Assertion (A): Point $P (0, 2)$ is the point of intersection of $y$-axis with the line $3x + 2y = 4$. Reason (R): The distance of point $P (0, 2)$ from $x$-axis is 2 units. (a) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A). (b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A). (c) Assertion (A) is true, but Reason (R) is false. (d) Assertion (A) is false, but Reason (R) is true.
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(b) Both Assertion (A) and Reason (R) are correct but Reason (R) is not the correct explanation of Assertion (A)
A$(-4, 5)$ and C$(8, 2)$ are the two opposite vertices of a parallelogram ABCD. Its diagonals intersect each other at P$(a, b)$. The relation between 'a' and 'b' is:
The coordinates of the centre of a circle are $(x-7, 2x)$. Find the value(s) of '$x$', if the circle passes through the point $(-9, 11)$ and has radius $5\sqrt{2}$ units.
If $\text{A}(a, 0)$, $\text{B}(1, 1)$ and $\text{C}(0, b)$ form a triangle, right angled at $\text{B}$ when joined, then establish a relation between $a$ and $b$.
Find a relation between $x$ and $y$ such that $P(x, y)$ is equidistant from the points $A(3, 5)$ and $B(7, 1)$. Hence, write the coordinates of the points on $x$-axis and $y$-axis which are equidistant from points $A$ and $B$.
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$PA = PB \implies PA^2 = PB^2$ $(x-3)^2 + (y-5)^2 = (x-7)^2 + (y-1)^2$ $\implies x - y = 2$ $\therefore$ Required point on $x$-axis is $(2, 0)$ $\&$ required point on $y$-axis is $(0, -2)$
A circle centered at $(2, 1)$ passes through the points A$(5, 6)$ and B($-3$, K). Find the value(s) of K. Hence find length of chord AB.
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Let centre be O$(2,1) \Rightarrow OA = OB$ (1 Mark) $\sqrt{(5 - 2)^2 + (6 - 1)^2} = \sqrt{(-3-2)^2 + (K - 1)^2}$ (1 Mark) $\Rightarrow 9 = (K - 1)^2$ $\Rightarrow K = -2,4$ (1/2 Mark for each value of K) For K = $-2$, AB = $\sqrt{128}$ or $8\sqrt{2}$ (1/2 Mark) For K = $4$, AB = $\sqrt{68}$ or $2\sqrt{17}$ (1/2 Mark)
If the point $P (x, y)$ is equidistant from the points $(3, 6)$ and $(-3, 4)$, obtain the relation between $x$ and $y$. Hence, find the coordinates of point $P$ if it lies on $x$-axis.
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$\sqrt{(x - 3)^2 + (y - 6)^2} = \sqrt{(x + 3)^2 + (y - 4)^2}$ (1 Mark) gives $-12x - 4y = -20$ or $3x + y = 5$ (1 Mark) As P lies on $x$ axis so $y = 0$ gives $x = \frac{5}{3}$ (1/2 Mark) Coordinates of P are $(\frac{5}{3},0)$ (1/2 Mark)
If the point D $(x, y)$ is equidistant from the points E $(0, 3)$ and F $(3, 0)$, prove that $x = y$. Hence, find the $x$ coordinate of the point D, if $\triangle DEF$ is an equilateral triangle.
Show that the quadrilateral ABCD with vertices A(0, 3), B(-2, 0), C(0, -5) and D(2, 0) is a kite. Also, find the length of each diagonal of the kite ABCD.
The vertices of a rhombus ABCD are A$(-3, -4)$, B$(5, -3)$, C$(1, 4)$ and D$(-7, 3)$. Find the length of both the diagonals. Hence, find area of the rhombus ABCD.
If the point A $(x, y)$ is equidistant from the points B $(-2, 0)$ and C $(2, 0)$, prove that the point A lies on y-axis. Also, find the coordinates of the point A, if $\triangle ABC$ is an equilateral triangle.
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$\sqrt{(x + 2)^2 + (y - 0)^2} = \sqrt{(x - 2)^2 + (y - 0)^2}$ (1 Mark) getting $x = 0$ (1/2 Mark) As $x = 0$ so A lies on y axis (1/2 Mark) As triangle is equilateral, $AB = BC = AC$ gives $\sqrt{(0 + 2)^2 + (y - 0)^2} = \sqrt{(-2 - 2)^2 + (0 - 0)^2}$ (1/2 Mark) which gives $y^2 = 12$ (1/2 Mark) Thus, coordinates of point A are $(0, \pm 2\sqrt{3})$ (1 Mark)
Assertion (A): The point $(-2, 4)$ divides the line segment joining the points $(-4, 8)$ and $(5, -10)$ in the ratio $2 : 7$ internally. Reason (R): If three points $P$, $Q$ and $R$ are collinear, then $PQ + QR = PR$.
(a)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(b)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
(c)Assertion (A) is true, but Reason (R) is false.
(d)Assertion (A) is false, but Reason (R) is true.
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(B) Both Assertion(A) and Reason(R) are true, but Reason(R) is not the correct explanation of Assertion(A).
In the given figure, $\triangle ABC$ is an equilateral triangle. $AD$ is a median of the triangle joining the points $A \left(0, \frac{5\sqrt{3}}{2}\right)$, $D(0, 0)$. Points $B$ and $C$ are (in same order) :
Show that the points $(-3, -3)$, $(3, 3)$ and $(-3\sqrt{3}, 3\sqrt{3})$ are the vertices of an equilateral triangle.
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Let $A (-3, -3)$, $B (3, 3)$ and $C (-3\sqrt{3}, 3\sqrt{3})$ be the given points. Using distance formula $AB = \sqrt{(3 + 3)^2 + (3 + 3)^2} = 6\sqrt{2}$ units $BC = \sqrt{(-3\sqrt{3}- 3)^2 + (3\sqrt{3} - 3)^2} = 6\sqrt{2}$ units $CA = \sqrt{(-3 + 3\sqrt{3})^2 + (-3 - 3\sqrt{3})^2} = 6\sqrt{2}$ units As $AB = BC = CA$, so the given points are the vertices of an equilateral triangle.
Show that the points $(-2, 3)$, $(8, 3)$ and $(6, 7)$ are the vertices of a right-angled triangle.
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Let the given points be A $(-2, 3)$, B $(8, 3)$ and C $(6, 7)$ Then, AB = $10$, BC = $\sqrt{4 + 16} = \sqrt{20}$, AC = $\sqrt{64 + 16} = \sqrt{80}$ $\therefore AB^2 = BC^2 + AC^2$ $\therefore$ the given points are the vertices of a right angled triangle.
Do the points $P (1, 0)$, $Q (-5, 0)$ and $R (-2, 5)$ form a triangle ? If so, name the type of triangle formed.
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$PQ = \sqrt{(-5 - 1)^2 + (0 – 0)^2} = 6$ (I) (1 Mark) $QR = \sqrt{(-2 + 5)^2 + (5 – 0)^2} = \sqrt{34}$ or $5.8$ $PR = \sqrt{(-2 - 1)^2 + (5 – 0)^2} = \sqrt{34}$ or $5.8$ Since sum of any two sides is greater than the third side, $\therefore$ Points $P, Q$ and $R$ form a triangle. (II) (1/2 Mark) $QR = PR \Rightarrow PQR$ forms an isosceles triangle. (III) (1/2 Mark)
Diagonals AC and BD of square ABCD intersect at P. Coordinates of points B and D are $(9, -2)$ and $(1, 6)$ respectively. (i) Find the co-ordinates of point P. (ii) Find the length of the side of the square.
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(i) Coordinates of P are $(\frac{9+1}{2}, \frac{-2+6}{2}) = (5,2)$ (1 Mark) (ii) $2 AB^2 = BD^2$ (1/2 Mark) $\Rightarrow 2 AB^2 = (9 – 1)^2 + (-2 – 6)^2$ (1/2 Mark) $\Rightarrow AB = 8$ (1/2 Mark) Hence, the length of the side of square is $8$ units.
Show that the points $(a, a)$, $(-a, -a)$ and $(-\sqrt{3}a, \sqrt{3}a)$ are the vertices of an equilateral triangle.
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Let A $(a, a)$, B $(–a, –a)$ and C $(-\sqrt{3}a, \sqrt{3}a)$ be the given points. AB = $\sqrt{(-a - a)^2 + (-a - a)^2} = \sqrt{8a^2}$ or $2\sqrt{2} a$ BC = $\sqrt{(-\sqrt{3}a + a)^2 + (\sqrt{3}a + a)^2} = \sqrt{8a^2}$ or $2\sqrt{2} a$ CA = $\sqrt{(-\sqrt{3}a – a)^2 + (\sqrt{3}a – a)^2} = \sqrt{8a^2}$ or $2\sqrt{2} a$ Since AB = BC = CA Therefore, $\triangle$ ABC is an equilateral triangle.
A line segment joining the points $P(-5, 11)$ and $Q$ is divided internally by the point $M(2, - 3)$ such that $PM: MQ = 7 : 2$. The coordinates of $Q$ are :
Assertion (A): The point which divides the line segment joining the points A $(1, 2)$ and B$(-1, 1)$ internally in the ratio $1: 2$ is $(\frac{-1}{3}, \frac{5}{3})$ Reason (R): The coordinates of the point which divides the line segment joining the points A $(x_1, y_1)$ and B$(x_2, y_2)$ in the ratio $m_1: m_2$ are $(\frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \frac{m_1y_2 + m_2y_1}{m_1 + m_2})$
(a)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(b)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(c)Assertion (A) is true, but Reason (R) is false.
(d)Assertion (A) is false, but Reason (R) is true.
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(D) Assertion (A) is false, but Reason(R) is true.
901 Mark · 🔁 March 2024 & March 2026 · Standardopen ↗
Assertion (A): Mid-point of a line segment divides the line segment in the ratio $1:1$. Reason (R): The ratio in which the point $(-3, k)$ divides the line segment joining the points $(-5, 4)$ and $(-2, 3)$ is $1: 2$.
(a)Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(b)Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(c)Assertion (A) is true, but Reason (R) is false.
(d)Assertion (A) is false, but Reason (R) is true.
Find the ratio in which the point $(-1, k)$ divides the line segment joining the points $(-3, 10)$ and $(6,-8)$. Hence, find the value of $k$.
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Let $C (-1, k)$ be divides the line segment joining the points $A (-3, 10)$ and $B (6, -8)$ in the ratio $m : 1$. Using section formula $-1 = \frac{-3+6m}{m+1}$ $\Rightarrow m = \frac{2}{7}$ Hence, required ratio is $2 : 7$ $k = \frac{10\times7-8\times2}{2+7} = 6$
In the given figure, point D divides the side BC of $\triangle ABC$ in the ratio $1:2$. Find length AD.
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Coordinates of point D = $(\frac{1\times 4+2\times (-2)}{1+2}, \frac{1\times 2+2\times 1}{1+2})$ i.e. $(0,\frac{4}{3})$ (1 Mark) AD = $\sqrt{(1-0)^2 + (5-\frac{4}{3})^2} = \frac{\sqrt{130}}{3}$ units (1 Mark)
Find the ratio in which point $P(-1, m)$ divides the line segment joining the points $A(2, 5)$ and $B(-5, -2)$. Hence, find the value of $m$.
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Let the required ratio be $k:1$ $-1 = \frac{-5k + 2}{k + 1}$ [$1$ mark] $\Rightarrow k = \frac{3}{4}$ [$\frac{1}{2}$ mark] $\Rightarrow$ ratio is $3:4$ $\Rightarrow m = 2$ [$\frac{1}{2}$ mark]
This section comprises $5$ Very Short Answer (VSA) type questions of $2$ marks each. Find the coordinates of the point which divides the line segment joining the points $A (-6, 10)$ and $B (3, - 8)$ in the ratio $2: 7$.
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Let the coordinates of point be $(x, y)$ $x = \frac{2\times3+7\times(-6)}{9} = -4$, $y = \frac{2\times(-8)+7\times10}{9} = 6$ (1+1 Mark) Coordinates of the point are $(-4, 6)$
Find the coordinates of the point which divides the line segment joining the points $P(-1, 1)$ and $Q(5, -7)$ in the ratio $2: 3$.
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Let the coordinates of the point be $(x, y)$ $x = \frac{2\times5+3\times(-1)}{5} = \frac{7}{5}$, $y = \frac{2\times(-7)+3\times1}{5} = -\frac{11}{5}$ (1+1 Mark) Coordinates of the point are $(\frac{7}{5}, -\frac{11}{5})$
The vertices of a $\triangle ABC$ are A$(-1, 3)$, B$(2, -3)$ and C$(4, 5)$. Find the coordinates of a point P on median AD such that AP: PD = $2:3$.
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Coordinates of D are $(\frac{4+2}{2}, \frac{5-3}{2})$ i.e. $(3, 1)$ (1 Mark) Coordinates of P are $(\frac{2\times3+3\times(-1)}{2+3}, \frac{2\times1 + 3\times3}{2+3})$ i.e. $(\frac{3}{5}, \frac{11}{5})$ (1 Mark)
Find the ratio in which the line segment joining the points $(5, 3)$ and $(-1, 6)$ is divided by Y-axis.
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Let the line segment divides y-axis at $(0, y)$. Let the required ratio be $k:1$ $\therefore 0 = \frac{(-1)k+5(1)}{k+1}$ $\Rightarrow k = 5$ Hence ratio is $5 : 1$
P$(-2, 5)$ and Q$(3, 2)$ are two points. Find the coordinates of the point R on line segment PQ such that $PR = 2QR$.
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Let coordinates of R be $(x, y)$. $PR : RQ = 2 : 1$ $x = \frac{2(3)+1(-2)}{2+1} = \frac{4}{3}$ $y = \frac{2(2)+1(5)}{2+1} = 3$ $\therefore$ Coordinates of the point R $(\frac{4}{3}, 3)$
In what ratio does the X-axis divides the line segment joining the points$(2, -3)$ and $(5, 6)$? Also, find the coordinates of the point of intersection.
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Let the co ordinate of the point of intersection be $(x, 0)$. Let ratio be $k:1$ $\frac{6k-3}{k+1} = 0$ $\Rightarrow k = \frac{1}{2}$ $\therefore$ required ratio is $1: 2$ $x = \frac{5\times1+2\times2}{1+2} = \frac{9}{3} = 3$ $\therefore$ the co ordinate of the point of intersection is $(3,0)$
Find the ratio in which the x-axis divides the line segment joining the points $(-6, 5)$ and $(-4, -1)$. Also, find the point of intersection.
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Let coordinates of $P$ be $(x, 0)$ and $P$ divides the line segement $AB$ in the ratio $k: 1$ (I) (1/2 Mark) $(\frac{-4k-6}{k+1}, \frac{-k+5}{k+1}) = (x, 0)$ (II) (1 Mark) $\frac{-k+5}{k+1} = 0 \Rightarrow k = 5$ (III) (1/2 Mark) Hence the required ratio is $5: 1$ (IV) (1/2 Mark) $\therefore$ Coordinates of $P$ are $(\frac{-4\times5-6}{5+1}, 0) = (\frac{-13}{3}, 0)$ (V) (1/2 Mark)
Parthi and Alisha found a treasure that is exactly on the straight line joining their locations. Parthi's location is at point $(-6, -5)$ and Alisha's location is at point $(10, 11)$. The distance from the treasure to Parthi's location is three times that of the distance to Alisha's location. Find the coordinates of the location of the treasure.
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Let the points P, A and T represent the location of Parthi, Alisha and Treasure respectively. Let the coordinates of T be $(x, y)$ P $(-6,-5)$ T $(x, y)$ A $(10, 11)$ $PT = 3 AT$ (1 Mark) $\frac{PT}{AT} = \frac{3}{1}$ (1/2 Mark) Coordinates of T are $(\frac{10\times3-6}{3+1}, \frac{11\times3-5}{3+1})$ (1 Mark) $= (6,7)$ (1/2 Mark)
Find the coordinates of the points of trisection of the line segment joining the points A($-1, 4$) and B($-3,-2$).
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A $(-1,4)$ P Q B $(-3,-2)$ Let P and Q be the points of trisection of AB, therefore AP :PB = 1:2 Coordinates of P $= (\frac{1(-3)+2(-1)}{3}, \frac{1(-2)+2(4)}{3})$ (1 Mark) $= (\frac{-5}{3}, 2)$ (1/2 Mark) Q is the mid-point of PB (1 Mark) Coordinates of Q $= (\frac{\frac{-5}{3}+(-3)}{2}, \frac{2+(-2)}{2})$ (1/2 Mark) $= (\frac{-7}{3}, 0)$
Prove that the point P dividing the line segment joining the points A($-1, 7$) and B($4, -3$) in the ratio $3:2$, lies on the line $x - 3y = -1$. Also find length of PA and PB.
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AP: PB = $3:2$ Coordinates of P = $(\frac{3\times 4+2\times (-1)}{3+2}, \frac{3\times (-3)+2\times 7}{3+2}) = (2,1)$ (1 Mark) Substituting $x = 2$ and $y = 1$ in the given equation L. H. S. = $x - 3y$ = $2 - 3(1)$ = $-1$ = R. H. S. $\therefore$ P lies on the given line (1 Mark) PA = $\sqrt{(2 + 1)^2 + (1 - 7)^2} = \sqrt{45}$ or $3\sqrt{5}$ (1/2 Mark) PB = $\sqrt{(2 - 4)^2 + (1 + 3)^2} = \sqrt{20}$ or $2\sqrt{5}$ (1/2 Mark)
Find the coordinates of the points of trisection of the line segment joining the points P$(5, -4)$ and Q$(-4, 2)$.
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Let the points A and B trisect the line segment joining P and Q. $\therefore$ PA : AQ = $1:2$. (1 Mark) Coordinates of A = $(\frac{1 \times (-4) + 2 \times 5}{1+2}, \frac{1 \times 2 + 2 \times (-4)}{1+2})$ ($\frac{1}{2}$ Mark) $= (2, -2)$ ($\frac{1}{2}$ Mark) Now, B is mid point of the line segment joining A and Q. Coordinates of B = $(\frac{2 - 4}{2}, \frac{-2 + 2}{2})$ ($\frac{1}{2}$ Mark) $= (-1, 0)$ ($\frac{1}{2}$ Mark) The line segment joining P and Q is trisected at $(2, -2)$ and $(-1, 0)$.
Find the ratio in which the point P $(6, k)$ divides the line segment joining the points M $(4, 2)$ and N $(8, -4)$. Hence find the value of $k$.
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Let the point P divides the line segment joining M and N in the ratio $r : 1$. Then, Coordinates of P = $(\frac{r \times 8 + 1 \times 4}{r+1}, \frac{r \times (-4) + 1 \times 2}{r+1}) = (6, k)$ (1 Mark) $\Rightarrow \frac{8r + 4}{r+1} = 6$ ($\frac{1}{2}$ Mark) $\Rightarrow r = 1$ ($\frac{1}{2}$ Mark) Thus, the point P divides the line segment joining M and N in the ratio $1:1$. ($\frac{1}{2}$ Mark) i.e. Coordinates of P = $(\frac{4+8}{2}, \frac{2-4}{2}) = (6, k)$ ($\frac{1}{2}$ Mark) The value of $k$ is $-1$. ($\frac{1}{2}$ Mark)
The line segment joining the points A $(-5, 1)$ and B $(7, 6)$ is trisected at the points P and Q such that P is nearer to A. If P lies on the line $x + y = k$, then find the value of $k$.
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AP : PB = $1 : 2$ (½) Coordinates of P are $(\frac{1\times 7+2\times(-5)}{1+2}, \frac{1\times 6+2\times 1}{1+2})$ i.e. $(-1, \frac{8}{3})$ (1½) $\therefore$ P lies on $x + y = k$ therefore $k = -1 + \frac{8}{3} = \frac{5}{3}$ (1)
If A(-2,-1), B(a, 0), C(4, b) and D(1, 2) are the vertices of a parallelogram ABCD, then find the values of a and b.
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Coordinates of the mid-point of AC = Coordinates of the mid-point of BD $(\frac{-2+4}{2}, \frac{-1+b}{2}) = (\frac{a+1}{2}, \frac{0+2}{2})$ $\therefore \frac{-2+4}{2} = \frac{a+1}{2} \Rightarrow a=1$ (1/2 Mark) and $\frac{-1+b}{2} = \frac{0+2}{2} \Rightarrow b=3$ (1/2 Mark)
The three vertices of a parallelogram ABCD, taken in order, are A(-1, 0), B(3, 1) and C(2, 2). Find the coordinates of the fourth vertex D.
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Let the coordinates of fourth vertex D be $(x, y)$ Coordinates of the mid-point of AC = Coordinates of the mid-point of BD $(\frac{-1+2}{2}, \frac{0+2}{2}) = (\frac{3+x}{2}, \frac{1+y}{2})$ $\therefore \frac{-1+2}{2} = \frac{3+x}{2} \Rightarrow x=-2$ (1/2 Mark) and $\frac{0+2}{2} = \frac{1+y}{2} \Rightarrow y=1$ (1/2 Mark)
If $\text{P}$ is the mid-point of the line segment forming the points $\text{A}(-2, 8)$ and $\text{B}(-6, -4)$, then the coordinates of $\text{P}$ are:
Assertion (A): If the points $A(4, 3)$ and $B(x, 5)$ lie on a circle with centre $O(2, 3)$, then the value of $x$ is $2$. Reason (R): Centre of a circle is the mid-point of each chord of the circle. (a) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A). (b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A). (c) Assertion (A) is true, but Reason (R) is false. (d) Assertion (A) is false, but Reason (R) is true.
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(c) Assertion (A) is true, but Reason (R) is false
If the points A$(4, 5)$, B$(m, 6)$, C$(4, 3)$ and D$(1, n)$ taken in this order are the vertices of a parallelogram ABCD, then find the values of $m$ and $n$.
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Coordinates of mid-point of AC = Coordinates of mid-point of BD (1 Mark) $\frac{4+4}{2}$, $\frac{5+3}{2}$) = ($\frac{1+m}{2}$, $\frac{n+6}{2}$) (1 Mark) m = 7 (1/2 Mark) n = 2 (1/2 Mark)
If points $A(-5, y)$, $B(2, -2)$, $C(8, 4)$ and $D(x, 5)$ taken in order, form a parallelogram ABCD, then find the values of $x$ and $y$. Hence, find lengths of sides of the parallelogram.
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ABCD is a parallelogram $\therefore$ Coordinates of mid pt. of BD = Coordinates of mid pt. of AC $(\frac{2+x}{2}, \frac{-2+5}{2}) = (\frac{8-5}{2}, \frac{4+y}{2})$ [$1$ mark] Getting $x = 1$ and $y = -1$ [$\frac{1}{2} + \frac{1}{2}$ mark] $AB = \sqrt{7^2 + (-1)^2} = \sqrt{50}$ or $5\sqrt{2}$ [$\frac{1}{2}$ mark] $BC = \sqrt{6^2 + 6^2} = \sqrt{72}$ or $6\sqrt{2}$ [$\frac{1}{2}$ mark]
$A(6, -3)$, $B(0, 5)$ and $C(-2, 1)$ are vertices of $\Delta ABC$. Points $P(3, 1)$ and $Q(2, -1)$ lie on sides AB and AC respectively. Check whether $\frac{AP}{PB} = \frac{AQ}{QC}$.
ABC is a triangle. B lies on x-axis at a distance of 4 units from y-axis at its right. C is on y-axis and it is 3 units away from the origin. If the coordinates of A are (0, 0), the perimeter of $\triangle ABC$ is :
The line represented by $\frac{x}{4} + \frac{y}{6} = 1$, intersects $x$-axis and $y$-axis respectively at P and Q. The coordinates of the mid-point of line segment PQ are:
The vertices of a $\Delta ABC$ are A$(-2, 4)$, B$(4, 3)$ and C$(1, -6)$. Find length of the median BD.
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Mid-point of AC is D $(\frac{-1}{2}, -1)$ (1) Length of median BD $= \sqrt{(4+\frac{1}{2})^2+(3+1)^2} = \sqrt{\frac{145}{4}}$ or $\frac{\sqrt{145}}{2}$ (1)
Points $A(-1, y)$ and $B(5, 7)$ lie on a circle with centre $O(2, -3y)$ such that $AB$ is a diameter of the circle. Find the value of $y$. Also, find the radius of the circle.
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Centre $O (2, -3y)$ is the mid point of $AB$ $\frac{-1+5}{2} = 2$, $\frac{y+7}{2} = -3y$ $\Rightarrow y = -1$ Radius = $OB = \sqrt{(5 - 2)^2 + (7 - (-1))^2} = 5$
The coordinates of the centre of a circle are $(2a, a - 7)$. Find the value(s) of '$a$' if the circle passes through the point $(11, -9)$ and has diameter $10\sqrt{2}$ units.
Find the length of the median through the vertex B of $\triangle ABC$ with vertices A($9$, $-2$), B($-3$, $7$) and C($-1$, $10$).
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Mid point of AC = $(4,4)$ Length of median from B to AC = $\sqrt{(4+3)^2 + (4-7)^2}$ $= \sqrt{7^2 + (-3)^2} = \sqrt{49+9} = \sqrt{58}$ Hence the length of median is $\sqrt{58}$ units
The centre of a circle is $(2a, a-7)$. Find the values of '$a$' if the circle passes through the point $(11,-9)$. Radius of the circle is $5\sqrt{2}$ cm.
$P (x, y)$, $Q (-2, – 3)$ and $R (2, 3)$ are the vertices of a right triangle PQR right angled at P. Find the relationship between $x$ and $y$. Hence, find all possible values of $x$ for which $y = 2$.
If the points $A(6, 1)$, $B(p, 2)$, $C(9, 4)$ and $D(7, q)$ are the vertices of a parallelogram $ABCD$, then find the values of $p$ and $q$. Hence, check whether $ABCD$ is a rectangle or not.
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Diagonals of a parallelogram bisect each other. $\therefore$ Co-ordinates of mid point of diagonal $AC$ = Co-ordinates of mid-point of diagonal $BD$. $(\frac{6+9}{2}, \frac{1+4}{2}) = (\frac{p+7}{2}, \frac{2+q}{2})$ (1 mark). $\Rightarrow \frac{p+7}{2} = \frac{15}{2}$ and $\frac{2+q}{2} = \frac{5}{2}$. $\therefore p=8$ and $q=3$ ($\frac{1}{2}$ mark). Diagonal $AC = \sqrt{3^2 + 3^2} = 3\sqrt{2}$ ($\frac{1}{2}$ mark). Diagonal $BD = \sqrt{(-1)^2 + 1^2} = \sqrt{2}$ ($\frac{1}{2}$ mark). $AC \neq BD \therefore ABCD$ is not a rectangle ($\frac{1}{2}$ mark).
The three vertices of a rhombus PQRS are P(2, 3), Q(6, 5) and R(-2, 1). Find the coordinates of the fourth vertex S and coordinates of the point where both the diagonals PR and QS intersect.
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Let the coordinates of fourth vertex S be $(x,y)$ Coordinates of mid-point of PR = Coordinates of mid-point of QS $(\frac{x+6}{2}, \frac{y+5}{2}) = (\frac{2-2}{2}, \frac{3+1}{2})$ (1 Mark) x = -6 (1/2 Mark) y = -7 (1/2 Mark) $\therefore$ coordinates of S = (-6, -7) mid- point of diagonal PR = (0, -1) (1 Mark)
Morning assembly is an integral part of every school's schedule. In the assembly, students always stand in rows and columns and this makes a coordinate system. In a school, there are $200$ students and they all assemble for prayer in $10$ rows. $4$ students are at A, B, C and D with the following positions of the coordinate system : A $(3, 4)$, B $(6, 7)$, C $(9, 4)$ and D $(6, 1)$. Based on the above, answer the following questions : (a) Find the distance between A and B. (b) Find the distance between C and D. (c) Show that ABCD forms a parallelogram. OR (c) Find the mid-point of the line segments AC and BD.
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(a) Distance between A $(3,4)$ and B $(6,7)$: $AB = \sqrt{(6-3)^2 + (7-4)^2} = \sqrt{3^2 + 3^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt{2}$ units. (b) Distance between C $(9,4)$ and D $(6,1)$: $CD = \sqrt{(6-9)^2 + (1-4)^2} = \sqrt{(-3)^2 + (-3)^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt{2}$ units. (c) (i) To show ABCD forms a parallelogram, we can show opposite sides are equal or diagonals bisect each other. Using distance formula for all sides: $AB = 3\sqrt{2}$ (from part a) $CD = 3\sqrt{2}$ (from part b) Distance between B $(6,7)$ and C $(9,4)$: $BC = \sqrt{(9-6)^2 + (4-7)^2} = \sqrt{3^2 + (-3)^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt{2}$ units. Distance between A $(3,4)$ and D $(6,1)$: $AD = \sqrt{(6-3)^2 + (1-4)^2} = \sqrt{3^2 + (-3)^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt{2}$ units. Since all sides are equal ($AB=BC=CD=DA=3\sqrt{2}$), ABCD is a rhombus, which is a type of parallelogram. OR (c) (ii) Mid-point of AC: A $(3,4)$, C $(9,4)$ Mid-point $= (\frac{3+9}{2}, \frac{4+4}{2}) = (\frac{12}{2}, \frac{8}{2}) = (6,4)$. Mid-point of BD: B $(6,7)$, D $(6,1)$ Mid-point $= (\frac{6+6}{2}, \frac{7+1}{2}) = (\frac{12}{2}, \frac{8}{2}) = (6,4)$. Since the mid-points of AC and BD are the same, the diagonals bisect each other. Therefore, ABCD is a parallelogram.
Jagdish has a field which is in the shape of a right angled triangle AQC. He wants to leave a space in the form of a square PQRS inside the field for growing wheat and the remaining for growing vegetables (as shown in the figure). In the field, there is a pole marked as O. Based on the above information, answer the following questions : (i) Taking O as origin, coordinates of P are $(-200, 0)$ and of Q are $(200, 0)$. PQRS being a square, what are the coordinates of R and S? (ii) (a) What is the area of square PQRS ? OR (b) What is the length of diagonal PR in square PQRS ? (iii) If S divides CA in the ratio K:1, what is the value of K, where point A is $(200, 800)$ ?
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(i) R$(200, 400)$, S$(-200, 400)$(ii) (a) side PQ = $(200+200)$ m = $400$ m Area of square PQRS = $400 \times 400$ $= 160000$ sq. units OR (ii) (b) Diagonal PR = $\sqrt{(400)^2 + (400)^2}$ $= \sqrt{3200}$ or $400\sqrt{2}$ (iii) C$(-600,0)$; A$(200,800)$; S$(-200,400)$ S divides CA in the ratio k: $1$ $-200 = \frac{k(200)+1(-600)}{k+1}$ $\Rightarrow k = 1$
Partha, a software engineer, lives in Jerusalem for his work. He lives in the most convenient area of the city from where bank, hospital, post office and supermarket can be easily accessed. In the graph, the bank is plotted as A$(9, 5)$, hospital as B$(-3,-1)$ and supermarket as C$(5,-5)$ such that A, B, C form a triangle. Based on the above given information, answer the following questions : (i) Find the distance between the bank and the hospital. (ii) In between the bank and the supermarket, there is a post office plotted at E which is their mid-point. Find the coordinates of E. (iii) (a) In between the hospital and the supermarket, there is a bus stop plotted as D, which is their mid-point. If Partha wants to reach the bus stand from the bank, then how much distance does he need to cover ? OR (b) P and Q are two different garment shops lying between the bank and the hospital, such that BP = PQ = QA. If the coordinates of P and Q are $(1, a)$ and $(b, 3)$ respectively, then find the values of 'a' and 'b'.
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(i) Distance between bank and hospital $= \sqrt{(-3-9)^2 + (-1 - 5)^2}$ $= \sqrt{180}$ units or $6\sqrt{5}$ units (ii) Coordinates of E are $(\frac{9+5}{2}, \frac{5+(-5)}{2}) = (7,0)$ (iii) (a) Coordinates of D are $(\frac{-3+5}{2}, \frac{-1+(-5)}{2}) = (1, -3)$ Distance Partha need to cover $= \sqrt{(9-1)^2 + (5 - (-3))^2}$ $= \sqrt{128}$ units or $8\sqrt{2}$ units OR (iii) (b) P is mid-point of BQ $\therefore a = \frac{-1+3}{2} = 1$ Q is mid-point of PA $\therefore b = \frac{1+9}{2} = 5$
Ryan, from a very young age, was fascinated by the twinkling of stars and the vastness of space. He always dreamt of becoming an astronaut one day. So he started to sketch his own rocket designs on the graph sheet. One such design is given below : Based on the above, answer the following questions : (i) Find the mid-point of the segment joining F and G. (ii) (a) What is the distance between the points A and C? OR (b) Find the coordinates of the point which divides the line segment joining the points A and B in the ratio $1:3$ internally. (iii) What are the coordinates of the point D?
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(i) Mid point of FG is $(\frac{-3+1}{2}, \frac{0+4}{2}) = (-1,2)$ (ii) (a) $$\begin{aligned}& AC = \sqrt{(1-3)^2 + (-2 - 4)^2} \\ & = \sqrt{52}\end{aligned}$$ or $2\sqrt{13}$ OR (ii) (b) The coordinates of required point are $$\begin{aligned}& (\frac{1\times3+3\times3}{1+3}, \frac{1\times2+3\times4}{1+3}) \\ & \text{i.e. } (3,\frac{7}{2}) \\ & (iii)\end{aligned}$$D(-2, -5)
A garden is in the shape of a square. The gardener grew saplings of Ashoka tree on the boundary of the garden at the distance of $1 \text{ m}$ from each other. He wants to decorate the garden with rose plants. He chose a triangular region inside the garden to grow rose plants. In the above situation, the gardener took help from the students of class $10$. They made a chart for it which looks like the given figure. Based on the above, answer the following questions : (i) If A is taken as origin, what are the coordinates of the vertices of $\triangle PQR$? (ii) (a) Find distances PQ and QR. OR (b) Find the coordinates of the point which divides the line segment joining points P and R in the ratio $2: 1$ internally. (iii) Find out if $\triangle PQR$ is an isosceles triangle.
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(i) $$\begin{aligned}& P (4,6), Q (3, 2), R (6, 5) \\ & (ii) (a) PQ = \sqrt{(4-3)^2 + (6 - 2)^2} = \sqrt{17} \\ & QR = \sqrt{(3 - 6)^2 + (2 - 5)^2} = \sqrt{18}\ \text{OR} \\ & (b) \text{ The coordinate of required point are } \left( \frac{6\times 2+1\times 4}{3}, \frac{5\times 2+1\times 6}{3} \right) \\ & \text{i.e. } \left( \frac{16}{3}, \frac{16}{3} \right) \\ & (iii) PQ = \sqrt{(4-3)^2 + (6-2)^2} = \sqrt{17} \\ & QR = \sqrt{(3 - 6)^2 + (2 - 5)^2} = \sqrt{18} \\ & PR = \sqrt{(4 - 6)^2 + (6-5)^2} = \sqrt{5} \\ & PQ \neq QR \neq PR \\ & \triangle PQR \text{ is not isosceles}\end{aligned}$$
Trees act the natural filters. By planting trees in and around school premises, we create cleaner and healthier air for students and local residents, reducing respiratory problems. A school in Noida has proposed and organised a community drive on tree plantation under the title "Save Earth, Plant Trees". Students of that school have planted saplings in the field such that it formed a quadrilateral as shown in the figure ABCD. Based on the information given above, answer the following questions : (i) Find the distance between the two saplings at A and D. (ii) (a) One student plants one sapling at the mid-point of AD. Then he moves along a straight line parallel to DB and sows another sapling on AB. What are the coordinates of the positions of these two new saplings? OR (ii) (b) A new sapling is kept at a point M on DB such that DM : MB = $3:1$. Find the coordinates of M. (iii) The line segments AC and BD bisect each other at P$(-2, 2)$. Find the coordinates of C.
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(i) Observing the graph, coordinators of points are A $(-5, 9)$ and D $(-6, 1)$ AD = $\sqrt{(-6 + 5)^2 + (1 - 9)^2} = \sqrt{65}$ (ii) (a) Mid point of AD = $\left(\frac{-6-5}{2}, \frac{1+9}{2}\right)$ i.e. $\left(-\frac{11}{2}, 5\right)$ Student will sow another sapling at mid point of AB. Point B is $(2, 3)$ Mid point of AB = $\left(\frac{2-5}{2}, \frac{3+9}{2}\right)$ i.e. $\left(-\frac{3}{2}, 6\right)$ OR (b) Coordinates of M = $\left(\frac{3 \times 2 + 1 \times (-6)}{3+1}, \frac{3 \times 3 + 1 \times 1}{3+1}\right)$ i.e. $\left(0, \frac{5}{2}\right)$ (iii) Coordinates of C are $(1, -5)$
Carom board is a very popular game. The board is a square of side length $65$ cm. It has circular pockets in each corner. Ansh strikes a disc, kept at position $P$ with a striker. The disc, hits the boundary of the board at $R$ and goes straight to pocket at corner $C$. It is given that $PS = 9$ cm, $PQ = 35$ cm, $BR = x$, $\angle PRQ = \alpha$ and $\angle CRB = \theta$. Based on the above information, answer the following questions: (i) Using law of reflection i.e. $\angle PRT = \angle CRT$, prove that $\theta = \alpha$. (ii) Prove that $\triangle PQR \sim \triangle CBR$ given that $PQ$ is perpendicular to $AB$. (iii) (a) Find the value of $x$ using similarity of triangles. OR (b) If $\frac{\text{Area } \triangle PQR}{\text{Area } \triangle CBR} = \frac{PQ^2}{CB^2}$, then find the value of $x$.
Observe the map of Jaipur city placed on a Cartesian plane. Taking Rambagh Palace as origin, the location of some places are given below : Point A: $(-4, 2)$ Rajasthan High Court Point B: $(4,-4)$ Birla Mandir Point C: $(4, 3)$ Heera Bagh Point D: $(-5, -2)$ Amar Jawan Jyoti Based on the above, answer the following questions : (i) Advocate Rehana stays at Heera Bagh. How much distance she has to cover daily to go to the court and coming back home? (ii) There is a crossing on X-axis which divides AD in a certain ratio. Find the ratio. (iii) (a) Is Birla Mandir equidistant from Heera Bagh and Amar Jawan Jyoti? Justify your answer. OR (b) Using section formula, show that points A, O and B are not collinear.
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(i) Distance travelled $= 2 AC$ $= 2\sqrt{(-4-4)^2 + (2 - 3)^2}$ (I) ($\frac{1}{2}$ Mark) $= 2\sqrt{64 + 1}$ (II) ($\frac{1}{2}$ Mark) $= 2\sqrt{65}$ Hence, required distance is $2\sqrt{65}$ units. (ii) Let the point $P(x, 0)$ divides $AD$ in the ratio $K : 1$ $\therefore AP: PD = K : 1$ Here, $0 = \frac{-2K+2}{K+1}$ (I) ($\frac{1}{2}$ Mark) $\Rightarrow K = 1$ (II) ($\frac{1}{2}$ Mark) $\therefore$ The required ratio is $1 : 1$ (iii) (a) $BC = \sqrt{(4 - 4)^2 + (-4 - 3)^2} = 7$ units (I) (1 Mark) $BD = \sqrt{(4 + 5)^2 + (-4 + 2)^2} = \sqrt{85}$ units (II) (1 Mark) $\therefore BC \neq BD$ $\therefore$ Birla Mandir is not equidistant from Heera Bagh and Amar Jawan Jyoti. OR (b) Let us assume that points $A, O, B$ are collinear and $AO : OB = K : 1$ Here, $0 = \frac{4K-4}{K+1}$ (I) ($\frac{1}{2}$ Mark) $\Rightarrow K = 1$ (II) ($\frac{1}{2}$ Mark) Also, $0 = \frac{-4K+2}{K+1}$ (III) ($\frac{1}{2}$ Mark) $\Rightarrow K = \frac{1}{2}$ (IV) ($\frac{1}{2}$ Mark) Since the value of $K$ is different in the above two cases, so points $A, O$ and $B$ are not collinear.
In the school garden, Arun (A), Babu (B), Chandra (C) and Daya (D) planted flower plants of Sunflower, Rose, Champa and Jasmine respectively at point A$(2, 8)$, B$(7, 8)$, C$(9, 3)$ and D$(2, 3)$ respectively. Based on the above, answer the following questions : (i) Find the distances AB and AD. (ii) Find BC – CD. (iii) (a) If Varun wants to plant his flower plant at a point M such that DM: MC = $3:2$, find the coordinates of M. OR (b) If N divides the line segment AC in the ratio $2: 3$, find the coordinates of N.
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(i) $AB = \sqrt{(7-2)^2+(8-8)^2} = 5$ (0.5 Mark) $AD = \sqrt{(2-2)^2+(3-8)^2} = 5$ (0.5 Mark) (ii) $BC – CD = \sqrt{(9 – 7)^2 + (3 – 8)^2} - \sqrt{(9 – 2)^2 + (3 – 3)^2} = \sqrt{29} – 7$ (1 Mark) (iii) (a) Let the coordinates of M be $(x, y)$ $x = \frac{3\times9+2\times2}{3+2} = \frac{31}{5}$ (1 Mark) $y = \frac{3\times3+2\times3}{3+2} = 3$ $\therefore$ Coordinates of M are $(\frac{31}{5}, 3)$ (1 Mark) OR (b) Let the coordinates of N be $(x, y)$ $x = \frac{2\times9+3\times2}{2+3} = \frac{24}{5}$ (1 Mark) $y = \frac{2\times3+3\times8}{2+3} = 6$ $\therefore$ Coordinates of N are $(\frac{24}{5}, 6)$ (1 Mark)
