Question

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 $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

Hint: 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 $2\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
\[a\left( {ax} \right) + a\left( {by} \right) + {\text{a}}\left( c \right)-b\left( {bx} \right) - b\left( {ay} \right) - b\left( c \right) = 0\]
\[ \Rightarrow \]\[{a^2}x + aby + {\text{a}}c-{b^2}x - bay - bc = 0\]
\[ \Rightarrow \]\[x\left( {{a^2}-{b^2}} \right) + aby - aby + c\left( {a - b} \right) = 0\]
\[ \Rightarrow \]\[x\left( {{a^2}-{b^2}} \right) + c\left( {a - b} \right) = 0\]
As we know that \[\left( {{a^2}-{b^2}} \right) = \left( {a - b} \right)\left( {a + b} \right)\]
Therefore, \[x\left( {a-b} \right)\left( {a + b} \right) + c\left( {a - b} \right) = 0\]
\[ \Rightarrow \]\[\left( {a-b} \right)\left( {x\left( {a + b} \right) + c} \right) = 0\]
\[ \Rightarrow \]\[x\left( {a + b} \right) + c = 0\]
\[ \Rightarrow \]\[x = \dfrac{{ - c}}{{\left( {a + b} \right)}}\]
Substituting the value of x in the equation 1 we get
\[a\dfrac{{ - c}}{{\left( {a + b} \right)}}\; + by + c = 0\]
\[ \Rightarrow \]\[\dfrac{{ - ca + by\left( {a + b} \right) + c\left( {a + b} \right)}}{{\left( {a + b} \right)}} = 0\]
\[ \Rightarrow \]\[ - ca + bay + {b^2}y + ca + cb = 0\]
\[ \Rightarrow \]\[bay + {b^2}y + cb = 0\]
\[ \Rightarrow \]\[b\left( {ay + by + c} \right) = 0\]
\[ \Rightarrow \]\[y\left( {a + b} \right) + c = 0\]
\[ \Rightarrow \]\[y = \dfrac{{ - c}}{{\left( {a + b} \right)}}\]
Therefore, (x, y) = \[\left( {\dfrac{{ - c}}{{\left( {a + b} \right)}},\dfrac{{ - c}}{{\left( {a + b} \right)}}} \right)\]
Since, we know that distance between the intersection point of 2 line and (1, 1) is less than $2\sqrt 2 $ i.e. d < $2\sqrt 2 $
Also we know that distance between two points is given by $d = \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
\[\sqrt {{{\left( {1 - \left( {\dfrac{{ - c}}{{a + b}}} \right)} \right)}^2} + {{\left( {1 - \left( {\dfrac{{ - c}}{{a + b}}} \right)} \right)}^2}} < 2\sqrt 2 \]
\[ \Rightarrow \]\[\sqrt {{{\left( {\dfrac{{a + b + c}}{{a + b}}} \right)}^2} + {{\left( {\dfrac{{a + b + c}}{{a + b}}} \right)}^2}} < 2\sqrt 2 \]
\[ \Rightarrow \]\[\sqrt {2{{\left( {\dfrac{{a + b + c}}{{a + b}}} \right)}^2}} < 2\sqrt 2 \]
\[ \Rightarrow \]\[\sqrt 2 \left( {\dfrac{{a + b + c}}{{a + b}}} \right) < 2\sqrt 2 \]
\[ \Rightarrow \]\[a + b + c < 2a + 2b\]
\[ \Rightarrow \]\[0 < 2a + 2b - a - b - c\]
\[ \Rightarrow \]\[a + b - c > 0\]
Therefore, a + b – c > 0 when the distance between point of intersection of lines and (1, 1) is less than $2\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 = \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.