If the points with the co-ordinates {a, ma}, {b, (m + 1)b}, {c, (m + 2)c} are collinear, then which of the following is correct ? -Maths 9th

1 Answer

Answer :

(d) (7, –2)Let the co-ordinates of R be (x, y). As can be easily seen, it is a point of external division Also, PR = 2QR⇒ R divides the join of P and Q externally in the ratio 2:1. ∴ x = \(rac{2 imes2-1 imes-3}{2-1}\), y = \(rac{2 imes1-1 imes4}{2-1}\)⇒ x = 4 + 3 = 7 and y = 2 – 4 = –2. ∴ Co-ordinates of R are (7, –2).

Related questions

Description : Draw the graph of the equation 3x + 4y = 12 and find the co-ordinates of the points of intersection of the equation with the co-ordinate axes. -Maths 9th

Last Answer : Solution :-

Description : Find the co-ordinates of the points on the x-axis which are at a distance of 10 units from the point (– 4, 8)? -Maths 9th

Last Answer : (a) Internal division: If P(x, y) divides the line segment formed by the joining of the points A (x1, y1) and B (x2, y2) internally in the ratio m1 : m2. Then\(x=rac{m_1x_2+m_2x_1}{m_1+m_2}\) and \(y=rac ... : 1, the co-ordinates of the mid-point are \(\bigg(rac{x_1+x_2}{2},rac{y_1+y_2}{2}\bigg)\).

Description : Let the opposite angular points of a square be (3, 4) and (1, – 1), Find the co-ordinates of the remaining angular points. -Maths 9th

Last Answer : P ≡ (-3, 2), Q ≡ (-5, -5), R ≡ (2, -3), S ≡ (4, 4)∴ PQ = \(\sqrt{(-5+3)^2+(-5-2)^2}\) = \(\sqrt{4+49}\) = \(\sqrt{53}\)QR = \(\sqrt{(2+5)^2+(-3+5)^2}\) = \(\sqrt{49+4}\) = \ ... of rhombus = \(rac{1}{2}\) x (Product of length of diagonals) = \(rac{1}{2}\) x \(5\sqrt2\) x \(9\sqrt2\) = 45 sq. units.

Description : The co-ordinates of mid-points of sides of a triangle are (1, 2), (0, –1) and (2, –1). Find its centroid. -Maths 9th

Last Answer : ABCD is a parallelogram, if the mid-points of diagonals AC and BD have the same co-ordinates (∵ Diagonals of a parallelogram bisect each other)Co-ordinates of mid-point of AC are \(\bigg(rac{a+2}{2},rac{-11+15}{2}\bigg)\) = \(\bigg(rac ... \(rac{a+2}{2}\) = 3 and 2 = \(rac{b+1}{2}\) ⇒ a = 4, b = 3.

Description : If the co-ordinates of the mid-points of the sides of a triangle are (1, 1), (2, –3), (3, 4), find its incentre. -Maths 9th

Last Answer : (b) 3 : 2 ; m = \(-rac{2}{5}\)Let P(m, 6) divides AB in the ratio k : 1. Then co-ordinates of P are \(\bigg(\)\(rac{2k-4}{k+1}\), \(rac{8k+3}{k+1}\)\(\bigg)\)Given, co-ordinates of P are (m, 6) ⇒\(rac{2k-4}{k+1} ... 2}-4}{rac{3}{2}+1}\) = \(rac{3-4}{rac{5}{2}}\) = \(rac{-2}{5}\)∴ m = \(rac{-2}{5}\).

Description : If the points A(1, 2), B(0, 0) and C(a, b) are collinear, then -Maths 9th

Last Answer : (a) - 2For three points to be collinear, area of the triangle formed by the three points should be equal to zero, i.e.\(rac{1}{2}\) [k(3k - 1) + 2k(1 - 2k) + 3(2k - 3k)] = 0⇒ \(rac{1}{2}\) [3k2 - k + ... = 0 or -2 Neglecting k = 0, as then (k, 2k) and (2k, 3k) will be the same point, we take k = -2.

Description : If the three points (k, 2k), (2k, 3k) and (3, 1) are collinear then k is equal to -Maths 9th

Last Answer : (d) 3Let (x, y) be the co-ordinates of the third vertex of the triangle. Then\(rac{0+2+x}{3}\) = 1 and \(rac{0+0+y}{3}\) = 1⇒ 2 + \(x\) = 3 and y = 3 ⇒ \(x\) = 1, y = 3. ∴ Co-ordinates of vertices of the triangle ... - y3) + x2 (y2 - y3) + x3(y1 - y2)]= \(rac{1}{2}\) [0+6+0] = \(rac{6}{2}\) = 3.

Description : If A(3, 5), B(– 5, – 4), C(7, 10) are the vertices of a parallelogram taken in order, then the co-ordinates of the fourth vertex are: -Maths 9th

Last Answer : (c) RhombusCo-ordinates of P are \(\bigg(rac{-1-1}{2},rac{-1+4}{2}\bigg)\)i.e, \(\big(-1,rac{3}{2}\big)\)Co-ordinates of Q are \(\bigg(rac{-1+5}{2},rac{4+4}{2}\bigg)\)i.e, (2, 4)Co-ordinates of R ... \sqrt{(2-2)^2+(4+1)^2}\) = \(\sqrt{25}\) = 5⇒ PR ≠ SQ ⇒ Diagonals are not equal ⇒ PQRS is a rhombus.

Description : Plot the following points and check whether they are collinear or not -Maths 9th

Last Answer : (i) Plotting the points P (1, 3), Q (-1, -1) and R (-2, - 3) on the graph paper and join these points, we get a straight line. Hence, these points are collinear. (ii) Plotting the points ... 6 (5, 5)on the graph paper and join these points, we get a straight line. Hence, given points are collinear.

Description : Plot the following points and check whether they are collinear or not -Maths 9th

Last Answer : (i) Plotting the points P (1, 3), Q (-1, -1) and R (-2, - 3) on the graph paper and join these points, we get a straight line. Hence, these points are collinear. (ii) Plotting the points ... 6 (5, 5)on the graph paper and join these points, we get a straight line. Hence, given points are collinear.

Description : Plot the following points and check whether they are collinear or not: -Maths 9th

Last Answer : Solution :-

Description : There are 18 points in a plane such that no three of them are in the same line except five points which are collinear. -Maths 9th

Last Answer : answer:

Description : Show that the points (a, b + c), (b, c + a), (c, a + b) are collinear. -Maths 9th

Last Answer : Let A(x1, y1) ≡ (1, 3), B(x2, y2) ≡ (2, 4), C(x3, y3) ≡ (5, 6) be the vertices of ΔABCArea of ΔABC = \(rac{1}{2}\) |{\(x_1\)(y2 – y3) + \(x_2\)(y3 – y1) + \(x​​_3\)(y1 – y2)}|= \(rac{1}{2}\) |{1(4 – 6) + 2(6 – 3) + 5(3 – 4)}| = \(rac{1}{2}\) |{–2 + 6 – 5}| = \(rac{1}{2}\) sq. units.

Description : If the points (x, 1), (1, 2) and (0, y + 1) are collinear show that -Maths 9th

Last Answer : Two lines are parallel if their slopes are equal∴ \(rac{0-(-8)}{3-(-5)}\) = \(rac{a-3}{4-6}\) ⇒ \(rac{8}{8}\) = \(rac{a-3}{-2}\) ⇒ a – 3 = –2 ⇒ a = 1.

Description : If the points A(1, 2), B(2, 4) and C(3, a) are collinear, what is the length of BC ? -Maths 9th

Last Answer : (c) √5 units Area of Δ ABC = 0 for collinearity of A, B, C.⇒ \(rac{1}{2}\)[1(4 – a) + 2(a – 2) + 3(2 – 4)] = 0 ⇒ 4 – a + 2a – 4 + 6 – 12 = 0 ⇒ a – 6 = 0 ⇒ a = 6. ∴ Point C ≡ (3, 6)⇒ BC = \(\sqrt{(3-2)^2+(6-4)^2}\) = \(\sqrt{1+4}\) = √5 units .

Description : Find the relation between x and y if points (2, 1), (x, y) and (7, 5) are collinear. -Maths 9th

Last Answer : answer:

Description : What do you mean by Co-ordinates (Abscissa and Ordinate)? Explain with figure. -Maths 9th

Last Answer : Co-ordinate Geometry is that branch of geometry in which two numbers, called co-ordinates are used to indicate the position of a point in a plane and which make use of algebraic methods in the study of geometric figures.

Description : Distance Formula for co-ordinates: -Maths 9th

Last Answer : The position of each point of the plane is determined with reference to the rectangular axes by means of a pair of numbers called co-ordinates which are distances of the point from the respective axes. The distance of the ... are (x, 0) and the co-ordinates of any point of the y-axis are (0, y)

Description : Section formula for co-ordinates: -Maths 9th

Last Answer : The co-ordinate axes separate the plane into four regions called the quadrants. By custom the quadrants are numbered I, II, III, IV, in the counter clockwise direction as shown in the figure. (i) For distances along the x- ... +)X′OY′3rd quadrantx < 0, y < 0(-, -)XOY′4th quadrantx > 0, y < 0(+, -)

Description : Find the co-ordinates of the circumcentre of the triangle whose vertices are (3, 0), (–1, –6) and (4, –1). Also find its circum-radius. -Maths 9th

Last Answer : Let A ≡ (2, - 2), B ≡ (-2, 1), C ≡ (5, 2 ). Then,AB = \(\sqrt{(-2-2)^2+(1+2)^2}\) = \(\sqrt{16+9}\) = \(\sqrt{25}\) = 5BC = \(\sqrt{(5+2)^2+(2-1)^2}\) = \(\sqrt{49+1}\) = \(\sqrt{50}\) = \( ... of ΔABC = \(rac{1}{2}\) x base x height = \(rac{1}{2}\) x AB x AC = \(rac{1}{2}\)x 5 x 5 = 12.5 sq. units.

Description : In the adjoining figure, P and Q have co-ordinates (4, 6)and (0, 3) respectively. Find (i) the co-ordinates of R (ii) Area of quadrilateral OAPQ. -Maths 9th

Last Answer : Let the line 2x + 3y - 30 = 0 divide the join of A(3, 4) and B(7, 8) at point C(p, q) in the ratio k : 1. Then,p = \(rac{7k+3}{k+1}\), q = \(rac{8k+4}{k+1}\)As the point C lies on the line 2x + 3y - 30 ... {3}{2}+1},rac{8 imesrac{3}{2}+4}{rac{3}{2}+1}\bigg)\) = \(\big(rac{27}{5},rac{32}{5}\big)\).

Description : Find the co-ordinates of the in-centre of the triangle whose vertices are (–36, 7), (20, 7) and (0, –8). -Maths 9th

Last Answer : Let A(1, 2), B(0, -1) and C(2, -1) be the mid-points of the sides PQ, QR and RP of the triangle PQR. Let the co-ordinates of P, Q and R be (x1, y1), (x2, y2) and (x3 , y3) respectively. Then, by the mid- ... ordinates of centroid of ΔPQR = \(\bigg(rac{3+(-1)+1}{3},rac{2+2+(-4)}{3}\bigg)\) = (1, 0).

Description : (–2, –1) and (4, –5) are the co-ordinates of vertices B and D respectively of rhombus ABCD. Find the equation of the diagonal AC. -Maths 9th

Last Answer : 3\(x\) - 2y + 5 = 0 ⇒ -2y = -3\(x\) - 5 ⇒ y = \(rac{3}{2}\)\(x\) + \(rac{5}{2}\)On comparing with y = m\(x\) + c, we see that slope of given line = \(rac{3}{2}\)As the required line is perpendicular to the given line, ... - 4)⇒ 3(y - 5) = - 2\(x\) + 8 ⇒ 3y - 15 = -2\(x\) + 8 ⇒ 3y + 2\(x\) - 23 = 0

Description : The line x – 4y = 6 is the perpendicular bisector of the segment AB and the co-ordinates of B are (1, 3). Find the co-ordinates of A. -Maths 9th

Last Answer : Co-ordinates of A are \(\bigg(rac{3 imes9+1 imes5}{3+1},rac{3 imes6+1 imes-2}{3+1}\bigg)\) = \(\bigg(rac{32}{4},rac{16}{4}\bigg)\), i.e. (8, 4)Now, \(x\) - 3y + 4 = 0 ⇒ -3y = -\(x\) - 4 ⇒ y = \(rac{x}{3}+rac{4} ... 8) [Using, y - y1 = m (x - x1)]⇒ 3y - 12 = \(x\) - 8 ⇒ 3y - \(x\) = 4.

Description : What is the equation of the straight line passing through the point (4, 3) and making intercepts on the co-ordinates axes whose sum is –1 ? -Maths 9th

Last Answer : Diagonals of a rhombus bisect each other at right angles ⇒ Co-ordinates of mid-points of AC and BD are equal∴ 0 = \(\bigg(rac{4+(-2)}{2},rac{-5+(-1)}{2}\bigg)\) = (1, -3)Slope of BD = \(rac{-5+1}{4+2}\) = \(rac{-4}{6}\) ... (rac{3}{2}\) isy + 3 = \(rac{3}{2}\) (x - 1)⇒ 2y + 6 = 3x - 3 ⇒ 2y = 3x - 9.

Description : What are the co-ordinates of the foot of the perpendicular from the point (2, 3) on the line x + y – 11 = 0 ? -Maths 9th

Last Answer : (d) 2x + 9y + 7 = 0PS being the median of ΔPQR, S is the mid-point of QR, i.e., Coordinates of S ≡ \(\bigg(rac{6+7}{2},rac{-1+3}{2}\bigg)\) = \(\bigg(rac{13}{2},1\bigg)\)Slope of line parallel to PS = Slope of PS= \(rac{1 ... y + 1) = \(rac{-2}{9}\)(x - 1), i.e., 9y + 9 = - 2x + 2 ⇒ 2x + 9y + 7 = 0.

Description : The co-ordinates of P and Q are (–3, 4) and (2, 1) respectively. -Maths 9th

Last Answer : (b) \(rac{\pi}{4}\)\(x\) cos θ + y sin θ = 2 ⇒ y sin θ = 2 - \(x\) cos θ ⇒ y = - \(x\) cot θ + 2 ⇒ Slope of this line = - cot θ Also, given \(x\) - y = 3 ⇒ y = \(x\) + 3 ⇒ Slope of this ... lines are perpendicular, - cot θ x 1 = -1⇒ cot θ = 1 ⇒ cot θ = cot \(rac{\pi}{4}\) ⇒ θ = \(rac{\pi}{4}\)

Description : For drawing a frequency polygon of a continuous frequency distribution, we plot the points whose ordinates are the frequencies of the respective classes and abscissae are, respectively -Maths 9th

Last Answer : NEED ANSWER

Description : For drawing a frequency polygon of a continuous frequency distribution, we plot the points whose ordinates are the frequencies of the respective classes and abscissae are, respectively -Maths 9th

Last Answer : (c) Class marks i.e., the mid-point of the classes are abscissa of the points, which we plot for frequency polygon.

Description : Two circles intersect at A and B. AC and AD are respectively the diameters of the circles. Prove that C, B and D are collinear. -Maths 9th

Last Answer : Join CB, BD and AB, Since, AC is a diameter of the circle with centre O. ∴ ∠ABC = 90° [angle in semi circle] ---- (i) Also, AD is a diameter of the circle with center O . ∴ ∠ABD = 90° [angle in ... ⇒ ∠ABC + ∠ABD = 180° So. CBD is a straight line. Hence C, B and D are collinear . Hence proved.

Description : Two circles intersect at A and B. AC and AD are respectively the diameters of the circles. Prove that C, B and D are collinear. -Maths 9th

Last Answer : Join CB, BD and AB, Since, AC is a diameter of the circle with centre O. ∴ ∠ABC = 90° [angle in semi circle] ---- (i) Also, AD is a diameter of the circle with center O . ∴ ∠ABD = 90° [angle in ... ⇒ ∠ABC + ∠ABD = 180° So. CBD is a straight line. Hence C, B and D are collinear . Hence proved.

Description : ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that (i) D is the mid-point of AC (ii) MD ⊥ AC (iii) CM = MA = ½ AB -Maths 9th

Last Answer : Solution: (i) In ΔACB, M is the midpoint of AB and MD || BC , D is the midpoint of AC (Converse of mid point theorem) (ii) ∠ACB = ∠ADM (Corresponding angles) also, ∠ACB = 90° , ∠ADM = 90° and MD ⊥ AC (iii ... SAS congruency] AM = CM [CPCT] also, AM = ½ AB (M is midpoint of AB) Hence, CM = MA = ½ AB

Description : (i) Find the co-ordinates of the points on the curve xy = 16 at which the normal drawn meet at origin. (ii) Find the points on the curve`4x^(2)+9 y^(2

Last Answer : (i) Find the co-ordinates of the points on the curve xy = 16 at which the normal drawn meet at ... ` at which the normal drawn is parallel to X-axis.

Description : If PQRS is trapezium such that PQ > RS and L, M are the mid-points of the diagonals PR and QS respectively then what is LM equal to? -Maths 9th

Last Answer : if pqrs is a trapezium so pq and RS are parallel is you draw a diagonal you will divide the trapezium into two parts such that two equal triangle so

Description : On plotting the points 0(0, 0), A(3, 0), 5(3, 4), C(0, 4) and joining OA, AB, BC and CO. -Maths 9th

Last Answer : (b) Here, point 0 (0, 0) is the origin. A(3, 0) lies on positive direction of X-axis, B (3, 4) lies in 1st quadrant and C (0, 4) lines on positive direction of Y-axis. On joining OA AB, BC and CO the figure obtained is a rectangle.

Description : On plotting the points 0(0, 0), A(3, 0), 5(3, 4), C(0, 4) and joining OA, AB, BC and CO. -Maths 9th

Last Answer : (b) Here, point 0 (0, 0) is the origin. A(3, 0) lies on positive direction of X-axis, B (3, 4) lies in 1st quadrant and C (0, 4) lines on positive direction of Y-axis. On joining OA AB, BC and CO the figure obtained is a rectangle.

Description : Draw the bar graph of linear equation whose solution are represented by the points having difference the co-ordinate as 25 units -Maths 9th

Last Answer : This answer was deleted by our moderators...

Description : ABCD is a parallelogram. The diagonals AC and BD intersect at the point O. If E, F, G and H are the mid-points of AO, DO, CO and BO respectively -Maths 9th

Last Answer : answer:

Description : If the sum of the squares of the distances of the point (x, y) from the points (a, 0) and (–a, 0) be 2b^2, then which of the following is correct ? -Maths 9th

Last Answer : (d) (0, 0)Let the vertices of Δ ABC be given as: A(0, 0), B(3, 0) and C(0, 4) The orthocentre O is the point of intersection of the altitudes drawn from the vertices of Δ ABC on the opposite sides. Slope of BC = \(rac{ ... ⇒ y = 0 ...(ii) From (i), \(x\) = 0 Hence, the orthocentre is (0, 0).

Description : Who Four points are always coplaner if:A. They lie on different planesB. They lie on different linesC. They line in the same planeD. They are collinear?

Last Answer : C. They lie in the same planeD. They are collinear

Description : If the angles of elevation of the top of tower from three collinear points `A`, `B` and `C`, on a line leading to the foot of the tower, are `30^(@)`,

Last Answer : If the angles of elevation of the top of tower from three collinear points `A`, `B` and `C`, on a line leading to ... )` C. `1 : sqrt(3)` D. `2 : 3`

Description : In the given figure, equal chords AB and CD of a circle with centre O cut at right angles at E. If M and N are the mid-points of AB and CD respectively, prove that OMEN is a square. -Maths 9th

Last Answer : Join OE. In ΔOME and ΔONE, OM =ON [equal chords are equidistant from the centre] ∠OME = ∠ONE = 90° OE =OE [common sides] ∠OME ≅ ∠ONE [by SAS congruency] ⇒ ME = NE [by CPCT] In quadrilateral OMEN, ... =ON , ME = NE and ∠OME = ∠ONE = ∠MEN = ∠MON = 90° Hence, OMEN is a square. Hence proved.

Description : In the given figure, equal chords AB and CD of a circle with centre O cut at right angles at E. If M and N are the mid-points of AB and CD respectively, prove that OMEN is a square. -Maths 9th

Last Answer : Join OE. In ΔOME and ΔONE, OM =ON [equal chords are equidistant from the centre] ∠OME = ∠ONE = 90° OE =OE [common sides] ∠OME ≅ ∠ONE [by SAS congruency] ⇒ ME = NE [by CPCT] In quadrilateral OMEN, ... =ON , ME = NE and ∠OME = ∠ONE = ∠MEN = ∠MON = 90° Hence, OMEN is a square. Hence proved.

Description : In Fig. 8.40, points M and N are taken on opposite sides AB and CD, respectively of a parallelogram ABCD such that AM = CN. Show that AC and MN bisect each other. -Maths 9th

Last Answer : Solution :-

Description : PQ and RS are two equal and parallel line segments.Any points M not lying on PQ or RS is joined to Q and S and lines through P parallel to SM meet at N.Prove that line segments MN and PQ are equal and parallel to each other. -Maths 9th

Last Answer : hope its clear

Description : Let O be any point inside a triangle ABC. Let L, M and N be the points on AB, BC and CA respectively, -Maths 9th

Last Answer : answer:

Description : Any convenient co-ordinate system or Cartesian co-ordinates which can be usedto define the picture is called a.spherical co-ordinates b.vector co-ordinates c.viewport co-ordinates d.world co-ordinates

Last Answer : d.world co-ordinates

Description : Co-ordinates are ranging according to the screen resolution. a.TRUE b.FALSE c. d.

Last Answer : a.TRUE

Description : Which co-ordinates allow common vector operations such as translation, rotation,scaling and perspective projection to be represented as a matrix by which the vector is multiplied? a.vector co-ordinates b.3D co-ordinates c.affine co-ordinates d.homogenous co-ordinates

Last Answer : d.homogenous co-ordinates