Question

For a > b > c > 0, the distance between (1 , 1) and the point of intersection of the lines ax + by + c = 0 and bx + ay + c = 0 is less than $2\sqrt{2}$. Then

• A. a + b - c > 0
• B. a - b + c < 0
• C. a - b + c > 0
• D. a + b - c < 0

Answer 1

Answer: (A)

a + b - c > 0

$a>b>c>0ax+by+c=0=>x=\frac{-by-c}{a}bx+ay+c=0=>b\left(\frac{-by-c}{a}\right)+ay+c=0=>-b^{2}y-bc+a^{2}y+ac=0=>y=\frac{bc-ac}{a^2-b^2}=>y=\frac{c(b-a)}{(a+b)(a-b)}=>y=\frac{-c}{a+b}x=\frac{-by-c}{a}=\frac{-b\left(\frac{-c}{a+b}\right)-c}{a}=\frac{bc-ac-bc}{a(a+b)}=\frac{-c}{a+b}$

Required point $(\frac{-c}{a+b},\frac{-c}{a+b})$

According to the question

$2\sqrt{2}-\sqrt{{(\frac{-c}{a+b}-1)}^2+{(\frac{-c}{a+b}-1)}^2}=2\sqrt{2}-\sqrt{{\left(\frac{-c-a-b}{a+b}\right)}^2+{\left(\frac{-c-a-b}{a+b}\right)}^2}=2\sqrt{2}-\sqrt{\frac{{2(-c-a-b)}^2}{{(a+b)}^2}}=2\sqrt{2}-\sqrt{2}\frac{(-c-a-b)}{a+b}=2\sqrt{2}+\sqrt{2}\frac{a+b+c}{a+b}=2\sqrt{2}a+2\sqrt{2}b+\sqrt{2}a+\sqrt{2}b+\sqrt{2}ca+b-c>0$

Proved