Let R be the interior region between the line | vedclass

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

Question

(B) Let $P = (a^2, a+1)$.For line $L_1: 3x-y+1=0$,the origin $(0,0)$ gives $L_1(0,0) = 3(0)-0+1 = 1 > 0$.Since $P$ lies on the same side as the origin

Vedclass · Accepted Answer

$(-3,0) \cup (\frac{1}{3}, 1)$

Vedclass · Answer

(B) Let $P = (a^2, a+1)$.For line $L_1: 3x-y+1=0$,the origin $(0,0)$ gives $L_1(0,0) = 3(0)-0+1 = 1 > 0$.Since $P$ lies on the same side as the origin,$L_1(a^2, a+1) > 0$.$3(a^2) - (a+1) + 1 > 0$ $\Rightarrow 3a^2 - a > 0$ $\Rightarrow a(3a-1) > 0$.Thus,$a \in (-\infty, 0) \cup (\frac{1}{3}, \infty) \dots (i)$.For line $L_2: x+2y-5=0$,the origin $(0,0)$ gives $L_2(0,0) = 0+2(0)-5 = -5 Since $P$ lies on the same side as the origin,$L_2(a^2, a+1) $a^2 + 2(a+1) - 5 Thus,$a \in (-3, 1) \dots (ii)$.Taking the intersection of $(i)$ and $(ii)$:$a \in (-3, 0) \cup (\frac{1}{3}, 1)$.

Let $R$ be the interior region between the lines $3x-y+1=0$ and $x+2y-5=0$ containing the origin. The set of all values of $a$,for which the points $(a^2, a+1)$ lie in $R$,is :

Similar Questions

Find the length of the perpendicular drawn from the origin to the line $\frac{x}{a} + \frac{y}{b} = 1$.

The distance between the lines $3x + 4y = 9$ and $6x + 8y = 15$ is

The base of an equilateral triangle is represented by the equation $2x - y - 1 = 0$ and its vertex is $(1, 2)$. Then,the length (in units) of the side of the triangle is:

If the points $(1, 2)$ and $(3, 4)$ lie on the same side of the straight line $3x - 5y + a = 0$,then $a$ lies in the set

The length of the perpendicular drawn from the origin $(0, 0)$ to the line joining the points $(x', y')$ and $(x'', y'')$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module