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.
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)
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).
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.
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 :
(a)$(4,-7)$
(b)$(27.5, -52)$
(c)$(-7, 4)$
(d)$(\frac{4}{9}, \frac{1}{9})$
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(A) $(4,-7)$
321 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)
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)
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)
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)
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) $$\begin{aligned}& P (4,6), Q (3, 2), R (6, 5) \\ & (ii) (a) PQ = \sqrt{(4-3)^2 + (6 - 2)^2} = \sqrt{17} \\ & QR = \sqrt{(3 - 6)^2 + (2 - 5)^2} = \sqrt{18}\ \text{OR} \\ & (b) \text{ The coordinate of required point are } \left( \frac{6\times 2+1\times 4}{3}, \frac{5\times 2+1\times 6}{3} \right) \\ & \text{i.e. } \left( \frac{16}{3}, \frac{16}{3} \right) \\ & (iii) PQ = \sqrt{(4-3)^2 + (6-2)^2} = \sqrt{17} \\ & QR = \sqrt{(3 - 6)^2 + (2 - 5)^2} = \sqrt{18} \\ & PR = \sqrt{(4 - 6)^2 + (6-5)^2} = \sqrt{5} \\ & PQ \neq QR \neq PR \\ & \triangle PQR \text{ is not isosceles}\end{aligned}$$
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 $x$.
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)
