Let C (α, β) be the circumcenter of the triangle formed by the lines4x + 3y = 69,4y – 3x = 17, andx + 7y = 61.Then (α

Engineering

Mathematics

Section Formulae and Centres of a Triangle

Question

Let C (α, β) be the circumcenter of the triangle formed by the lines
4x + 3y = 69,
4y – 3x = 17, and
x + 7y = 61.
Then (α – β)² + α + β is equal to

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Triangle is right angle at vertex B. Hence circum centre is mid point of AC
For A solve x + 7y = 61 and 4x + 3y = 69

4x + 28y = 244

– 25y = – 175

y = 7

x = 12

A(12, 7)
For C solve x + 7y = 61 and 3x – 4y = – 17

3x + 21y = 183

– 25y = – 200

y = 8

x = 5

C(5, 8)

$Curcumcenter \equiv (\frac{5 + 12}{2}, \frac{7 + 8}{2}) \equiv (\frac{17}{2}, \frac{15}{2})$

$α = \frac{17}{2}, β = \frac{15}{2}$

Hence (α – β)² + α + β = 1 + 16 = 17