Question

Let $AB: x+2y-3=0$, $BC: 2x-y+5=0$ and $AC: x-2=0$ be sides of $\triangle ABC$. Match the entries of List-I with the correct entries of List-II.

List-I	List-II
(P) Circumcentre of $\triangle ABC$, is	(1) $\left(-\frac{7}{5}, \frac{11}{5}\right)$
(Q) Centroid of $\triangle ABC$, is	(2) $\left(\frac{13}{5}, \frac{131}{30}\right)$
(R) Orthocentre of $\triangle ABC$, is	(3) $\left(2, \frac{19}{4}\right)$
(S) Vertex $B$, is	(4) $(1,1)$
	(5) The point which divides the distance between orthocentre and circumcentre in the ratio 2:1 internally

• A. P$\rightarrow$(1); Q$\rightarrow$(5); R$\rightarrow$(4); S$\rightarrow$(3)
• B. P$\rightarrow$(4); Q$\rightarrow$(5); R$\rightarrow$(2); S$\rightarrow$(3)
• C. P$\rightarrow$(3); Q$\rightarrow$(5); R$\rightarrow$(1); S$\rightarrow$(1)
• D. P$\rightarrow$(2); Q$\rightarrow$(2); R$\rightarrow$(3); S$\rightarrow$(1)

Answer 1

Answer: (C)

P$\rightarrow$(3); Q$\rightarrow$(5); R$\rightarrow$(1); S$\rightarrow$(1)

Answer 2

The vertices are A(2, 1/2), B(-7/5, 11/5), and C(2, 9). Since $m_{AB} = -1/2$ and $m_{BC} = 2$, AB is perpendicular to BC. Thus, $\triangle ABC$ is right-angled at B. Orthocentre (R) is the vertex B: $(-7/5, 11/5)$, which matches (1). Circumcentre (P) is the midpoint of hypotenuse AC: $((2+2)/2, (1/2+9)/2) = (2, 19/4)$, which matches (3). Centroid (Q) is the point dividing the segment joining orthocentre and circumcentre in the ratio 2:1 internally. This is the definition given in (5). Vertex B (S) is $(-7/5, 11/5)$, which matches (1). Thus, P$\rightarrow$(3); Q$\rightarrow$(5); R$\rightarrow$(1); S$\rightarrow$(1).