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Consider the equation as shown:

(x-a)(y-b) = (x-2a)(y- b/2)

x(x + 1/2b) + y(y + a/2) -2xy = 5+(x-y)²

On comparing the coefficient, a student says these pairs of equations is consistent. Is he/she correct? Which of these explains why?
(a) Yes; because they are parallel lines.
(b) No; because they are interesting lines.
(c) Yes; because they are parallel lines.
(d) No; because they are interesting lines.​

User Lior Iluz
by
6.7k points

1 Answer

4 votes


\large\underline{\sf{Solution-}}

First equation is


\rm \longmapsto\:(x - a)(y - b) = (x - 2a)\bigg(y -(b)/(2)\bigg)

can be further simplified to


\rm \longmapsto\:xy - bx - ay + ab = xy - (bx)/(2) - 2ay + ab


\rm \longmapsto\: - bx - ay = - (bx)/(2) - 2ay


\rm \longmapsto\: - bx + (bx)/(2) = - 2ay + ay


\rm \longmapsto\: - (bx)/(2) = - ay


\rm \longmapsto\: (bx)/(2) = ay


\rm \longmapsto\:bx = 2ay


\rm:\longmapsto \:\boxed{\tt{ bx - 2ay = 0}} - - - - (1)

Second equation is


\rm \longmapsto\:x\bigg(x + (1)/(2b) \bigg) + y \bigg(y + (a)/(2)\bigg) - 2xy = 5 + {(x - y)}^(2)


\rm \longmapsto\: {x}^(2) + (x)/(2b) + {y}^(2) + (ay)/(2) - 2xy = 5 + {x}^(2) + {y}^(2) - 2xy


\rm \longmapsto\: (x)/(2b) + (ay)/(2) = 5


\rm \longmapsto\: (x + aby)/(2b) = 5


\rm :\longmapsto\:\boxed{\tt{ x + aby = 10b}} - - - (2)

So, we have two equations in simplest form as


\rm \longmapsto\:bx - 2ay = 0

and


\rm \longmapsto\:x + aby = 10b

Now, Consider


\rm \longmapsto\:(a_1)/(a_2) = (b)/(1) = b


\rm \longmapsto\:(b_1)/(b_2) = ( - 2a)/(ab) = - (2)/(b)


\bf\implies \:(a_1)/(a_2) \: \\e \: (b_1)/(b_2)

This implies, System of equations is consistent having unique solution.

So, The student is correct as lines are intersecting.

So, option (b) is correct.

User TheLifeOfSteve
by
6.6k points
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