Two points P(a, 2) and Q(1, b) lie on either | vedclass

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

Question

(B) Since $P(a, 2)$ and $Q(1, b)$ lie on either side of the line $2x - 3y + 1 = 0$,the product of the values obtained by substituting the points into

Vedclass · Accepted Answer

$(-\infty, 1)$

Vedclass · Answer

(B) Since $P(a, 2)$ and $Q(1, b)$ lie on either side of the line $2x - 3y + 1 = 0$,the product of the values obtained by substituting the points into the line equation must be negative: $(2a - 3(2) + 1)(2(1) - 3b + 1) $(2a - 5)(3 - 3b) $(2a - 5)(1 - b) Since $P(a, 2)$ is the intersection of $4x + 3y + k = 0$ and $3x + 4y + k = 0$: $4a + 3(2) + k = 0 \Rightarrow 4a + 6 + k = 0$ $3a + 4(2) + k = 0 \Rightarrow 3a + 8 + k = 0$ Subtracting the two equations: $(4a + 6 + k) - (3a + 8 + k) = 0$ $\Rightarrow a - 2 = 0$ $\Rightarrow a = 2$. Substituting $a = 2$ into the inequality: $(2(2) - 5)(1 - b) $(4 - 5)(1 - b) $-(1 - b) $b - 1 Thus,$b \in (-\infty, 1)$.

Two points $P(a, 2)$ and $Q(1, b)$ lie on either side of the line $2x - 3y + 1 = 0$. If $P$ is the point of intersection of the lines $4x + 3y + k = 0$ and $3x + 4y + k = 0$,then the range of $b$ is

Similar Questions

If $O$ is the origin and $P, Q$ are points on the line $3x + 4y + 15 = 0$ such that $OP = OQ = 9$, then the area of $\triangle OPQ$ is (in $\sqrt{2}$)

$A$ straight line makes an intercept on the $Y$-axis twice as long as that on the $X$-axis and is at a unit distance from the origin. Then the line is represented by the equations:

The distance between the two parallel lines $y = 2x + 7$ and $y = 2x + 5$ is:

The number of points on the line $x+y=4$ which are at a unit distance from the line $2x+2y=5$ is

Find the distance of the point $(-1, 1)$ from the line $12(x + 6) = 5(y - 2)$. (in $units$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module