If the point (1, a) lies between the lines x | vedclass

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

Question

(A) The given lines are $L_1: x + y - 1 = 0$ and $L_2: 2x + 2y - 3 = 0$. For the point $(1, a)$ to lie between these two lines,the expressions $L_1(1,

Vedclass · Accepted Answer

$(0, 1/2)$

Vedclass · Answer

(A) The given lines are $L_1: x + y - 1 = 0$ and $L_2: 2x + 2y - 3 = 0$. For the point $(1, a)$ to lie between these two lines,the expressions $L_1(1, a)$ and $L_2(1, a)$ must have opposite signs. Let $f(x, y) = x + y - 1$ and $g(x, y) = 2x + 2y - 3$. At $(1, a)$,$f(1, a) = 1 + a - 1 = a$. At $(1, a)$,$g(1, a) = 2(1) + 2(a) - 3 = 2a - 1$. Since the point lies between the lines,$f(1, a) \cdot g(1, a) Therefore,$a(2a - 1) This inequality holds when $a$ is between the roots of $a(2a - 1) = 0$,which are $a = 0$ and $a = 1/2$. Thus,$a \in (0, 1/2)$.

If the point $(1, a)$ lies between the lines $x + y = 1$ and $2(x + y) = 3$,then in which interval does $a$ lie?

Similar Questions

The equation of the line passing through $(1, 2)$ and having a distance equal to $7$ units from the point $(8, 9)$ is:

The positions of the points $(3, 4)$ and $(2, -6)$ with respect to the line $3x - 4y = 8$ are:

The line $L$ given by $\frac{x}{5} + \frac{y}{b} = 1$ passes through the point $(13, 32)$. The line $K$ is parallel to $L$ and its equation is $\frac{x}{c} + \frac{y}{3} = 1$. Find the distance between $L$ and $K$.

The distance between the lines $3x + 4y = 9$ and $6x + 8y = 15$ is (in $\text{ units}$)

The perpendicular distance of the straight line $12x + 5y = 7$ from the origin is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module