Straight Lines
Straight Line
Allen Star Batch
Grade 11
Question:
MATCH THE FOLLOWING:
(A) $P(3, 1), Q(6, 5)$ and $R(x, y)$ are three points such that angle $PRQ$ is right angle and the area of $\triangle PRQ$ is $7$, then number of such points $R$ is.
(B) Let $ABCD$ is a square with sides of unit length. Points $E$ and $F$ are taken on sides $AB$ and $AD$ respectively so that $AE=AF$. The maximum possible area of quadrilateral $CDFE$ is
(C) Let $A=(0,0), B=(5,0), C=(5,3)$ and $D=(0,3)$ are the vertices of rectangle $ABCD$. If $P$ is a variable point lying inside the rectangle $ABCD$ and $d(P, L)$ denote perpendicular distance of point $P$ from line $L$. If $d(P, AB)=\min\{d(P, BC), d(P, AD), d(P, CD)\}$, then area of the region in which $P$ lies is:
(D) The slope of one of lines given by $ax^2+2hxy+by^2=0$ be the square of the slope of the other, if $\phi(a+b)+\alpha bh+\beta h^2=0$, then $\alpha+\beta$ equals:
Step-by-Step Solution
Key Concept: A point inscribed in a circle with a diameter as one side forms a right angle, limiting the triangle's altitude to the radius.
(A) For a triangle inscribed in a circle with diameter $PQ$ where $P(3,1)$ and $Q(6,5)$, we find $PQ = 5$ and the maximum possible radius is $\frac{5}{2}$. Since the computed area constraint gives $h = \frac{14}{5} > \frac{5}{2}$, no such triangle can exist. (B) For the area of quadrilateral $CDEF$, we express $A(x) = 1 - \frac{1}{2}x^2 - \frac{1}{2}(1-x) = \frac{1+x-x^2}{2}$, which has maximum $A_{max} = \frac{5}{8}$ at $x = \frac{1}{2}$. (C) Point $P$ must satisfy three distance constraints simultaneously, lying in the intersection region with area $\frac{21}{4}$. (D) From $a + 2hm + bm^2 = 0$ with $m = \frac{y}{x}$, substitution and algebraic manipulation yields $a + b = 8$.
Correct Answer: [A-r] [B-s] [C-q] [D-p]