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How to find an equation perpendicular to

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3x + 4y=12 that passes through (7,1)?

asked Nov 6, 2014 in ALGEBRA 2 by anonymous

1 Answer

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The line equation 3x + 4y = 12

Write the equation in slope intercept form y = mx + b

Where m is slope and b is y intercept.

4y = - 3x + 12

y = (- 3x + 12)/4

y = (-3/4)x + 3

Compare it to slope intercept form.

m = -3/4, b = 3

Required line perpendicular to 3x + 4y = 12.

We know that perpendicular lines slopes are negative reciprocal to each other.

Thus, the slope of the required line is 4/3.

Now, the line equation is y = (4/3)x + b.

Find the y - intercept by substituting the point in the line equation say (x, y) = (7, 1).

1 = (4/3)7 + b

b = 1 - (28/3)

b = (3 - 28)/3

b = - 25/3

The line equation in slope - intercept - form is y = (4/3)x - (25/3).

answered Nov 6, 2014 by david Expert

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