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What is the equation of the line that intesects 3x + 4y = 8 at (0, 2) and is...
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The product of the slope of two perpendicular lines is equal to -1. The line 3x + 4y = 8 can be written in the slope intercept form as y = (-3/4)x + 2
As the slope of this line is -3/4, the slope of the required line that is perpendicular to it is 4/3. As the required line intersects at (0, 2), the equation of the line is (y - 2)/x = 4/3
=> 3y - 6 = 4x
=> 4x - 3y + 6 = 0
The line 4x - 3y + 6 = 0 is perpendicular to 3x + 4y = 8 and intersects it at (0, 2).
Posted by justaguide on June 29, 2012 at 4:58 PM (Answer #1)
High School Teacher
The given line AB is 3x + 4y = 8
Or, 4y = -3x + 8
Or, y= (-3/4) x + 8
Therefore the slope of the line m1 = -3/4 [since, y = mx + c ]
It is given that the require lie is perpendicular to the line AB. i.e. m1*m2 = -1 [ where m2 is the slope of the require line ]
m2 = (-1)/m1 = (-1)/(-3/4) = 4/3
m2 = 4/3 --------(2)
Next It is stated that the require line intersect the given line AB at (0,2) Which means y-interxept (c) of the require line is 2 i.e. c=2
therefore the equation of the require line ----->
y = (4/3)x + 2 [ substituting the value of the slope m and the y- intercept (c) in the equation : y = mx + c ]
Or, y = (4x + 6)/3
Or, 3y = 4x + 6
Or, 3y - 4x -6 =0
Hence equation of the require line --->3y - 4x - 6 = 0 Answer
Posted by vaaruni on June 30, 2012 at 6:59 AM (Answer #2)
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