The product of the perpendicular distances fr | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given the equation of the pair of straight lines: $12x^2 + 25xy + 12y^2 + 10x + 11y + 2 = 0$ $(i)$ First,factorize the homogeneous part: $12x^2 +

Vedclass · Accepted Answer

$\frac{2}{25}$

Vedclass · Answer

(B) Given the equation of the pair of straight lines: $12x^2 + 25xy + 12y^2 + 10x + 11y + 2 = 0$ $(i)$ First,factorize the homogeneous part: $12x^2 + 25xy + 12y^2 = (3x + 4y)(4x + 3y) = 0$. Let the two lines be $(3x + 4y + c_1) = 0$ and $(4x + 3y + c_2) = 0$. Their product is $(3x + 4y + c_1)(4x + 3y + c_2) = 12x^2 + 25xy + 12y^2 + (4c_1 + 3c_2)x + (3c_1 + 4c_2)y + c_1c_2 = 0$. Comparing this with equation $(i)$: $4c_1 + 3c_2 = 10$ $3c_1 + 4c_2 = 11$ Solving these equations: Multiply the first by $4$ and second by $3$: $16c_1 + 12c_2 = 40$ $9c_1 + 12c_2 = 33$ Subtracting gives $7c_1 = 7 \Rightarrow c_1 = 1$. Substituting $c_1 = 1$ into $4(1) + 3c_2 = 10$ $\Rightarrow 3c_2 = 6$ $\Rightarrow c_2 = 2$. The lines are $3x + 4y + 1 = 0$ and $4x + 3y + 2 = 0$. The perpendicular distances from the origin $(0,0)$ are: $p_1 = \frac{|0 + 0 + 1|}{\sqrt{3^2 + 4^2}} = \frac{1}{5}$ $p_2 = \frac{|0 + 0 + 2|}{\sqrt{4^2 + 3^2}} = \frac{2}{5}$ The product of the distances is $p_1 \cdot p_2 = \frac{1}{5} \cdot \frac{2}{5} = \frac{2}{25}$.

The product of the perpendicular distances from the origin to the pair of straight lines $12x^2 + 25xy + 12y^2 + 10x + 11y + 2 = 0$ is

Similar Questions

The lines represented by the equation $x^2-y^2-x+3y-2=0$ are :

The equation $(x - 5)^2 + (x - 5)(y - 6) - 2(y - 6)^2 = 0$ represents

The joint equation of two lines passing through $(-2, 3)$ and parallel to the bisectors of the angle between the co-ordinate axes is

If $4ab = 3h^2$,then the ratio of slopes of the lines represented by $ax^2 + 2hxy + by^2 = 0$ is

The equations of the lines represented by the equation $ax^2 + (a + b)xy + by^2 + x + y = 0$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module