If u = a_1x + b_1y + c_1 = 0,v = a_2x + b_2y | vedclass

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

Question

(A) Given $u = a_1x + b_1y + c_1 = 0$ and $v = a_2x + b_2y + c_2 = 0$.Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = \lambda$ (where $\l

Vedclass · Accepted Answer

The same straight line $u$

Vedclass · Answer

(A) Given $u = a_1x + b_1y + c_1 = 0$ and $v = a_2x + b_2y + c_2 = 0$.Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = \lambda$ (where $\lambda$ is a constant),we have $a_1 = \lambda a_2$,$b_1 = \lambda b_2$,and $c_1 = \lambda c_2$.Substituting these into the equation $u + kv = 0$:$(a_1x + b_1y + c_1) + k(a_2x + b_2y + c_2) = 0$$(\lambda a_2x + \lambda b_2y + \lambda c_2) + k(a_2x + b_2y + c_2) = 0$$(\lambda + k)(a_2x + b_2y + c_2) = 0$Since $a_2x + b_2y + c_2 = 0$ represents the line $v$,the equation $u + kv = 0$ simplifies to a multiple of the original line $v$ (or $u$),which is the same straight line.

If $u = a_1x + b_1y + c_1 = 0$,$v = a_2x + b_2y + c_2 = 0$ and $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$,then the curve $u + kv = 0$ is

Similar Questions

The equation of the line passing through $(a, b)$ and parallel to the line $\frac{x}{a} + \frac{y}{b} = 1$ is

If the normal form of the equation of a straight line $4x + 3y + 2 = 0$ is $x \cos \alpha + y \sin \alpha = p$ and its intercept form is $\frac{x}{a} + \frac{y}{b} = 1$,then $\frac{p \sec \alpha}{ab} = $

If the line $6x - 7y + 8 + \lambda(3x - y + 5) = 0$ is parallel to the $y$-axis,then $\lambda$ is equal to

The equation of the line which cuts off an intercept $3$ units on $OX$ and an intercept $-2$ units on $OY$ is:

$A$ line passes through $P(-4, 1)$ and meets the coordinate axes at points $A$ and $B$. If $P$ divides the segment $AB$ internally in the ratio $1:2$,then the equation of the line is

Vedclass Test Series

Exam Paper Generator

Online Exam Module