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Question

Let ABC,be a triangle such that the coordinates of the vertex A are (3,1).Equation of the median through B is 2x+y3=0 and equation of the angular bisector of C is 7x4y1=0.Find the slope of line BC.

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Solution

Consider the coordinates of C as (a,b), and since the point C lies on the angular bisector of C, therefore the equation becomes,

7a4b=1 (1)

The midpoint of AC will be (a32,b+12) and as the median lies on the midpoint of B, therefore the equation becomes,

2(a32)+(b+12)=3

2a6+b+1=6

2a+b=11 (2)

From equation (1) and (2),

a=3,b=5

Slope of AC is,

m1=513(3)

=23

The slope of bisector is,

m2=Coeff.ofxCoeff.ofy

=74

The angle between AC and the bisector will be,

tanα=m1m21+m1m2

=∣ ∣ ∣ ∣23741+(23)(74)∣ ∣ ∣ ∣

=12

Let m be the slope of BC.

The angle between BC and bisector will be equal to the angle between AC and bisector.

mm21+mm2=12

∣ ∣ ∣m741+m(74)∣ ∣ ∣=12

4m77m+4=±12

m=23;m=18

When the slope of BC is 23, then A,B,C will be collinear and it will form no triangle.

Therefore, the slope of BC is 18.


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