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Question: If a + b + c = 0 and \(p \ne 0\) the lines \(ax + (b + c)y = p,bx + (c + a)y = p\) and \(cx + (a + b...

If a + b + c = 0 and p0p \ne 0 the lines ax+(b+c)y=p,bx+(c+a)y=pax + (b + c)y = p,bx + (c + a)y = p and cx+(a+b)y=pcx + (a + b)y = p
A). Do not intersect
B). Intersect
C). Are concurrent
D). None of these

Explanation

Solution

First, we can solve the given equations into generalized forms, then we compare the equations and try to simplify them. After that, we use the definitions of intersecting, do not intersect and concurrent to solve the requirements.
From the given we have a+b+c=0a + b + c = 0 and p0p \ne 0 are the two pieces of information.

Complete step-by-step solution:
Since from the problem given that we have three equations, which are ax+(b+c)y=p,bx+(c+a)y=pax + (b + c)y = p,bx + (c + a)y = p and cx+(a+b)y=pcx + (a + b)y = p
We see that all the equations are equal to pp only, so it is easy to solve and compare.
Take equation one, ax+(b+c)y=pax + (b + c)y = p then giving the terms to inside then we have ax+(b+c)y=pax+by+cy=pax + (b + c)y = p \Rightarrow ax + by + cy = p
Take equation two, bx+(c+a)y=pbx + (c + a)y = p then giving the terms to inside then we have bx+(c+a)y=pbx+cy+ay=pbx + (c + a)y = p \Rightarrow bx + cy + ay = p
Take equation three, cx+(a+b)y=pcx + (a + b)y = p then giving the terms to inside then we have cx+(a+b)y=pcx+ay+by=pcx + (a + b)y = p \Rightarrow cx + ay + by = p
Hence, we have three simplified equations, which are ax+by+cy=p,bx+cy+ay=p,cx+ay+by=pax + by + cy = p,bx + cy + ay = p,cx + ay + by = p
Now add all these equations we get ax+by+cy+bx+cy+ay+cx+ay+by=p+p+pax + by + cy + bx + cy + ay + cx + ay + by = p + p + p
Further solving on taking the common terms, we get ax+by+cy+bx+cy+ay+cx+ay+by=p+p+p(a+b+c)(x+y+y)=3pax + by + cy + bx + cy + ay + cx + ay + by = p + p + p \Rightarrow (a + b + c)(x + y + y) = 3p
Since we know that a+b+c=0a + b + c = 0 then we get (a+b+c)(x+y+y)=3p0.(x+y+y)=3pp=0(a + b + c)(x + y + y) = 3p \Rightarrow 0.(x + y + y) = 3p \Rightarrow p = 0 but since from the given we have p0p \ne 0.
A set of three or more lines are termed concurrent when they pass through one common point or coincide exactly at one point The common point where all coincide lines is known as the point of concurrency. Thus, its determinant or the values of the coordinate points are non-zero.
Thus, the lines are concurrent.
Therefore, option C) is concurrent is correct.

Note: Not-intersecting lines are similar to the parallel lines as they will never meet and also, they have the same distance from one another.
One point is acting to the other points is known as the point intersect.
Since we are also able to solve the given problem, by finding its determinant values with the help of a substituting method. As the determinant will be only zero and also it fails the p0p \ne 0 and thus it is concurrent.
b+c=ab + c = - a then axay=pax - ay = p and similarly find all the terms and taking the 3×33 \times 3 determinant will get zero.