JEE Advanced 2013 · Paper 1 · Mathematics · Straight Lines
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 222 \sqrt{2}22.Then
A. a+b−c>0a + b - c > 0a+b−c>0
(a−b)x+(b−a)y=0\left(\text{a} - \text{b}\right) \text{x} + \left(\text{b} - \text{a}\right) \text{y} = 0(a−b)x+(b−a)y=0⇒x=y⇒ \text{x} = \text{y}⇒x=y⇒Point of intersection(−ca+b,−ca+b)⇒ \text{Point\text{ of } intersection} \left(\dfrac{- \text{c}}{\text{a} + \text{b}} \text{,} \dfrac{- \text{c}}{\text{a} + \text{b}}\right)⇒Point of intersection(a+b−c,a+b−c)Now using distance formula between two points d=(x1−x2)2+(y1−y2)2d = \sqrt{{\left({x}_{1} - {x}_{2}\right)}^{2} + {\left({y}_{1} - {y}_{2}\right)}^{2}}d=(x1−x2)2+(y1−y2)2
⇒(1+ca+b)2+(1+ca+b)2<22⇒ \sqrt{{\left(1 + \dfrac{\text{c}}{\text{a} + \text{b}}\right)}^{2} + {\left(1 + \dfrac{\text{c}}{\text{a} + \text{b}}\right)}^{2}} < 2\sqrt{2}⇒(1+a+bc)2+(1+a+bc)2<22⇒2(∣a+b+ca+b∣)<22⇒ \sqrt{2}\left(\left|\dfrac{\text{a} + \text{b} + \text{c}}{\text{a} + \text{b}}\right|\right) < 2\sqrt{2}⇒2(a+ba+b+c)<22
⇒(a+b+ca+b)<2⇒ \left(\dfrac{\text{a} + \text{b} + \text{c}}{\text{a} + \text{b}}\right) < 2⇒(a+ba+b+c)<2
⇒a+b+c<2(a+b)\text{⇒a} + \text{b} + \text{c} < 2\left(\text{a} + \text{b}\right)⇒a+b+c<2(a+b)⇒a+b−c>0⇒ \text{a} + \text{b} - \text{c} > 0⇒a+b−c>0
