If line A contains Q(5,1) and is parallel to line MN with M(-2,4) and N(2,1), which ordered pair would be on the perpendicular t

Question

hichkok12 [17]

1 year ago

7

If line A contains Q(5,1) and is parallel to line MN with M(-2,4) and N(2,1), which ordered pair would be on the perpendicular t

ransversal if its equation is y= 4/3x + 1/3

Mathematics

2 answers:

vivado [14] · Answer 1 · 2023-10-26T10:10:36+03:00

Your question does not say what were your options, therefore I will answer generically: in order to understand if a point (ordered pair) is contained in a line, you need to substitute the x-component of the pair in the equation of the line and see if the calculations give you the y-component of the pair.

Example:
Your line is <span> y = 4/3x + 1/3
Let's see if <span>(0, 0) and (2, 3) </span>belong to this line

y</span> = <span>4/3·0 + 1/3 = 1/3 </span>≠ 0
Therefore, the line does not contain (0, 0)

y = 4/3·2 + 1/3 = 9/3 = 3
Therefore, the line contains (2, 3)

Luden [163] · Answer 2 · 2023-10-26T11:32:42+03:00

Point -1, -1 are on the line equation y = 4/3x + 1/3

<h3>Further explanation</h3>

Straight-line equations are mathematical equations that are described in the plane of cartesian coordinates

General formula

$\large{\boxed{\bold{y-y1=m(x-x1)}}}$

or

y = mx + c

Where

m = straight-line gradient which is the slope of the line

x1, y1 = the Cartesian coordinate that is crossed by the line

c = constant

The formula for a gradient (m) between 2 points in a line

m = Δy / Δx

$\large{\boxed{\bold{m=\frac{y_2-y_1}{x_2-x_1}}}}$

There are several gradient properties that a line has

1. parallel to the x-axis

the gradient that is parallel to the x-axis is = 0

2. parallel to the y-axis

the gradient that is parallel to the y-axis is undefined

3. 2 parallel lines

the gradient of 2 parallel lines are of equal value

4. 2 perpendicular lines

the gradient of 2 perpendicular lines, if both of them are multiplied , the value will be = -1

line MN with M (-2,4) and N (2,1)

the gradient:

$\frac{1-4}{2+2} = -\frac{3}{4}$

Then the equation of the line: (through the point -2.4)

$y-4=-\frac{3}{4}(x + 2)$

$y=-\frac{3}{4}+\frac{5}{2}$

Line A with a line equation that is perpendicular to the line MN will have a gradient:

line gradient MN x line gradient A = -1

(-3/4) x gradient line A = -1

gradient line A = 4/3

So the line A with the equation y = 4/3x + 1/3 is appropriate

From the several answer options, we enter the value of these points into the equation y = 4/3x + 1/3, whether or not it meets. If it meets then the point is contained in the existing line equation