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Find the equation of a straight line passing through (-3;2) and(5;8) and calculate its...

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kamused | Student, College Freshman | eNoter

Posted February 16, 2011 at 11:16 PM via web

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Find the equation of a straight line passing through (-3;2) and(5;8) and calculate its gradient.

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giorgiana1976 | College Teacher | Valedictorian

Posted February 16, 2011 at 11:21 PM (Answer #1)

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The equation of the straight line, that passing through the given points, is:

y - y = m(x2 - x1), where x1,x2,y1,y2 are the coordinates of the given points and m represents the gradient of the line.

We'll determine the gradient:

m = (y2 - y1)/(x2 - x1)

m = (8 - 2) / [ 5- (-3)]

m = 6/8

m = 3/4

We'll write the equation:

y - 2 = ( 3/4)(x + 3)

The equation of the line is y = 3x/4 + 17/4 and the gradient is m = 3/4.

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted February 17, 2011 at 12:02 AM (Answer #2)

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We have to find the straight line passing through the points ( -3,2) and ( 5,8)

The equation of a line passing through the points ( x1, y1) and ( x2, y2) is given by ( y - y1) = [ ( y2 - y1)/(x2 - x1)]*( x - x1).

Here ( y2 - y1)/(x2 - x1) is the gradient of the line.

Substituting the values we have we get :

( y - y1) = [ ( y2 - y1)/(x2 - x1)]*( x - x1)

=> ( y - 2) = [ ( 8 - 2)/(5  + 3)]*( x + 3)

=> ( y - 2) = ( 6)/8)*( x + 3)

=> y - 2 = (3/4)*( x+ 3)

=> 4y - 8 = 3x + 9

=> 3x - 4y + 17 = 0

The gradient is 3/4

The required equation of the line is 3x - 4y + 17 = 0 and its gradient is 3/4.

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