3 - 7(x - x_1)\][/tex]
This means [tex]\(3\)[/tex] is [tex]\(y_1\)[/tex] and considering when [tex]\( x = x_1 \)[/tex]:
[tex]\[ y = 3 \][/tex]
Thus, [tex]\( y_1 = 3 \)[/tex].
For the equation [tex]\( y = 3 - 7(x - 7)\)[/tex]:
The [tex]\( x_1 = 7 \)[/tex]
Therefore, the point [tex]\((x_1, y_1)\)[/tex] Harold used is:
[tex]\((7, 3)\)[/tex].
So the point Harold correctly used in writing his equation is [tex]\( (7, 3) \)[/tex].
