NCERT Solutions · Class 12 Mathematics Relations and Functions

Let O denote the origin and write \(\displaystyle d(\mathrm{P})\) for the distance of a point P from O; thus \(\displaystyle (\mathrm{P},\mathrm{Q})\in\mathrm{R}\iff d(\mathrm{P})=d(\mathrm{Q})\). Each property follows from the corresponding property of equality of real numbers.Reflexive: \(\displaystyle d(\mathrm{P})=d(\mathrm{P})\) for every point P, so \(\displaystyle (\mathrm{P},\mathrm{P})\in\mathrm{R}\).Symmetric: if \(\displaystyle (\mathrm{P},\mathrm{Q})\in\mathrm{R}\), then \(\displaystyle d(\mathrm{P})=d(\mathrm{Q})\), hence \(\displaystyle d(\mathrm{Q})=d(\mathrm{P})\), i.e. \(\displaystyle (\mathrm{Q},\mathrm{P})\in\mathrm{R}\).Transitive: if \(\displaystyle d(\mathrm{P})=d(\mathrm{Q})\) and \(\displaystyle d(\mathrm{Q})=d(\mathrm{S})\), then \(\displaystyle d(\mathrm{P})=d(\mathrm{S})\), i.e. \(\displaystyle (\mathrm{P},\mathrm{S})\in\mathrm{R}\).Hence R is an equivalence relation.Now fix \(\displaystyle \mathrm{P}=(x_{0},y_{0})\ne(0,0)\) and put \(\displaystyle r=d(\mathrm{P})=\sqrt{x_{0}^{2}+y_{0}^{2}}>0\). A point \(\displaystyle \mathrm{Q}=(x,y)\) is related to P precisely when \[\sqrt{x^{2}+y^{2}}=r \quad\Longleftrightarrow\quad x^{2}+y^{2}=r^{2}, \] squaring being reversible here because both sides are non-negative. This is exactly the equation of the circle with centre at the origin and radius \(\displaystyle r\). Since \(\displaystyle d(\mathrm{P})=r\), the point P itself satisfies it.Therefore the set of all points related to \(\displaystyle \mathrm{P}\ne(0,0)\) is the circle \(\displaystyle x^{2}+y^{2}=r^{2}\) with centre the origin, passing through P.

Let A be the set of all triangles and \(\displaystyle \mathrm{R}=\{(\mathrm{T}_{1},\mathrm{T}_{2}):\mathrm{T}_{1}\text{ is similar to }\mathrm{T}_{2}\}\).Reflexive: every triangle is similar to itself (corresponding sides in the ratio \(\displaystyle 1:1\)), so \(\displaystyle (\mathrm{T},\mathrm{T})\in\mathrm{R}\).Symmetric: if \(\displaystyle \mathrm{T}_{1}\sim\mathrm{T}_{2}\) with ratio of corresponding sides \(\displaystyle k\), then \(\displaystyle \mathrm{T}_{2}\sim\mathrm{T}_{1}\) with ratio \(\displaystyle 1/k\) (equivalently, the two triangles have equal corresponding angles, and that condition is symmetric). So \(\displaystyle (\mathrm{T}_{2},\mathrm{T}_{1})\in\mathrm{R}\).Transitive: if \(\displaystyle \mathrm{T}_{1}\sim\mathrm{T}_{2}\) with ratio \(\displaystyle k_{1}\) and \(\displaystyle \mathrm{T}_{2}\sim\mathrm{T}_{3}\) with ratio \(\displaystyle k_{2}\), then corresponding sides of \(\displaystyle \mathrm{T}_{1}\) and \(\displaystyle \mathrm{T}_{3}\) are in the ratio \(\displaystyle k_{1}k_{2}\), so \(\displaystyle \mathrm{T}_{1}\sim\mathrm{T}_{3}\).Hence R is an equivalence relation.Now test the three triangles by the SSS-similarity criterion — arrange the sides in increasing order and compare the ratios. \(\displaystyle \mathrm{T}_{1}:3,4,5\); \(\displaystyle \mathrm{T}_{2}:5,12,13\); \(\displaystyle \mathrm{T}_{3}:6,8,10\).\(\displaystyle \mathrm{T}_{1}\) and \(\displaystyle \mathrm{T}_{3}\): \[\frac{3}{6}=\frac{4}{8}=\frac{5}{10}=\frac{1}{2}, \] all three ratios equal, so \(\displaystyle \mathrm{T}_{1}\sim\mathrm{T}_{3}\).\(\displaystyle \mathrm{T}_{1}\) and \(\displaystyle \mathrm{T}_{2}\): \(\displaystyle \dfrac{3}{5}\ne\dfrac{4}{12}\), so they are not similar.\(\displaystyle \mathrm{T}_{2}\) and \(\displaystyle \mathrm{T}_{3}\): \(\displaystyle \dfrac{5}{6}\ne\dfrac{12}{8}\), so they are not similar.Being right-angled is not enough for similarity; only \(\displaystyle \mathrm{T}_{1}\) and \(\displaystyle \mathrm{T}_{3}\) are related.

Let \(\displaystyle n(\mathrm{P})\) denote the number of sides of a polygon P, so \(\displaystyle (\mathrm{P}_{1},\mathrm{P}_{2})\in\mathrm{R}\iff n(\mathrm{P}_{1})=n(\mathrm{P}_{2})\). The three properties again reduce to properties of equality of numbers.Reflexive: \(\displaystyle n(\mathrm{P})=n(\mathrm{P})\), so \(\displaystyle (\mathrm{P},\mathrm{P})\in\mathrm{R}\) for every polygon P.Symmetric: \(\displaystyle n(\mathrm{P}_{1})=n(\mathrm{P}_{2})\Rightarrow n(\mathrm{P}_{2})=n(\mathrm{P}_{1})\), so \(\displaystyle (\mathrm{P}_{1},\mathrm{P}_{2})\in\mathrm{R}\Rightarrow(\mathrm{P}_{2},\mathrm{P}_{1})\in\mathrm{R}\).Transitive: \(\displaystyle n(\mathrm{P}_{1})=n(\mathrm{P}_{2})\) and \(\displaystyle n(\mathrm{P}_{2})=n(\mathrm{P}_{3})\Rightarrow n(\mathrm{P}_{1})=n(\mathrm{P}_{3})\).Hence R is an equivalence relation.The triangle T with sides $\displaystyle 3$, $\displaystyle 4$, $\displaystyle 5$ has \(\displaystyle n(\mathrm{T})=3\). A polygon P is related to T precisely when \(\displaystyle n(\mathrm{P})=3\). So the required set is the set of all triangles in A — every three-sided polygon, of any shape or size, not merely the right-angled ones and not merely those similar to T.

L is the set of all lines in the XY-plane and \(\displaystyle \mathrm{R}=\{(\mathrm{L}_{1},\mathrm{L}_{2}):\mathrm{L}_{1}\parallel\mathrm{L}_{2}\}\), with the usual convention that a line is parallel to itself.Reflexive: every line is parallel to itself, so \(\displaystyle (\mathrm{L},\mathrm{L})\in\mathrm{R}\) for every line L in the plane.Symmetric: if \(\displaystyle \mathrm{L}_{1}\parallel\mathrm{L}_{2}\), then \(\displaystyle \mathrm{L}_{2}\parallel\mathrm{L}_{1}\), so \(\displaystyle (\mathrm{L}_{2},\mathrm{L}_{1})\in\mathrm{R}\).Transitive: if \(\displaystyle \mathrm{L}_{1}\parallel\mathrm{L}_{2}\) and \(\displaystyle \mathrm{L}_{2}\parallel\mathrm{L}_{3}\), then \(\displaystyle \mathrm{L}_{1}\parallel\mathrm{L}_{3}\) (all three have the same direction, i.e. the same slope). So \(\displaystyle (\mathrm{L}_{1},\mathrm{L}_{3})\in\mathrm{R}\).Hence R is an equivalence relation.Lines related to \(\displaystyle y=2x+4\): this line has slope \(\displaystyle m=2\). A non-vertical line \(\displaystyle y=mx+c\) is parallel to it exactly when its slope equals 2. Therefore the required set is \[\{\,y=2x+c\ :\ c\in\mathbf{R}\,\}, \] the family of all lines of slope $\displaystyle 2$, which includes the line \(\displaystyle y=2x+4\) itself (the case \(\displaystyle c=4\)).

\(\displaystyle \mathrm{A}=\{1,2,3,4\}\) and \[\mathrm{R}=\{(1,1),(1,2),(1,3),(2,2),(3,2),(3,3),(4,4)\} \] (the given seven pairs, rearranged).Reflexive: R must contain \(\displaystyle (1,1),(2,2),(3,3),(4,4)\). All four are present, so R is reflexive.Symmetric: \(\displaystyle (1,2)\in\mathrm{R}\), but \(\displaystyle (2,1)\notin\mathrm{R}\). So R is not symmetric. This already rules out (A), (C) and (D).Transitive: check every chain \(\displaystyle (a,b),(b,c)\) in R. \(\displaystyle (1,1),(1,2)\Rightarrow(1,2)\in\mathrm{R}\); \(\displaystyle (1,1),(1,3)\Rightarrow(1,3)\in\mathrm{R}\); \(\displaystyle (1,2),(2,2)\Rightarrow(1,2)\in\mathrm{R}\); \(\displaystyle (1,3),(3,2)\Rightarrow(1,2)\in\mathrm{R}\); \(\displaystyle (1,3),(3,3)\Rightarrow(1,3)\in\mathrm{R}\); \(\displaystyle (3,2),(2,2)\Rightarrow(3,2)\in\mathrm{R}\); \(\displaystyle (3,3),(3,2)\Rightarrow(3,2)\in\mathrm{R}\); and the chains built from \(\displaystyle (2,2)\) and \(\displaystyle (4,4)\) alone reproduce themselves. Every required pair is present, so R is transitive.Answer: (B) R is reflexive and transitive but not symmetric.