First slide

Cartesian plane

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

Moderate

A

$a + b - c > 0$
B

$a - b + c < 0$
C

$a - b + c > 0$
D

$a + b - c < 0$

Solution

Solving given lines for their point of intersection,
we get the point of intersection as $(- \frac{c}{a + b}, - \frac{c}{a + b})$ , Its distance from ( 1 , 1) is
$\sqrt{{(1 + \frac{c}{a + b})}^{2} + {(1 + \frac{c}{a + b})}^{2}} < 2 \sqrt{2}$
$\begin{array}{l} or (a + b + c)^{2} < 4 (a + b)^{2} or (a + b + c)^{2} - (2 a + 2 b)^{2} < 0 \\ or (c - a - b) (c + 3 a + 3 b) < 0 \\ Since a > b > c > 0, (c - a - b) < 0 or a + b - c > 0 \end{array}$

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