Solveeit Logo
Login

Question

Question: For a > b > c > 0, the distance between (1, 1) and the point of intersection of the line ax + by + c...

For a > b > c > 0, the distance between (1, 1) and the point of intersection of the line ax + by + c = 0 and bx + ay + c = 0 is less than 222\sqrt 2 , then
(A). a + b – c > 0
(B). a – b + c < 0
(C). a – b + c > 0
(D). a + b – c < 0

Explanation

Solution

Before attempting this question, one should have prior knowledge about the concept of intersection of line and also remember to multiply the one of the equation of line by a and another equation by b to find the value of x and y, use this information to approach the solution of the question.

Complete step-by-step answer :
According to the given information we have 2 lines ax + by + c = 0 and bx + ay + c = 0 intersecting each other at point where distance between (1, 1) and the intersection point is less than 222\sqrt 2
Taking ax + by + c = 0 as equation 1 and bx + ay + c = 0 as equation 2
Taking the intersection point as (x, y)
So to find the value of x and y we have to find the value of x and y from equation 1 and equation 2
Now subtracting equation 1 by equation 2 and multiplying equation 1 by a and equation 2 by b
a(ax)+a(by)+a(c)−b(bx)−b(ay)−b(c)=0a\left( {ax} \right) + a\left( {by} \right) + {\text{a}}\left( c \right)-b\left( {bx} \right) - b\left( {ay} \right) - b\left( c \right) = 0
\Rightarrow $$$${a^2}x + aby + {\text{a}}c-{b^2}x - bay - bc = 0
\Rightarrow $$$$x\left( {{a^2}-{b^2}} \right) + aby - aby + c\left( {a - b} \right) = 0
\Rightarrow $$$$x\left( {{a^2}-{b^2}} \right) + c\left( {a - b} \right) = 0
As we know that (a2−b2)=(a−b)(a+b)\left( {{a^2}-{b^2}} \right) = \left( {a - b} \right)\left( {a + b} \right)
Therefore, x(a−b)(a+b)+c(a−b)=0x\left( {a-b} \right)\left( {a + b} \right) + c\left( {a - b} \right) = 0
\Rightarrow $$$$\left( {a-b} \right)\left( {x\left( {a + b} \right) + c} \right) = 0
\Rightarrow $$$$x\left( {a + b} \right) + c = 0
\Rightarrow $$$$x = \dfrac{{ - c}}{{\left( {a + b} \right)}}
Substituting the value of x in the equation 1 we get
a−c(a+b)  +by+c=0a\dfrac{{ - c}}{{\left( {a + b} \right)}}\; + by + c = 0
\Rightarrow $$$$\dfrac{{ - ca + by\left( {a + b} \right) + c\left( {a + b} \right)}}{{\left( {a + b} \right)}} = 0
\Rightarrow $$$$ - ca + bay + {b^2}y + ca + cb = 0
\Rightarrow $$$$bay + {b^2}y + cb = 0
\Rightarrow $$$$b\left( {ay + by + c} \right) = 0
\Rightarrow $$$$y\left( {a + b} \right) + c = 0
\Rightarrow $$$$y = \dfrac{{ - c}}{{\left( {a + b} \right)}}
Therefore, (x, y) = (−c(a+b),−c(a+b))\left( {\dfrac{{ - c}}{{\left( {a + b} \right)}},\dfrac{{ - c}}{{\left( {a + b} \right)}}} \right)
Since, we know that distance between the intersection point of 2 line and (1, 1) is less than 222\sqrt 2 i.e. d < 222\sqrt 2
Also we know that distance between two points is given by d=(a−x)2+(b−y)2d = \sqrt {{{\left( {a - x} \right)}^2} + {{\left( {b - y} \right)}^2}} here d is the distance between two points and (x, y) is the coordinates of the first point and (a, b) is the coordinates of the 2 point
Therefore, substituting the values in the above formula we get
(1−(−ca+b))2+(1−(−ca+b))2<22\sqrt {{{\left( {1 - \left( {\dfrac{{ - c}}{{a + b}}} \right)} \right)}^2} + {{\left( {1 - \left( {\dfrac{{ - c}}{{a + b}}} \right)} \right)}^2}} < 2\sqrt 2
\Rightarrow $$$$\sqrt {{{\left( {\dfrac{{a + b + c}}{{a + b}}} \right)}^2} + {{\left( {\dfrac{{a + b + c}}{{a + b}}} \right)}^2}} < 2\sqrt 2
\Rightarrow $$$$\sqrt {2{{\left( {\dfrac{{a + b + c}}{{a + b}}} \right)}^2}} < 2\sqrt 2
\Rightarrow $$$$\sqrt 2 \left( {\dfrac{{a + b + c}}{{a + b}}} \right) < 2\sqrt 2
\Rightarrow $$$$a + b + c < 2a + 2b
\Rightarrow $$$$0 < 2a + 2b - a - b - c
\Rightarrow $$$$a + b - c > 0
Therefore, a + b – c > 0 when the distance between point of intersection of lines and (1, 1) is less than 222\sqrt 2
Hence, option A is the correct option.

Note : In the above solution we came across the two terms “line” where line can be defined as a straight figure which exists in one-dimension which have endless dimensions or a line is also the distance between the 2 points and the distance between the two ends points is given by d=(a−x)2+(b−y)2d = \sqrt {{{\left( {a - x} \right)}^2} + {{\left( {b - y} \right)}^2}} here d is the distance between two points and (x, y) is the coordinates of the first point and (a, b) is the coordinates of the 2 end points of the line.