Board Previous Year Questions

Board Previous Year Questions (PYQs)

These are exam-style questions modelled on CBSE and State Board papers from recent years. Each is fully solved with the reasoning a board examiner expects.

Scoring tip: In coordinate geometry, always write the formula first, then substitute. Examiners award method marks even if the arithmetic slips.

Work through all 26. They span 1-mark, 2-mark, 3-mark and 5-mark patterns.

PYQ 1 (1 mark): Find the distance between $(0, 0)$ and $(36, 15)$.

Solution:

1. $d = \sqrt{36^2 + 15^2} = \sqrt{1296 + 225} = \sqrt{1521} = 39$.

Final Answer: 39 units.

PYQ 2 (1 mark): Find the midpoint of $(-5, 7)$ and $(3, -1)$.

Solution:

1. $M = \left(\dfrac{-5+3}{2}, \dfrac{7-1}{2}\right) = (-1, 3)$.

Final Answer: $(-1, 3)$.

PYQ 3 (1 mark): Find the centroid of the triangle with vertices $(0, 6)$, $(8, 0)$, $(-2, 3)$.

Solution:

1. $G = \left(\dfrac{0+8-2}{3}, \dfrac{6+0+3}{3}\right) = (2, 3)$.

Final Answer: $(2, 3)$.

PYQ 4 (1 mark): In what ratio does the point $(-1, 6)$ divide the segment joining $(-3, 10)$ and $(6, -8)$?

Solution:

1. On x with ratio $k:1$: $-1 = \dfrac{6k - 3}{k+1} \Rightarrow -k - 1 = 6k - 3 \Rightarrow 7k = 2 \Rightarrow k = \tfrac27$.
2. Ratio $= 2 : 7$.

Final Answer: $2 : 7$.

PYQ 5 (2 marks): Find the value of $k$ for which the points $(2, 3)$, $(4, k)$, $(6, -3)$ are collinear.

Solution:

1. $2(k+3) + 4(-3-3) + 6(3-k) = 0$.
2. $2k + 6 - 24 + 18 - 6k = 0 \Rightarrow -4k = 0 \Rightarrow k = 0$.

Final Answer: $k = 0$.

PYQ 6 (2 marks): Find the area of the triangle whose vertices are $(2, 3)$, $(-1, 0)$, $(2, -4)$.

Solution:

1. Area $= \tfrac12|2(0+4) + (-1)(-4-3) + 2(3-0)|$.
2. $= \tfrac12|8 + 7 + 6| = \tfrac{21}{2} = 10.5$.

Final Answer: 10.5 square units.

PYQ 7 (2 marks): Find a point on the x-axis equidistant from $(2, -5)$ and $(-2, 9)$.

Solution:

1. Let $P(x, 0)$: $(x-2)^2 + 25 = (x+2)^2 + 81$.
2. $-4x + 29 = 4x + 85 \Rightarrow -8x = 56 \Rightarrow x = -7$.

Final Answer: $(-7, 0)$.

PYQ 8 (2 marks): Find the ratio in which the y-axis divides the segment joining $(5, -6)$ and $(-1, -4)$.

Solution:

1. On y-axis $x = 0$; ratio $k:1$: $0 = \dfrac{-k + 5}{k+1} \Rightarrow k = 5$.
2. Ratio $= 5 : 1$.

Final Answer: $5 : 1$.

PYQ 9 (2 marks): If the point $P(x, y)$ is equidistant from $A(5, 1)$ and $B(-1, 5)$, prove $3x = 2y$.

Solution:

1. $(x-5)^2 + (y-1)^2 = (x+1)^2 + (y-5)^2$.
2. $-10x + 25 - 2y + 1 = 2x + 1 - 10y + 25$.
3. $-10x - 2y = 2x - 10y \Rightarrow 8y = 12x \Rightarrow 3x = 2y$.

Final Answer: $3x = 2y$. Proved.

PYQ 10 (3 marks): Find the coordinates of the points of trisection of the segment joining $(4, -1)$ and $(-2, -3)$.

Solution:

1. $P$ ($1:2$): $\left(\dfrac{1(-2)+2(4)}{3}, \dfrac{1(-3)+2(-1)}{3}\right) = (2, -\tfrac53)$.
2. $Q$ ($2:1$): $\left(\dfrac{2(-2)+1(4)}{3}, \dfrac{2(-3)+1(-1)}{3}\right) = (0, -\tfrac73)$.

Final Answer: $(2, -\tfrac53)$ and $(0, -\tfrac73)$.

PYQ 11 (3 marks): Find the ratio in which the segment joining $(1, -5)$ and $(-4, 5)$ is divided by the x-axis. Also find the point of division.

Solution:

1. On x-axis $y = 0$; ratio $k:1$: $0 = \dfrac{5k - 5}{k+1} \Rightarrow 5k = 5 \Rightarrow k = 1$.
2. Ratio $= 1 : 1$; $x = \dfrac{-4 + 1}{2} = -\tfrac32$.

Final Answer: Ratio $1 : 1$; point $(-\tfrac32, 0)$.

PYQ 12 (3 marks): Show that the points $(1, 7)$, $(4, 2)$, $(-1, -1)$, $(-4, 4)$ are the vertices of a square.

Solution:

1. $AB = \sqrt{9+25} = \sqrt{34}$; $BC = \sqrt{25+9} = \sqrt{34}$; $CD = \sqrt{9+25} = \sqrt{34}$; $DA = \sqrt{25+9} = \sqrt{34}$ — all equal.
2. Diagonals: $AC = \sqrt{4+64} = \sqrt{68}$; $BD = \sqrt{64+4} = \sqrt{68}$ — equal.
3. All sides equal + equal diagonals ⇒ square.

Final Answer: Square. Proved.

PYQ 13 (3 marks): Find the area of the triangle formed by $(1, -1)$, $(-4, 6)$, $(-3, -5)$.

Solution:

1. Area $= \tfrac12|1(6+5) + (-4)(-5+1) + (-3)(-1-6)|$.
2. $= \tfrac12|11 + 16 + 21| = \tfrac{48}{2} = 24$.

Final Answer: 24 square units.

PYQ 14 (3 marks): Find the point which divides the segment joining $(-1, 7)$ and $(4, -3)$ in the ratio $2 : 3$.

Solution:

1. $x = \dfrac{2(4)+3(-1)}{5} = \dfrac{5}{5} = 1$.
2. $y = \dfrac{2(-3)+3(7)}{5} = \dfrac{15}{5} = 3$.

Final Answer: $(1, 3)$.

PYQ 15 (3 marks): If $A(-2, 1)$, $B(a, 0)$, $C(4, b)$, $D(1, 2)$ are vertices of a parallelogram $ABCD$, find $a$ and $b$.

Solution:

1. Diagonals bisect: midpoint of $AC$ = midpoint of $BD$.
2. $\dfrac{-2+4}{2} = \dfrac{a+1}{2} \Rightarrow 2 = a + 1 \Rightarrow a = 1$.
3. $\dfrac{1+b}{2} = \dfrac{0+2}{2} \Rightarrow 1 + b = 2 \Rightarrow b = 1$.

Final Answer: $a = 1$, $b = 1$.

PYQ 16 (3 marks): Find the value of $k$ if the points $A(2, 3)$, $B(4, k)$, $C(6, -3)$ form a triangle of area 5 square units.

Solution:

1. Area $= \tfrac12|2(k+3) + 4(-3-3) + 6(3-k)| = 5$.
2. $|{-4k}| = 10 \Rightarrow |k| = \tfrac52$.
3. $k = \tfrac52$ or $k = -\tfrac52$.

Final Answer: $k = \pm\tfrac52$.

PYQ 17 (3 marks): Find the distance between the points $(a \cos\theta, 0)$ and $(0, a \sin\theta)$.

Solution:

1. $d = \sqrt{(a\cos\theta)^2 + (a\sin\theta)^2} = \sqrt{a^2(\cos^2\theta + \sin^2\theta)}$.
2. $= \sqrt{a^2} = |a|$.

Final Answer: $|a|$ units.

PYQ 18 (5 marks): The vertices of $\triangle ABC$ are $A(4, 6)$, $B(1, 5)$, $C(7, 2)$. A line $DE$ with $D$ on $AB$ and $E$ on $AC$ divides them so that $\dfrac{AD}{AB} = \dfrac{AE}{AC} = \dfrac14$. Find the area of $\triangle ADE$ and compare with the area of $\triangle ABC$.

Solution:

1. $D$ divides $AB$ as $1:3$: $D = \left(\dfrac{1(1)+3(4)}{4}, \dfrac{1(5)+3(6)}{4}\right) = \left(\dfrac{13}{4}, \dfrac{23}{4}\right)$.
2. $E$ divides $AC$ as $1:3$: $E = \left(\dfrac{1(7)+3(4)}{4}, \dfrac{1(2)+3(6)}{4}\right) = \left(\dfrac{19}{4}, 5\right)$.
3. Area $\triangle ABC = \tfrac12|4(5-2)+1(2-6)+7(6-5)| = \tfrac12|12-4+7| = \tfrac{15}{2}$.
4. Area $\triangle ADE = \tfrac12|4(\tfrac{23}{4}-5) + \tfrac{13}{4}(5-6) + \tfrac{19}{4}(6-\tfrac{23}{4})|= \tfrac12|3 - \tfrac{13}{4} + \tfrac{19}{16}| = \tfrac{15}{32}$.
5. Ratio $= \dfrac{15/32}{15/2} = \dfrac{1}{16} = \left(\dfrac14\right)^2$.

Final Answer: Area$(ADE) = \tfrac{15}{32}$, which is $\tfrac{1}{16}$ of Area$(ABC)$. Proved consistent with the side ratio squared.

PYQ 19 (5 marks): Find the coordinates of the points which divide the segment joining $A(-2, 2)$ and $B(2, 8)$ into four equal parts.

Solution:

1. Midpoint $M = (0, 5)$.
2. Quarter point $P$ (midpoint of $A$ and $M$) $= (-1, 3.5)$.
3. Three-quarter point $Q$ (midpoint of $M$ and $B$) $= (1, 6.5)$.

Final Answer: $(-1, 3.5)$, $(0, 5)$, $(1, 6.5)$.

PYQ 20 (5 marks): The points $A(0, -1)$, $B(2, 1)$, $C(0, 3)$, $D(-2, 1)$ are the vertices of a quadrilateral. Show that it is a square and find its area.

Solution:

1. $AB = \sqrt{4+4} = 2\sqrt2$; $BC = \sqrt{4+4} = 2\sqrt2$; $CD = \sqrt{4+4} = 2\sqrt2$; $DA = \sqrt{4+4} = 2\sqrt2$.
2. Diagonals: $AC = \sqrt{0+16} = 4$; $BD = \sqrt{16+0} = 4$ — equal.
3. All sides equal + equal diagonals ⇒ square.
4. Area $= \text{side}^2 = (2\sqrt2)^2 = 8$.

Final Answer: Square; area 8 square units.

PYQ 21 (3 marks): Find the relation between $x$ and $y$ such that the point $(x, y)$ is equidistant from $(3, 6)$ and $(-3, 4)$.

Solution:

1. $(x-3)^2 + (y-6)^2 = (x+3)^2 + (y-4)^2$.
2. $-6x + 9 - 12y + 36 = 6x + 9 - 8y + 16$.
3. $-6x - 12y + 45 = 6x - 8y + 25 \Rightarrow -12x - 4y + 20 = 0 \Rightarrow 3x + y = 5$.

Final Answer: $3x + y = 5$.

PYQ 22 (3 marks): Find the ratio in which the point $P(\tfrac34, \tfrac{5}{12})$ divides the segment joining $A(\tfrac12, \tfrac32)$ and $B(2, -5)$.

Solution:

1. On x with ratio $k:1$: $\tfrac34 = \dfrac{2k + \tfrac12}{k+1}$.
2. $\tfrac34(k+1) = 2k + \tfrac12 \Rightarrow \tfrac34 k + \tfrac34 = 2k + \tfrac12 \Rightarrow \tfrac14 = \tfrac54 k \Rightarrow k = \tfrac15$.
3. Ratio $= 1 : 5$.

Final Answer: $1 : 5$.

PYQ 23 (3 marks): Prove that the points $(3, 0)$, $(6, 4)$, $(-1, 3)$ are the vertices of a right-angled isosceles triangle.

Solution:

1. $AB^2 = 9+16 = 25$; $BC^2 = 49+1 = 50$; $CA^2 = 16+9 = 25$.
2. $AB = CA$ ⇒ isosceles.
3. $AB^2 + CA^2 = 25+25 = 50 = BC^2$ ⇒ right-angled at $A$.

Final Answer: Right-angled isosceles triangle. Proved.

PYQ 24 (5 marks): If $A(-5, 7)$, $B(-4, -5)$, $C(-1, -6)$, $D(4, 5)$ are vertices of a quadrilateral, find its area.

Solution:

1. $\triangle ABC = \tfrac12|{-5(-5+6) -4(-6-7) -1(7+5)}| = \tfrac12|{-5 + 52 - 12}| = \tfrac{35}{2}$.
2. $\triangle ACD = \tfrac12|{-5(-6-5) -1(5-7) +4(7+6)}| = \tfrac12|55 + 2 + 52| = \tfrac{109}{2}$.
3. Total $= \tfrac{35}{2} + \tfrac{109}{2} = 72$.

Final Answer: 72 square units.

PYQ 25 (3 marks): Find the centroid of a triangle if two of its vertices are $(3, -5)$ and $(-7, 4)$ and its centroid lies at the origin. Find the third vertex.

Solution:

1. $\dfrac{3 - 7 + x}{3} = 0 \Rightarrow x = 4$.
2. $\dfrac{-5 + 4 + y}{3} = 0 \Rightarrow y = 1$.

Final Answer: Third vertex $(4, 1)$.

PYQ 26 (5 marks): Find the area of the triangle formed by joining the midpoints of the sides of the triangle whose vertices are $(0, -1)$, $(2, 1)$, $(0, 3)$. Find the ratio of this area to the area of the original triangle.

Solution:

1. Midpoints: $P$ of $(0,-1),(2,1) = (1, 0)$; $Q$ of $(2,1),(0,3) = (1, 2)$; $R$ of $(0,3),(0,-1) = (0, 1)$.
2. Area of midpoint triangle $= \tfrac12|1(2-1) + 1(1-0) + 0(0-2)| = \tfrac12|1+1| = 1$.
3. Area of original $= \tfrac12|0(1-3) + 2(3+1) + 0(-1-1)| = \tfrac12|8| = 4$.
4. Ratio $= 1 : 4$.

Final Answer: Midpoint triangle area 1; ratio $1 : 4$. (Always $1:4$, since the medial triangle has sides half the original.)