If p_1 and p_2 denote the lengths of perpendi | vedclass

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

Question

(C) The given equation of the pair of lines is $15 x^2 + 31 x y + 14 y^2 = 0$. Factoring the quadratic expression: $15 x^2 + 10 x y + 21 x y + 14 y^2

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) The given equation of the pair of lines is $15 x^2 + 31 x y + 14 y^2 = 0$. Factoring the quadratic expression: $15 x^2 + 10 x y + 21 x y + 14 y^2 = 0$ $\Rightarrow 5 x(3 x + 2 y) + 7 y(3 x + 2 y) = 0$ $\Rightarrow (3 x + 2 y)(5 x + 7 y) = 0$. Thus,the two lines are $L_1: 3 x + 2 y = 0$ and $L_2: 5 x + 7 y = 0$. The perpendicular distance from $(2,3)$ to $3 x + 2 y = 0$ is $d_1 = \frac{|3(2) + 2(3)|}{\sqrt{3^2 + 2^2}} = \frac{12}{\sqrt{13}}$. The perpendicular distance from $(2,3)$ to $5 x + 7 y = 0$ is $d_2 = \frac{|5(2) + 7(3)|}{\sqrt{5^2 + 7^2}} = \frac{31}{\sqrt{74}}$. Since $p_1 > p_2$,we have $p_1 = \frac{31}{\sqrt{74}}$ and $p_2 = \frac{12}{\sqrt{13}}$. Now,calculating $p_1^2 + \frac{1}{74} - p_2^2 + \frac{1}{13} = \frac{31^2}{74} + \frac{1}{74} - \frac{12^2}{13} + \frac{1}{13} = \frac{961+1}{74} - \frac{144-1}{13} = \frac{962}{74} - \frac{143}{13} = 13 - 11 = 2$.

If $p_1$ and $p_2$ denote the lengths of perpendiculars from $(2,3)$ onto the lines given by $15 x^2+31 x y+14 y^2=0$,and if $p_1 > p_2$,then $p_1^2 + \frac{1}{74} - p_2^2 + \frac{1}{13}$ is equal to

Similar Questions

If the slope of one of the lines in the pair of lines $8x^2 + axy + y^2 = 0$ is thrice the slope of the other line,then $a =$

If $ax^2 - y^2 + 4x - y = 0$ represents a pair of lines,then $a = $

If the equation $ax^{2} + by^{2} + cx + cy = 0$,$c \neq 0$ represents a pair of lines,then

The equation $2y^2 - xy - x^2 + 6x - 8 = 0$ represents

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module