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Question

In a triangle ABC, one of the vertex A is (3,4) and vertices B(x1,y1), C(x2,y2) lie on the line 3x+4y=5. The area of the triangle ABC is 20 sq. units. If x1+x2=10 and x1+y1x2+y2=3k, then the smaller value of [k], (where [.] is greatest integer function) is


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Solution



In triangle ABC, ADBC
Height AD=∣ ∣33+44532+42∣ ∣=4
Area of triangle ΔABC=12×AD×BC=20
BC=10=(x2x1)2+(y2y1)2
(x2x1)2+(y2y1)2=100....(i)

We know that B,C lies on the line 3x+4y=5
x1=54y13 and x2=54y23x2x1=4(y1y2)3....(ii)
From equation (i) and (ii), we get
y1y2=±6x2x1=±8
Also, x1+x2=10
So from above condition
x1=1,9 x2=9,1 and y1=12,112 y2=112,12

x1+y1x2+y2=1+129112=3k[k]=7
or
x1+y1x2+y2=95.51+.5=73=3k[k]=1
Hence the smaller value of [k] is 1

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