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Nady [450]
3 years ago
13

It doesn't matter which of the two points on a line you choose to call (x1, y1) and which you choose to call (x2, y2) to calcula

te the slope of the line.
A. True
B. False
Mathematics
2 answers:
stiv31 [10]3 years ago
8 0
ANSWER

A. True

EXPLANATION

Let

(x_1,y_1) = (1,2)



and

(x_2,y_2) = (2,3)


be two points on the straight line.

Then the slope is given by

m =  \frac{y_2-y_1}{x_2-x_1}


This implies that,

m =  \frac{3 - 2}{2 - 1}  =  \frac{1}{1}  = 1

Let us now choose it the other way round,


(x_1,y_1) = (2,3)
(x_2,y_2) = (1,2)


Then the slope is,

m =  \frac{2 - 3}{1 - 2}  =  \frac{ - 1}{ - 1}  = 1

We still had they same result. Hence it doesn't matter which one you choose to call

(x_1,y_1)
and which to call

(x_2,y_2)

frosja888 [35]3 years ago
3 0
The correct answer is true ~ Apex 
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Step-by-step explanation:

y = 3.25x + 1.25

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Q.6. The equation of the ellipse whose centre is at the origin and the x-axis, the major axis, which passes
azamat

<h3>Answer:</h3>

Equation of the ellipse = 3x² + 5y² = 32

<h3>Step-by-step explanation:</h3>

<h2>Given:</h2>

  • The centre of the ellipse is at the origin and the X axis is the major axis

  • It passes through the points (-3, 1) and (2, -2)

<h2>To Find:</h2>

  • The equation of the ellipse

<h2>Solution:</h2>

The equation of an ellipse is given by,

\sf \dfrac{x^2}{a^2} +\dfrac{y^2}{b^2} =1

Given that the ellipse passes through the point (-3, 1)

Hence,

\sf \dfrac{(-3)^2}{a^2} +\dfrac{1^2}{b^2} =1

Cross multiplying we get,

  • 9b² + a² = 1 ²× a²b²
  • a²b² = 9b² + a²

Multiply by 4 on both sides,

  • 4a²b² = 36b² + 4a²------(1)

Also by given the ellipse passes through the point (2, -2)

Substituting this,

\sf \dfrac{2^2}{a^2} +\dfrac{(-2)^2}{b^2} =1

Cross multiply,

  • 4b² + 4a² = 1 × a²b²
  • a²b² = 4b² + 4a²-------(2)

Subtracting equations 2 and 1,

  • 3a²b² = 32b²
  • 3a² = 32
  • a² = 32/3----(3)

Substituting in 2,

  • 32/3 × b² = 4b² + 4 × 32/3
  • 32/3 b² = 4b² + 128/3
  • 32/3 b² = (12b² + 128)/3
  • 32b² = 12b² + 128
  • 20b² = 128
  • b² = 128/20 = 32/5

Substituting the values in the equation for ellipse,

\sf \dfrac{x^2}{32/3} +\dfrac{y^2}{32/5} =1

\sf \dfrac{3x^2}{32} +\dfrac{5y^2}{32} =1

Multiplying whole equation by 32 we get,

3x² + 5y² = 32

<h3>Hence equation of the ellipse is 3x² + 5y² = 32</h3>
8 0
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