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2 years ago

Determine the gradient and the co-ordinates of the x and y intercept of line whose equation is 2y + 3x = 1

Physics

1 answer:

Savatey [412]2 years ago

4 0

Answer:

The x - intercept is 1/3

The y - intercept is 1/2

The gradient is -3/2

Explanation:

To find the x - intercept of the equation 2y + 3x = 1, we find the value of x when y = 0. So,

2y + 3x = 1

2(0) + 3x = 1

0 + 3x = 1

3x = 1

x = 1/3

So, the x - intercept is 1/3

To find the y - intercept of the equation 2y + 3x = 1, we find the value of y when x = 0. So,

2y + 3x = 1

2y + 3(0) = 1

2y + 0 = 1

2y = 1

y = 1/2

So, the y - intercept is 1/2

To find the gradient of the equation 2y + 3x = 1, we re-write it in gradient intercept form by making y subject of the formula.

So, 2y + 3x = 1

2y = -3x + 1

y = -3x/2 + 1/2

The coefficient of x which equals -3/2 is the gradient.

The gradient is -3/2

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Given that,

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Using formula of lens

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Put the value into the formula

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(b). We need to calculate the focal length

Using formula of power

$f =\dfrac{1}{P}$

Put the value into the formula

$f=-\dfrac{100}{3.00}$

$f=-33.33\ cm$

We need to calculate the image distance

Using formula of lens

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

Put the value into the formula

$\dfrac{1}{-33.33}=\dfrac{1}{v}-\dfrac{1}{\infty}$

$-\dfrac{1}{33.33}+\dfrac{1}{\infty}=\dfrac{1}{v}$

$\dfrac{1}{v}=-\dfrac{1}{33.33}$

$v=-33.33\ cm$

The eye's far point is 33.33 cm from the eye.

Hence, (a). The eye's near point is 68.98 cm from the eye.

(b). The eye's far point is 33.33 cm from the eye.

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Determine the gradient and the co-ordinates of the x and y intercept of line whose equation is 2y + 3x = 1​

Determine the gradient and the co-ordinates of the x and y intercept of line whose equation is 2y + 3x = 1