If two lines x+(a-1)y=1 and 2x+a^2y=1 (a in R | vedclass

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

Question

(D) Given equations of lines: $x+(a-1)y=1$ and $2x+a^2y=1$. Slope of $x+(a-1)y=1$ is $m_1 = \frac{-1}{a-1}$. Slope of $2x+a^2y=1$ is $m_2 = \frac{-2}{

Vedclass · Accepted Answer

$\sqrt{\frac{2}{5}}$

Vedclass · Answer

(D) Given equations of lines: $x+(a-1)y=1$ and $2x+a^2y=1$. Slope of $x+(a-1)y=1$ is $m_1 = \frac{-1}{a-1}$. Slope of $2x+a^2y=1$ is $m_2 = \frac{-2}{a^2}$. Since the lines are perpendicular,$m_1 	imes m_2 = -1$. $\frac{-1}{a-1} 	imes \frac{-2}{a^2} = -1$ $\frac{2}{a^2(a-1)} = -1$ $2 = -a^3 + a^2$ $a^3 - a^2 + 2 = 0$. Factoring the cubic equation: $(a+1)(a^2-2a+2) = 0$. Since $a^2-2a+2 = (a-1)^2 + 1 > 0$ for all real $a$,we must have $a+1=0$,so $a=-1$. Substituting $a=-1$ into the line equations: $x+(-1-1)y=1 \Rightarrow x-2y=1$. $2x+(-1)^2y=1 \Rightarrow 2x+y=1$. Solving the system: $x-2y=1$ $(i)$ $2x+y=1$ (ii) Multiply (ii) by $2$: $4x+2y=2$. Adding $(i)$ and (ii): $5x=3 \Rightarrow x=\frac{3}{5}$. Substituting $x=\frac{3}{5}$ into (ii): $2(\frac{3}{5})+y=1 \Rightarrow y=1-\frac{6}{5} = -\frac{1}{5}$. The point of intersection is $(\frac{3}{5}, -\frac{1}{5})$. The distance from the origin $(0,0)$ is $\sqrt{(\frac{3}{5})^2 + (-\frac{1}{5})^2} = \sqrt{\frac{9}{25} + \frac{1}{25}} = \sqrt{\frac{10}{25}} = \sqrt{\frac{2}{5}}$.

If two lines $x+(a-1)y=1$ and $2x+a^2y=1$ $(a \in R-\{0,1\})$ are perpendicular,then the distance of their point of intersection from the origin is

Similar Questions

Consider the straight lines:
$L_1 : x - y = 1$
$L_2 : x + y = 1$
$L_3 : 2x + 2y = 5$
$L_4 : 2x - 2y = 7$
The correct statement is:

The distance of a point $(2, 5)$ from the line $3x + y + 4 = 0$ measured along the lines $L_1$ and $L_2$ are the same. If the slope of line $L_1$ is $\frac{3}{4}$,then the slope of the line $L_2$ is

If the lines $y = 3x + 1$ and $2y = x + 3$ are equally inclined to the line $y = mx + 4$,then $m =$ ?

Let the angle between the lines $x-2y+3=0$ and $kx-y+2=0$ be $45^{\circ}$. If $k_1, k_2$ $(k_1 > k_2)$ are two distinct real values of $k$,then $k_1-2=$

If $A(-3,3), B(1,1), C(1,-1)$ and $D(-2,-2)$ are the vertices of a quadrilateral,the angle between the diagonals $AC$ and $BD$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module