The two pairs of straight lines 12x²+7xy-12y²=0 and 12x²+7xy-12y²-x+7y-1=0 constitute a

Correct Answer is : square of area 1251/25251 sq.units

rectangle of area 1/10 sq.units

rectangle of area 1/15 sq.units

Correct Answer is : square of area 1251/25251 sq.units

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AP EAPCET 06-Jul-2022 Shift 2 Paper

Section: Mathematics

Question : 45 of 160

Marks: +1, -0

12x^{2}+7xy-12y^{2}=0

12x^{2}+7xy-12y^{2}-x+7y-1=0

constitute a

Solution: 👈: Video Solution

12x^2+7xy-12y^2=0

12x^2+16xy-9xy-12y^2=0

4x(3x+4y)-3y(3x+4y)=0

(4x-3y)(3x+4y)=0

4x-3y=0

and

3x+4y=0

Slope of the two lines are

\frac{4}{3}

and

\frac{-3}{4}

, respectively.

The two lines are perpendicular.

12x^2+7xy-12y^2-x+7y-1=0

12x^2+x(7y-1)-12y^2+7y-1=0

x=-\frac{b+\sqrt{b^2-4ac}}{2a}

x=-\frac{(7y-1) \pm \sqrt{(7y-1)^2-4\times 12(-12y^2+7y-1)}}{2\times 12}

x=\frac{1-7y \pm \sqrt{49y^2+1-14y-336y+48+576y^2}}{24}

\Rightarrow 24x=1-7y \pm \sqrt{625y^2-350y+49}

\Rightarrow 24x=1-7y \pm (25y-7)

\Rightarrow 24x=1-7y+25y-7

24x=1-7y-25y+7

\Rightarrow 24x-18y+6=0

24x+32y-8=0

m_3=\frac{24}{18}

and

m_4=\frac{-24}{32}

\Rightarrow m_3=\frac{4}{3}

and

m_4=\frac{-3}{4}

Again,

m_3 \times m_4=-1

, the two lines are perpendicular.

The two lines

3x+4y=0

and

4x-3y=0

intersect at a point

(0,0)

d_1

, and

d_2

are the distance of a point

(0,0)

to the line

24x-18y+6=0

and

24x+32y-8=0

d_1=\frac{6}{\sqrt{(24)^2+(18)^2}}, d_2=\frac{8}{\sqrt{(24)^2+(32)^2}}

d_1=\frac{6}{30}=\frac{1}{5}, d_2=\frac{8}{40}=\frac{1}{5}

\therefore

The given figure is a square of side

\frac{1}{5}

unit. Area of square

=\frac{1}{25}

sq. units

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