MATCH THE FOLLOWING:
(A) Two perpendicular straight lines are drawn from the origin to make an isosceles triangle together with the line $2x + y = 5$ Then the area of triangle is
(B) Let the line $2x + y = 4$ meet $x$-axis at $A$ and $y$-axis at $B$, and the perpendicular bisector of $AB$ meets the horizontal line through $(0, -1)$ at $C$. Let $G$ be the centroid of the triangle $ABC$. Then perpendicular distance from $G$ to $AB$ equals
(C) The number of integral points inside the triangle made by the line $3x + 4y - 12 = 0$ with the coordinate axes which are equidistant from at least two sides is/are (an integral point is a point both of whose coordinates are integers)
(D) The line $x = c$ cuts the triangle with corners $(0, 0), (1, 1)$ and $(9, 1)$ into two regions. For the area of the two regions to be the same $c$ must be equal to:
Step-by-Step Solution
Key Concept: Use properties of isosceles triangles, centroids, and incentres to find geometric relationships and distances in coordinate geometry.
(A) For an isosceles triangle with $OD = AD = BD$, we have $OD = P = rac{|s|}{\sqrt{5}}$. The area of $ riangle OAB = rac{1}{2}(2P)P = P^2 = 5$. (B) The slope of $CD$ is $ rac{1}{2}$ and $C = (-5, -1)$. Since $G$ is the centroid, the perpendicular distance from $G$ to $AB$ is $ rac{1}{3}$ times the perpendicular distance from $C$ to $AB$, which equals $\sqrt{5}$. (C) Point $P$ must lie on at least one angle bisector. The incentre is calculated as $I = (1, 1)$, so $P \equiv (1, 1)$ is the only valid answer. (D) Images of $A$ with respect to $y = x$ and $y = 0$ lie on $BC$ which is $y = 3x - 5$, and the perpendicular distance from $A$ to $BC$ is $ rac{4}{\sqrt{10}}$, giving $\sqrt{10}d(A, BC) = 4$.
Correct Answer: [A-q] [B-p] [C-s] [D-r]