Solved on Sep 17, 2023

Find the value of $x$ such that the line segment $\overline{NO}$ is parallel to $\overline{PQ}$ , given $N(-1,-1), O(9,9), P(-8,3)$ , and $Q(x, 2)$ .

STEP 1

Assumptions1. Points N, O,, and Q are in a plane with coordinates N(-1,-1), O(9,9),(-8,3), and Q(x,) respectively. . Line segment NO is parallel to line segment PQ.

STEP 2

The slope of a line segment can be found using the formula $lope = \frac{y2 - y1}{x2 - x1}$

STEP 3

First, we need to find the slope of line segment NO using the coordinates of points N and O.
$lope_{NO} = \frac{9 - (-1)}{9 - (-1)}$

STEP 4

Calculate the slope of line segment NO.
$lope_{NO} = \frac{9 - (-1)}{9 - (-1)} =1$

STEP 5

Since line segment NO is parallel to line segment PQ, their slopes are equal. So, the slope of line segment PQ is also1.

STEP 6

Now, we can find the x-coordinate of point Q using the slope of line segment PQ and the coordinates of point. We can rearrange the slope formula to solve for x $x = y2 - Slope_{Q} \cdot (y1 - x1)$

STEP 7

Plug in the values for the slope of line segment PQ and the coordinates of point to calculate the x-coordinate of point Q.
$x =2 -1 \cdot (3 - (-))$

STEP 8

Calculate the x-coordinate of point Q.
$x =2 -1 \cdot (3 - (-8)) = -7$ So, the x-coordinate of point Q that makes line segment NO parallel to line segment PQ is -7.

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Find the value of $x$ such that the line segment $\overline{NO}$ is parallel to $\overline{PQ}$ , given $N(-1,-1), O(9,9), P(-8,3)$ , and $Q(x, 2)$ .

STEP 1

Assumptions1. Points N, O,, and Q are in a plane with coordinates N(-1,-1), O(9,9),(-8,3), and Q(x,) respectively. . Line segment NO is parallel to line segment PQ.

STEP 2

The slope of a line segment can be found using the formula $lope = \frac{y2 - y1}{x2 - x1}$

STEP 3

First, we need to find the slope of line segment NO using the coordinates of points N and O.
$lope_{NO} = \frac{9 - (-1)}{9 - (-1)}$

STEP 4

Calculate the slope of line segment NO.
$lope_{NO} = \frac{9 - (-1)}{9 - (-1)} =1$

STEP 5

Since line segment NO is parallel to line segment PQ, their slopes are equal. So, the slope of line segment PQ is also1.

STEP 6

Now, we can find the x-coordinate of point Q using the slope of line segment PQ and the coordinates of point. We can rearrange the slope formula to solve for x $x = y2 - Slope_{Q} \cdot (y1 - x1)$

STEP 7

Plug in the values for the slope of line segment PQ and the coordinates of point to calculate the x-coordinate of point Q.
$x =2 -1 \cdot (3 - (-))$

STEP 8

Calculate the x-coordinate of point Q.
$x =2 -1 \cdot (3 - (-8)) = -7$ So, the x-coordinate of point Q that makes line segment NO parallel to line segment PQ is -7.

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Find the value of xxx such that the line segment NO‾\overline{NO}NO is parallel to PQ‾\overline{PQ}PQ​, given N(−1,−1),O(9,9),P(−8,3)N(-1,-1), O(9,9), P(-8,3)N(−1,−1),O(9,9),P(−8,3), and Q(x,2)Q(x, 2)Q(x,2).

STEP 1

Assumptions1. Points N, O,, and Q are in a plane with coordinates N(-1,-1), O(9,9),(-8,3), and Q(x,) respectively. . Line segment NO is parallel to line segment PQ.

STEP 2

The slope of a line segment can be found using the formulalope=y2−y1x2−x1lope = \frac{y2 - y1}{x2 - x1}lope=x2−x1y2−y1​

STEP 3

First, we need to find the slope of line segment NO using the coordinates of points N and O.lopeNO=9−(−1)9−(−1)lope_{NO} = \frac{9 - (-1)}{9 - (-1)}lopeNO​=9−(−1)9−(−1)​

STEP 4

Calculate the slope of line segment NO.lopeNO=9−(−1)9−(−1)=1lope_{NO} = \frac{9 - (-1)}{9 - (-1)} =1lopeNO​=9−(−1)9−(−1)​=1

STEP 5

Since line segment NO is parallel to line segment PQ, their slopes are equal. So, the slope of line segment PQ is also1.

STEP 6

Now, we can find the x-coordinate of point Q using the slope of line segment PQ and the coordinates of point. We can rearrange the slope formula to solve for xx=y2−SlopeQ⋅(y1−x1)x = y2 - Slope_{Q} \cdot (y1 - x1)x=y2−SlopeQ​⋅(y1−x1)

STEP 7

Plug in the values for the slope of line segment PQ and the coordinates of point to calculate the x-coordinate of point Q.x=2−1⋅(3−(−))x =2 -1 \cdot (3 - (-))x=2−1⋅(3−(−))

STEP 8

Calculate the x-coordinate of point Q.x=2−1⋅(3−(−8))=−7x =2 -1 \cdot (3 - (-8)) = -7x=2−1⋅(3−(−8))=−7So, the x-coordinate of point Q that makes line segment NO parallel to line segment PQ is -7.

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Find the value of $x$ such that the line segment $\overline{NO}$ is parallel to $\overline{PQ}$ , given $N(-1,-1), O(9,9), P(-8,3)$ , and $Q(x, 2)$ .

The slope of a line segment can be found using the formula $lope = \frac{y2 - y1}{x2 - x1}$

First, we need to find the slope of line segment NO using the coordinates of points N and O.
$lope_{NO} = \frac{9 - (-1)}{9 - (-1)}$

Calculate the slope of line segment NO.
$lope_{NO} = \frac{9 - (-1)}{9 - (-1)} =1$

Now, we can find the x-coordinate of point Q using the slope of line segment PQ and the coordinates of point. We can rearrange the slope formula to solve for x $x = y2 - Slope_{Q} \cdot (y1 - x1)$

Plug in the values for the slope of line segment PQ and the coordinates of point to calculate the x-coordinate of point Q.
$x =2 -1 \cdot (3 - (-))$

Calculate the x-coordinate of point Q.
$x =2 -1 \cdot (3 - (-8)) = -7$ So, the x-coordinate of point Q that makes line segment NO parallel to line segment PQ is -7.