Question

A line meets x-axis and y-axis at A and B respectively and O is the origin.

$\begin{array}{cccccc}& \text{Column I}& & \text{Column 2}& & \text{Column 3}\\ & & & \text{Equation of AB}& & \text{Area of}\Delta OAB\\ \text{(I)}& \text{Centroid}\Delta OAB\text{is (1, 2)}& \text{(i)}& 2x+y=2& \text{(P)}& \text{6 sq. units}\\ \text{(II)}& \text{Circumcenter of}\Delta OAB\text{is (1, 2)}& \text{(ii)}& 3x+4y=12& \text{(Q)}& \text{9 sq. units}\\ \text{(III)}& \text{Distance of the orthocentre of}\Delta OAB& \text{(iii)}& 2x+y=6& \text{(R)}& \text{1 sq. units}\\ & \text{From A and B is 1 and 2 respectively}& & & & \\ \text{(IV)}& \text{Incenter of}\Delta OAB\text{is (1, 1)}& \text{(iv)}& 2x+y=4& \text{(S)}& \text{4 sq. units}\end{array}$

Which of the following is correct combination?

• A. (II), (i), (R)
• B. (II), (ii), (P)
• C. (II), (iii), (Q)
• D. (II), (iv), (S)

Answer 1

The correct option is D

(II), (iv), (S)

Let OA = a, OB = b $\therefore A\equiv (a,0)$ and $B\equiv (0,b)$

(I)

$\Delta OAB\equiv (\frac{0+a+0}{3},\frac{0+0+b}{3})\equiv (\frac{a}{3},\frac{b}{3})\equiv (1,2)$

$\therefore$ a = 3, b = 6

$\therefore$ Equation of AB $=\frac{x}{3}+\frac{y}{6}=1\Rightarrow 2x+y=6$

(II) Circumcenter of $\Delta OAB$ is the middle point of the hypotenuse AB

$\Rightarrow (\frac{a}{2},\frac{b}{2})=(1,2)\Rightarrow a=2,b=4$

$\therefore$ Equation of AB =$\frac{x}{2}+\frac{y}{4}=1\Rightarrow 2x+y=4$

(III) Orthocenter of the triangle OAB is O, so that OA = 1 and OB = 2

$\therefore$ a = 1, b = 2

$\therefore$ Equation of AB is $(\frac{x}{1}+\frac{y}{2})=1\Rightarrow 2x+y=2$

(IV) Distance of (1, 1) from OA or OB is 1

3x + 4y = 12 satisfy this condition as $\left|\frac{3\times 1+4\times 1-12}{\sqrt{3^2+4^2}}\right|=1$