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Kay [80]
1 year ago
14

Which line is perpendicular to the line 3y - 4x = 6

Mathematics
1 answer:
Scilla [17]1 year ago
6 0

The equation of the line that is perpendicular to the line 3y - 4x = 6 is: c. 4y + 3x = 8.

<h3>How to Determine the Equation of Lines that are Perpendicular?</h3>

The lines that are perpendicular to each other have slope value of their equation, m, that are negative reciprocals of each other or the product of their slopes is equal to -1.

Given the line 3y - 4x = 6, rewrite the equation in slope-intercept form, y = mx + b, and determine the value of m:

3y = 4x + 6

Divide both sides by 3

3y/3 = 4x/3 + 6/3

y = 4/3x + 2

The slope (m) = 4/3.

The slope of the line that would be perpendicular to the line 3y - 4x = 6, will have a slope of -3/4, which is the negative reciprocal of 4/3.

Rewrite 4y + 3x = 8 in slope-intercept form:

4y = -3x + 8

4y/4 = -3/4x + 8/4

y = -3/4x + 2

The slope of 4y + 3x = 8 is -3/4.

Therefore, the perpendicular line is: c. 4y + 3x = 8.

Learn more about perpendicular lines equation on:

brainly.com/question/7098341

#SPJ1

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Given below are the graphs of two lines, y=-0.5 + 5 and y=-1.25x + 8 and several regions and points are shown. Note that C is th
zalisa [80]
We have the following equations:

(1) \ y=-0.5x+5 \\ (2) \ y=-1.25x+8

So we are asked to write a system of equations or inequalities for each region and each point.

Part a)

Region Example A

y \leq -0.5x+5 \\ y \leq -1.25x+8

Region B.

Let's take a point that is in this region, that is:

P(0,6)

So let's find out the signs of each inequality by substituting this point in them:

y \ (?)-0.5x+5  \\ 6 \ (?) -0.5(0)+5 \\ 6 \ (?) \ 5 \\ 6\ \textgreater \ 5 \\  \\ y \ (?) \ -1.25x+8 \\ 6 \ (?) -1.25(0)+8 \\ 6 \ (?) \ 8 \\ 6\ \textless \ 8

So the inequalities are:

(1) \ y  \geq  -0.5x+5 \\  (2) \ y  \leq  -1.25x+8

Region C.

A point in this region is:

P(0,10)

So let's find out the signs of each inequality by substituting this point in them:

y \ (?)-0.5x+5 \\ 10 \ (?) -0.5(0)+5 \\ 10 \ (?) \ 5 \\ 10\ \textgreater \ 5 \\ \\ y \ (?) \ -1.25x+8 \\ 10 \ (?) -1.25(0)+8 \\ 10 \ (?) \ 8 \\ 10 \ \ \textgreater \  \ 8

So the inequalities are:

(1) \ y  \geq  -0.5x+5 \\ (2) \ y  \geq  -1.25x+8

Region D.

A point in this region is:

P(8,0)

So let's find out the signs of each inequality by substituting this point in them:

y \ (?)-0.5x+5 \\ 0 \ (?) -0.5(8)+5 \\ 0 \ (?) \ 1 \\ 0 \ \ \textless \  \ 1 \\ \\ y \ (?) \ -1.25x+8 \\ 0 \ (?) -1.25(8)+8 \\ 0 \ (?) \ -2 \\ 0 \ \ \textgreater \ \ -2

So the inequalities are:

(1) \ y  \leq  -0.5x+5 \\ (2) \ y  \geq  -1.25x+8

Point P:

This point is the intersection of the two lines. So let's solve the system of equations:

(1) \ y=-0.5x+5 \\ (2) \ y=-1.25x+8 \\ \\ Subtracting \ these \ equations: \\ 0=0.75x-3 \\ \\ Solving \ for \ x: \\ x=4 \\  \\ Solving \ for \ y: \\ y=-0.5(4)+5=3

Accordingly, the point is:

\boxed{p(4,3)}

Point q:

This point is the x-intercept of the line:

y=-0.5x+5

So let:

y=0

Then

x=\frac{5}{0.5}=10

Therefore, the point is:

\boxed{q(10,0)}

Part b) 

The coordinate of a point within a region must satisfy the corresponding system of inequalities. For each region we have taken a point to build up our inequalities. Now we will take other points and prove that these are the correct regions.

Region Example A

The origin is part of this region, therefore let's take the point:

O(0,0)

Substituting in the inequalities:

y \leq -0.5x+5 \\ 0 \leq -0.5(0)+5 \\ \boxed{0 \leq 5} \\ \\ y \leq -1.25x+8 \\ 0 \leq -1.25(0)+8 \\ \boxed{0 \leq 8}

It is true.

Region B.

Let's take a point that is in this region, that is:

P(0,7)

Substituting in the inequalities:

y \geq -0.5x+5 \\ 7 \geq -0.5(0)+5 \\ \boxed{7 \geq \ 5} \\ \\ y  \leq \ -1.25x+8 \\ 7 \ \leq -1.25(0)+8 \\ \boxed{7 \leq \ 8}

It is true

Region C.

Let's take a point that is in this region, that is:

P(0,11)

Substituting in the inequalities:

y \geq -0.5x+5 \\ 11 \geq -0.5(0)+5 \\ \boxed{11 \geq \ 5} \\ \\ y \geq \ -1.25x+8 \\ 11 \ \geq -1.25(0)+8 \\ \boxed{11 \geq \ 8}

It is true

Region D.

Let's take a point that is in this region, that is:

P(9,0)

Substituting in the inequalities:

y  \leq -0.5x+5 \\ 0 \leq -0.5(9)+5 \\ \boxed{0 \leq \ 0.5} \\ \\ y \geq \ -1.25x+8 \\ 0 \geq -1.25(9)+8 \\ \boxed{0 \geq \ -3.25}

It is true

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