Answer :
Let's solve this step by step.
1. Calculate the slope of the line passing through (66,33) and (99,51)
The formula for the slope [tex]\(m\)[/tex] between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is:
[tex]\[ m = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
Substituting the given points [tex]\((66, 33)\)[/tex] and [tex]\((99, 51)\)[/tex]:
[tex]\[ m = \frac{51 - 33}{99 - 66} = \frac{18}{33} = \frac{6}{11} \][/tex]
So, the slope [tex]\(m\)[/tex] is [tex]\( \boxed{\frac{6}{11}} \)[/tex].
2. Write the equation of this line in point-slope form
The point-slope form of the equation of a line is given by:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
Using one of the given points [tex]\((66, 33)\)[/tex] and the slope [tex]\(\frac{6}{11}\)[/tex]:
[tex]\[ y - 33 = \frac{6}{11}(x - 66) \][/tex]
Thus, the point-slope form is [tex]\( \boxed{y - 33 = \frac{6}{11}(x - 66)} \)[/tex].
3. Write the equation of this line in slope-intercept form
The slope-intercept form of a line is:
[tex]\[ y = mx + b \][/tex]
To find [tex]\(b\)[/tex], we use one of the given points and the slope. Again using the point [tex]\((66, 33)\)[/tex] and the slope [tex]\(\frac{6}{11}\)[/tex]:
Substitute [tex]\(x = 66\)[/tex], [tex]\(y = 33\)[/tex], and [tex]\(m = \frac{6}{11}\)[/tex] into the slope-intercept form:
[tex]\[ 33 = \frac{6}{11}(66) + b \][/tex]
[tex]\[ 33 = 6 \times 6 + b \][/tex]
[tex]\[ 33 = 36 + b \][/tex]
Solving for [tex]\(b\)[/tex], we get:
[tex]\[ b = 33 - 36 = -3 \][/tex]
Thus, the slope-intercept form is [tex]\( \boxed{y = \frac{6}{11}x - 3} \)[/tex].
4. Determine if the line passes through the point [tex]\((-11, -8)\)[/tex]
We need to check if the point [tex]\((-11, -8)\)[/tex] satisfies the equation of the line:
Using the slope-intercept form [tex]\(y = \frac{6}{11}x - 3\)[/tex], substitute [tex]\(x = -11\)[/tex] and see if [tex]\(y\)[/tex] equals [tex]\(-8\)[/tex]:
[tex]\[ y = \frac{6}{11}(-11) - 3 = -6 - 3 = -9 \][/tex]
Since [tex]\(-9 \neq -8\)[/tex], the point [tex]\((-11, -8)\)[/tex] does not lie on the line.
Therefore, the answer is [tex]\( \boxed{\text{A. No}} \)[/tex].
1. Calculate the slope of the line passing through (66,33) and (99,51)
The formula for the slope [tex]\(m\)[/tex] between two points [tex]\((x_1, y_1)\)[/tex] and [tex]\((x_2, y_2)\)[/tex] is:
[tex]\[ m = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
Substituting the given points [tex]\((66, 33)\)[/tex] and [tex]\((99, 51)\)[/tex]:
[tex]\[ m = \frac{51 - 33}{99 - 66} = \frac{18}{33} = \frac{6}{11} \][/tex]
So, the slope [tex]\(m\)[/tex] is [tex]\( \boxed{\frac{6}{11}} \)[/tex].
2. Write the equation of this line in point-slope form
The point-slope form of the equation of a line is given by:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
Using one of the given points [tex]\((66, 33)\)[/tex] and the slope [tex]\(\frac{6}{11}\)[/tex]:
[tex]\[ y - 33 = \frac{6}{11}(x - 66) \][/tex]
Thus, the point-slope form is [tex]\( \boxed{y - 33 = \frac{6}{11}(x - 66)} \)[/tex].
3. Write the equation of this line in slope-intercept form
The slope-intercept form of a line is:
[tex]\[ y = mx + b \][/tex]
To find [tex]\(b\)[/tex], we use one of the given points and the slope. Again using the point [tex]\((66, 33)\)[/tex] and the slope [tex]\(\frac{6}{11}\)[/tex]:
Substitute [tex]\(x = 66\)[/tex], [tex]\(y = 33\)[/tex], and [tex]\(m = \frac{6}{11}\)[/tex] into the slope-intercept form:
[tex]\[ 33 = \frac{6}{11}(66) + b \][/tex]
[tex]\[ 33 = 6 \times 6 + b \][/tex]
[tex]\[ 33 = 36 + b \][/tex]
Solving for [tex]\(b\)[/tex], we get:
[tex]\[ b = 33 - 36 = -3 \][/tex]
Thus, the slope-intercept form is [tex]\( \boxed{y = \frac{6}{11}x - 3} \)[/tex].
4. Determine if the line passes through the point [tex]\((-11, -8)\)[/tex]
We need to check if the point [tex]\((-11, -8)\)[/tex] satisfies the equation of the line:
Using the slope-intercept form [tex]\(y = \frac{6}{11}x - 3\)[/tex], substitute [tex]\(x = -11\)[/tex] and see if [tex]\(y\)[/tex] equals [tex]\(-8\)[/tex]:
[tex]\[ y = \frac{6}{11}(-11) - 3 = -6 - 3 = -9 \][/tex]
Since [tex]\(-9 \neq -8\)[/tex], the point [tex]\((-11, -8)\)[/tex] does not lie on the line.
Therefore, the answer is [tex]\( \boxed{\text{A. No}} \)[/tex].
