# What is the equation of the line that intesects 3x + 4y = 8 at (0, 2) and is perpendicular to it.

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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)**.

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**