Question

If ${a^2} + {b^2} - {c^2} - 2ab = 0$, then the given family of straight lines $ax + by + c = 0$, is concurrent at the points?
A. (-1,1)
B. (1,-1)
C. (1,1)
D. (-1,-1)

Answer 1

Here we are given with the general equation and a condition is given to solve the question, here first we need to solve with the condition given then combining both condition and equation we can reach the final result. We know that concurrent lines are the lines which exactly intersect at a point.

Complete step-by-step answer:
Here to solve the given equation first we need to solve for the condition given, on solving and simplifying we get:

\Rightarrow {(a - b)^2} - {c^2} = 0 $$ Here we get the equation in the general form of: $$ \Rightarrow {a^2} - {b^2} = (a - b)(a + b)$$ On solving by this method we get: $$ \Rightarrow (a - b - c)(a - b + c) = 0 \\\ \Rightarrow a = b + c\,or\,b = a + c $$ Substituting it in the equation of the line we get: Putting the value of “a”, in the equation of straight line we get: $$ \Rightarrow ax + by + c = 0 \\\ \Rightarrow (b + c)x + by + c = 0 \\\ \Rightarrow bx + cx + by + c = 0 \\\ \Rightarrow b(x + y) + c(x + 1) = 0 $$ Here we can see that: Either, $$x + y = 0,\,or\,x + 1 = 0$$ Now from here we get that, $$ \Rightarrow x = - y$$ which is the equation for a straight line with negative slope. Hence the concurrency at the points will be, $$ \Rightarrow (1, - 1)$$ because it will satisfy the equation of the straight line we get here. **Note:** Here we see that the pair of straight lines that we see here need to be solved by adjusting the equations and the condition given to us, and similarly the questions related to the coordinate geometry are needed to be solved by the relation given.