Question

if 2x-3y-5=0 is the perpendicular bisector of the line segment (3,-4) and ( alpha, beta), then find the value of alpha+beta​

Answer 1

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Answer 2

Alpha+Beta is 1.

Given:

$2x-3y-5=0$ is the perpendicular bisector

The line segment joins (3,-4) and ( alpha, beta)

To Find:

alpha+beta

Solution:

$2x-3y-5=0$ can be written as

$y=\frac{2x}{3} -\frac{5}{3}$

Therefore the slope of this line is $\frac{2}{3}$.

Hence the slope of the line perpendicular to this is -1.5.

Hence the line segment will be

$y=-1.5x +c$

As we know that (3,-4) satisfies this line,

$c= y+1.5x=-4+4.5 =0.5$

Therefore the point of intersection of the two lines is the midpoint of the two points given.

$2x-3y-5=0\\1.5x+y-0.5=0\\4.5x+3y-1.5=0$

Adding the first and the third equation we get

$6.5x=6.5\\x=1\\y=-1$

Therefore

$alpha = 2*1-3 = -1\\beta = 2*-1 +4 =2$

Hence alpha + beta = 1.

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