Straight Lines
Straight Line
Allen Star Batch
Grade 11
Question:
MATCH THE FOLLOWING:
(A) If $P\left(1+\frac{t}{\sqrt{2}}, 2+\frac{t}{\sqrt{2}}\right)$ be any point on a line then value of $t$ for which the point $P$ lies between parallel lines $x+2y=1$ and $2x+4y=15$ is
(B) If the point $(2x_1-x_2+t(x_1-x_2), 2y_1-y_2+t(y_2-y_1))$ divides the join of $(x_1,y_1)$ and $(x_2,y_2)$ internally, then $t$ lies in
(C) If the point $(1, 1)$ always remains in the interior of the triangle formed by the lines $y=x, y=0$ and $x+y=4$, then $t$ lies in
(D) Set of values of '$t$' for which the point $P(t, t^2-2)$ lies inside the triangle formed by lines $x+y=1, y=x+1$ and $y=-1$ is
Step-by-Step Solution
Key Concept: Use parametric representation of a point on a line and apply geometric/algebraic constraints from multiple conditions simultaneously to determine valid parameter ranges.
Point $P$ lies on line $y = x + 1$ and must satisfy conditions relative to two other lines. For option (A), we find the coordinates of $P$ as $P(1+\frac{t}{\sqrt{2}}, 2+\frac{t}{\sqrt{2}})$ where $t$ is a parameter. Using the condition that $P$ and origin are on the same side of line $QR$ but opposite sides of line $ST$, we obtain two inequalities that together give $t \in (\frac{-4\sqrt{2}}{3}, \frac{5\sqrt{2}}{6})$. For option (B), point $P$ divides the segment joining $(x_1, y_1)$ and $(x_2, y_2)$ internally in ratio $(t-1):(2-t)$, requiring $(t-1)(2-t) > 0$, which means $t \in (1,2)$. Option (C) shows from the figure that $t \in (0,1)$. Option (D) finds intersections of $y = x + 1$ with the parabola $x^2 - 2 = x + 1$, giving $x = \frac{1 \pm \sqrt{13}}{2}$ and thus $t \in (\frac{1-\sqrt{13}}{2}, -1) \cup (\frac{\sqrt{13}-1}{2})$.
Correct Answer: [A-r] [B-p] [C-s] [D-q]