zhang25
contestada

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.​

Respuesta :

SalmonWorldTopper

[tex]\large\underline{\sf{Solution-}}[/tex]

First equation is

[tex]\rm \longmapsto\:(x - a)(y - b) = (x - 2a)\bigg(y -\dfrac{b}{2}\bigg) [/tex]

can be further simplified to

[tex]\rm \longmapsto\:xy - bx - ay + ab = xy - \dfrac{bx}{2} - 2ay + ab[/tex]

[tex]\rm \longmapsto\: - bx - ay = - \dfrac{bx}{2} - 2ay [/tex]

[tex]\rm \longmapsto\: - bx + \dfrac{bx}{2} = - 2ay + ay[/tex]

[tex]\rm \longmapsto\: - \dfrac{bx}{2} = - ay[/tex]

[tex]\rm \longmapsto\: \dfrac{bx}{2} = ay[/tex]

[tex]\rm \longmapsto\:bx = 2ay[/tex]

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

Second equation is

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

[tex]\rm \longmapsto\: {x}^{2} + \dfrac{x}{2b} + {y}^{2} + \dfrac{ay}{2} - 2xy = 5 + {x}^{2} + {y}^{2} - 2xy[/tex]

[tex]\rm \longmapsto\: \dfrac{x}{2b} + \dfrac{ay}{2} = 5[/tex]

[tex]\rm \longmapsto\: \dfrac{x + aby}{2b} = 5[/tex]

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

So, we have two equations in simplest form as

[tex]\rm \longmapsto\:bx - 2ay = 0 [/tex]

and

[tex]\rm \longmapsto\:x + aby = 10b[/tex]

Now, Consider

[tex]\rm \longmapsto\:\dfrac{a_1}{a_2} = \dfrac{b}{1} = b[/tex]

[tex]\rm \longmapsto\:\dfrac{b_1}{b_2} = \dfrac{ - 2a}{ab} = - \dfrac{2}{b} [/tex]

[tex]\bf\implies \:\dfrac{a_1}{a_2} \: \ne \: \dfrac{b_1}{b_2} [/tex]

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

So, The student is correct as lines are intersecting.

So, option (b) is correct.

Otras preguntas