The product of the perpendicular distances fr | vedclass

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

Question

(C) The given pair of lines is $x^2 - 4xy + y^2 = 0$. Comparing this with the general form $ax^2 + 2hxy + by^2 = 0$,we have $a = 1$,$h = -2$,and $b =

Vedclass · Accepted Answer

$\frac{3}{2}$

Vedclass · Answer

(C) The given pair of lines is $x^2 - 4xy + y^2 = 0$. Comparing this with the general form $ax^2 + 2hxy + by^2 = 0$,we have $a = 1$,$h = -2$,and $b = 1$. The perpendicular distance from a point $(x_1, y_1)$ to the lines $L_1: a_1x + b_1y + c_1 = 0$ and $L_2: a_2x + b_2y + c_2 = 0$ is given by $d_1 = \frac{|a_1x_1 + b_1y_1 + c_1|}{\sqrt{a_1^2 + b_1^2}}$ and $d_2 = \frac{|a_2x_1 + b_2y_1 + c_2|}{\sqrt{a_2^2 + b_2^2}}$. The product of distances is $d_1 d_2 = \frac{|ax_1^2 + 2hx_1y_1 + by_1^2|}{\sqrt{(a-b)^2 + 4h^2}}$. Substituting the values $x_1 = 1, y_1 = -1, a = 1, h = -2, b = 1$: $d_1 d_2 = \frac{|1(1)^2 + 2(-2)(1)(-1) + 1(-1)^2|}{\sqrt{(1-1)^2 + 4(-2)^2}} = \frac{|1 + 4 + 1|}{\sqrt{0 + 16}} = \frac{6}{4} = \frac{3}{2}$.

The product of the perpendicular distances from $(1, -1)$ to the pair of lines $x^2 - 4xy + y^2 = 0$ is:

Similar Questions

If the equation $3x^{2}+10xy+3y^{2}+16y+k=0$ represents a pair of lines,then the value of $k$ is

The equation $2x^2 + 4xy - ky^2 + 4x + 2y - 1 = 0$ represents a pair of lines. The value of $k$ is

The equation of one of the lines represented by the equation $x^2 + 2xy \cot \theta - y^2 = 0$ is

The area of the triangle formed by the pair of lines $23x^2 - 48xy + 3y^2 = 0$ with the line $2x + 3y + 5 = 0$ is

If $x^2+\alpha y^2+2 \beta y=a^2$ represents a pair of perpendicular lines,then $\beta$ equals to

Vedclass Test Series

Exam Paper Generator

Online Exam Module