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

If the points (x, 1), (1, 2) and (0, y + 1) are collinear show that -Maths 9th
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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.
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Find the relation between x and y if points (2, 1), (x, y) and (7, 5) are collinear. -Maths 9th

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

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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.

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

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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.

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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.

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Plot the points (x, y) given by the following table. -Maths 9th

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Description : X and Y are points on the side LN of the triangle LMN such that LX = XY = YN. -Maths 9th

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In figure X and Y are the mid-points of AC and AB respectively, QP || BC and CYQ and BXP are straight lines. -Maths 9th

Description : In figure X and Y are the mid-points of AC and AB respectively, QP || BC and CYQ and BXP are straight lines. -Maths 9th

Last Answer : Given X and Y are the mid-points of AC and AB respectively. Also, QP|| BC and CYQ, BXP are straight lines. To prove ar (ΔABP) = ar (ΔACQ) Proof Since, X and Y are the mid-points of AC and AB respectively. So, ... ar (ΔBYX) + ar (XYAP) = ar (ΔCXY) + ar (YXAQ) ⇒ ar (ΔABP) = ar (ΔACQ) Hence proved.

In quadrilateral ABCD of the given figure, X and Y are points on diagonal AC such that AX = CY and BXDY ls a parallelogram. -Maths 9th

Description : In quadrilateral ABCD of the given figure, X and Y are points on diagonal AC such that AX = CY and BXDY ls a parallelogram. -Maths 9th

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

X and y are points on the side LN of the triangle LMN , such that LX = XY = YN . Through X, a line is drawn parallel to LM to meet MN at Z. -Maths 9th

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Last Answer : Here, △XZM and △XZL are on the same base (XZ) and lie between the same parallels (XZ || LM). ∴ ar(△XZL) = ar( △XZM) Adding ar(△XZY) on both sides , we have ar(△XZL) + ar(△XZY) = ar(△XZM) + ar(△XZY) ⇒ ar(△LZY) = ar(quad.MZYX)

Plot the points (x, y) given by the following table. -Maths 9th

Description : Plot the points (x, y) given by the following table. -Maths 9th

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X and Y are points on the side LN of the triangle LMN such that LX = XY = YN. -Maths 9th

Description : X and Y are points on the side LN of the triangle LMN such that LX = XY = YN. -Maths 9th

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In figure X and Y are the mid-points of AC and AB respectively, QP || BC and CYQ and BXP are straight lines. -Maths 9th

Description : In figure X and Y are the mid-points of AC and AB respectively, QP || BC and CYQ and BXP are straight lines. -Maths 9th

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Description : Draw the graph of the linear equation 3x + 4y = 6. At what points, does the graph cut the x-axis and the y-axis? -Maths 9th

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Let ABCD be a quadrilateral. Let X and Y be the mid-points of AC and BD respectively and the lines through X and Y respectively -Maths 9th

Description : Let ABCD be a quadrilateral. Let X and Y be the mid-points of AC and BD respectively and the lines through X and Y respectively -Maths 9th

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X, Y are the mid-points of opposite sides AB and DC of a parallelogram ABCD. AY and DX are joined intersecting in P. CX and BY are joined -Maths 9th

Description : X, Y are the mid-points of opposite sides AB and DC of a parallelogram ABCD. AY and DX are joined intersecting in P. CX and BY are joined -Maths 9th

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Write the coordinates of two points on X-axis and two points on Y-axis which are at equal distances from the origin. Connect all these points and make them as vertices of quadrilateral. Name the quadrilateral thus formed. -Maths 9th

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Last Answer : Let a be the equal distance from origin on both axes. Now, the coordinates of two points on equal distance 'a'on x-axis are Pla, 0) and R(-a, 0). Also, the coordinates of two points on equal distance 'a' on Y-axis are Q(0, a) and S(0, -a). Join all the four points on the graph.

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6. Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other. -Maths 9th

Description : 6. Show that the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other. -Maths 9th

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5. In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see Fig. 8.31). Show that the line segments AF and EC trisect the diagonal BD. -Maths 9th

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Last Answer : . Solution: Given that, ABCD is a parallelogram. E and F are the mid-points of sides AB and CD respectively. To show, AF and EC trisect the diagonal BD. Proof, ABCD is a parallelogram , AB || CD also, ... (i), DP = PQ = BQ Hence, the line segments AF and EC trisect the diagonal BD. Hence Proved.

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Description : 3. ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus. -Maths 9th

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Description : 2. ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle. -Maths 9th

Last Answer : Solution: Given in the question, ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. To Prove, PQRS is a rectangle. Construction, Join AC and BD. Proof: In ΔDRS and ... , In PQRS, RS = PQ and RQ = SP from (i) and (ii) ∠Q = 90° , PQRS is a rectangle.

ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig 8.29). AC is a diagonal. Show that: (i) SR || AC and SR = 1/2 AC (ii) PQ = SR (iii) PQRS is a parallelogram. -Maths 9th

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Description : Points P and Q have been taken on opposite sides AB and CD, respectively of a parallelogram ABCD such that AP = CQ. Show that AC and PQ bisect each other. -Maths 9th

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Points P and Q have been taken on opposite sides AB and CD, respectively of a parallelogram ABCD such that AP = CQ. Show that AC and PQ bisect each other. -Maths 9th

Description : Points P and Q have been taken on opposite sides AB and CD, respectively of a parallelogram ABCD such that AP = CQ. Show that AC and PQ bisect each other. -Maths 9th

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In Fig. 8.37, ABCD is a parallelogram and P, Q are the points on the diagonal BD such that BQ = DP. Show what APCQ is a parallelogram. -Maths 9th

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Last Answer : Solution :-

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

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ABCD is a parallelogram in which P and Q are the mid-points of opposite sides AB and CD (Fig. 8.48). If AQ intersects DP at S and BQ intersects CP at R, show that -Maths 9th

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Last Answer : Solution :-

ABCD is a rectangle and p q r s are the mid points of the side AB BC CD AND DA respectively. Show that the quadrilateral PQRS is a rhombus -Maths 9th

Description : ABCD is a rectangle and p q r s are the mid points of the side AB BC CD AND DA respectively. Show that the quadrilateral PQRS is a rhombus -Maths 9th

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

Let P(–3, 2), Q(–5, –5), R(2, –3) and S(4, 4) be four points in a plane. Then show that PQRS is a rhombus. Is it a square ? -Maths 9th

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Last Answer : Let P(1, -1), Q \(\big(rac{-1}{2},rac{1}{2}\big)\) and R(1,2) be the vertices of the ΔPQR.Then, PQ = \(\sqrt{\big(rac{-1}{2}-1\big)^2+\big(rac{1}{2}+1\big)^2}\) = \(\sqrt{rac{9}{4}+rac{9}{4}} ... {3\sqrt2}{2}\)PR = \(\sqrt{(1-1)^2+(2+1)^2}\) = \(\sqrt9\) = 3∵ PQ = QR, the triangle PQR is isosceles.

A rectangle is formed by joining the mid-points of the sides of a rhombus. Show that the area of rectangle is half the area of rhombus. -Maths 9th

Description : A rectangle is formed by joining the mid-points of the sides of a rhombus. Show that the area of rectangle is half the area of rhombus. -Maths 9th

Last Answer : hope its clear

If x+y=10 and x=z then show that z+y=10 by using appropriate eculids axioms? -Maths 9th

Description : If x+y=10 and x=z then show that z+y=10 by using appropriate eculids axioms? -Maths 9th

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

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Description : If x+y =10 and x=z then show that z+y =10 -Maths 9th

Last Answer : It is given that x+y =10 Also x= z Therefore, x+y =10 Z+y =10 ( x = z)

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Description : If x+y=10 and x=z then show that z+y=10 by using appropriate eculids axioms? -Maths 9th

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

If x+y =10 and x=z then show that z+y =10 -Maths 9th

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Last Answer : It is given that x+y =10 Also x= z Therefore, x+y =10 Z+y =10 ( x = z)