Question

4.angle between 2 straight lines
Show that straight lines form isosceles triangle
4x - 3y -18 = 0, 3x-4y +16= 0, x + y - 2 = 0​

Answer 1

Answer:

y^2 + xy - 12x^2 = 0y

2

+xy−12x

2

=0

\Rightarrow y^2 + 4xy - 3xy - 12y^2 = 0⇒y

2

+4xy−3xy−12y

2

=0

\Rightarrow y(y + 4x) - 3x(y + 4x) = 0⇒y(y+4x)−3x(y+4x)=0

\Rightarrow (y + 4x)(y - 3x) = 0⇒(y+4x)(y−3x)=0

So, y = 3xy=3x or y = -4xy=−4x are the two straight lines represented by the given equation.

We consider two cases, the first being when y = 3xy=3x is common to both.

Then, we get ax^2 + 2hx(3x) + b(3x)^2 = 0ax

2

+2hx(3x)+b(3x)

2

=0

i.e. ax^2 + 6hx^2 + 9bx^2 = 0ax

2

+6hx

2

+9bx

2

=0

a = -6h - 9b = -3(2h + 3b)a=−6h−9b=−3(2h+3b)

The second case being line y = -4xy=−4x is common to both, which gives

ax^2 + 2hx (-4x) + b(-4x)^2 = 0ax

2

+2hx(−4x)+b(−4x)

2

=0

i.e. ax^2 - 8hx^2 + 16bx^2 = 0ax

2

−8hx

2

+16bx

2

=0

a = 8h - 16b = 8(h - 2b)a=8h−16b=8(h−2b)

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Step-by-step explanation: