QuestionFind $x$ so that the line through $(x, 3)$ and $(5, 9)$ has a horizontal slope.

Studdy Solution

STEP 1

Assumptions1. The line passes through the points $(x,3)$ and $(5,9)$ . . The line has a horizontal slope.

STEP 2

The slope of a line passing through two points $(x1, y1)$ and $(x2, y2)$ is given by the formula $lope = \frac{y2 - y1}{x2 - x1}$

STEP 3

We know that the slope of a horizontal line is0. So, we can set up the equation for the slope of the line passing through the points $(x,3)$ and $(5,9)$ and set it equal to0.
$0 = \frac{9 -3}{5 - x}$

STEP 4

olve the equation for $x$ .
$0 = \frac{6}{ - x}$

STEP 5

To solve for $x$ , we can multiply both sides of the equation by $(5 - x)$ to get rid of the fraction.
$0 =$

STEP 6

This equation is not possible, as0 cannot equal6. Therefore, there is no value of $x$ that will make the line passing through the points $(x,3)$ and $(5,9)$ have a horizontal slope.
There is no solution for $x$ .

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QuestionFind $x$ so that the line through $(x, 3)$ and $(5, 9)$ has a horizontal slope.

Studdy Solution

STEP 1

Assumptions1. The line passes through the points $(x,3)$ and $(5,9)$ . . The line has a horizontal slope.

STEP 2

The slope of a line passing through two points $(x1, y1)$ and $(x2, y2)$ is given by the formula $lope = \frac{y2 - y1}{x2 - x1}$

STEP 3

We know that the slope of a horizontal line is0. So, we can set up the equation for the slope of the line passing through the points $(x,3)$ and $(5,9)$ and set it equal to0.
$0 = \frac{9 -3}{5 - x}$

STEP 4

olve the equation for $x$ .
$0 = \frac{6}{ - x}$

STEP 5

To solve for $x$ , we can multiply both sides of the equation by $(5 - x)$ to get rid of the fraction.
$0 =$

STEP 6

This equation is not possible, as0 cannot equal6. Therefore, there is no value of $x$ that will make the line passing through the points $(x,3)$ and $(5,9)$ have a horizontal slope.
There is no solution for $x$ .

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QuestionFind xxx so that the line through (x,3)(x, 3)(x,3) and (5,9)(5, 9)(5,9) has a horizontal slope.

Studdy Solution

STEP 1

Assumptions1. The line passes through the points (x,3)(x,3)(x,3) and (5,9)(5,9)(5,9). . The line has a horizontal slope.

STEP 2

The slope of a line passing through two points (x1,y1)(x1, y1)(x1,y1) and (x2,y2)(x2, y2)(x2,y2) is given by the formulalope=y2−y1x2−x1lope = \frac{y2 - y1}{x2 - x1}lope=x2−x1y2−y1​

STEP 3

We know that the slope of a horizontal line is0. So, we can set up the equation for the slope of the line passing through the points (x,3)(x,3)(x,3) and (5,9)(5,9)(5,9) and set it equal to0.0=9−35−x0 = \frac{9 -3}{5 - x}0=5−x9−3​

STEP 4

olve the equation for xxx.0=6−x0 = \frac{6}{ - x}0=−x6​

STEP 5

To solve for xxx, we can multiply both sides of the equation by (5−x)(5 - x)(5−x) to get rid of the fraction.0=0 =0=

STEP 6

This equation is not possible, as0 cannot equal6. Therefore, there is no value of xxx that will make the line passing through the points (x,3)(x,3)(x,3) and (5,9)(5,9)(5,9) have a horizontal slope.There is no solution for xxx.

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QuestionFind $x$ so that the line through $(x, 3)$ and $(5, 9)$ has a horizontal slope.

Assumptions1. The line passes through the points $(x,3)$ and $(5,9)$ . . The line has a horizontal slope.

The slope of a line passing through two points $(x1, y1)$ and $(x2, y2)$ is given by the formula $lope = \frac{y2 - y1}{x2 - x1}$

We know that the slope of a horizontal line is0. So, we can set up the equation for the slope of the line passing through the points $(x,3)$ and $(5,9)$ and set it equal to0.
$0 = \frac{9 -3}{5 - x}$

olve the equation for $x$ .
$0 = \frac{6}{ - x}$

To solve for $x$ , we can multiply both sides of the equation by $(5 - x)$ to get rid of the fraction.
$0 =$

This equation is not possible, as0 cannot equal6. Therefore, there is no value of $x$ that will make the line passing through the points $(x,3)$ and $(5,9)$ have a horizontal slope.
There is no solution for $x$ .