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

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

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 + 2k – 4k2 – 3k] = 0⇒ k2 + 2k = 0 ⇒ 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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Description : If the three points (k, 2k), (2k, 3k) and (3, 1) are collinear then k is equal to -Maths 9th

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Description : If the coordinates of the two points are P(-2, 3) and Q(-3, 5), then (Abscissa of P) – (Abscissa of Q) is -Maths 9th

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Description : If the coordinates of the two points are P(-2, 3) and Q(-3, 5), then (Abscissa of P) – (Abscissa of Q) is -Maths 9th

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If AB = 12 cm, BC = 16 cm and AB is perpendicular to BC, then the radius of the circle passing through the points A, B and C is -Maths 9th

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Description : if A,Band c are three points on a line and B lies between A and C then prove that AB+BC=AC -Maths 9th

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if A,Band c are three points on a line and B lies between A and C then prove that AB+BC=AC -Maths 9th

Description : if A,Band c are three points on a line and B lies between A and C then prove that AB+BC=AC -Maths 9th

Last Answer : AB=AC-BC BC =AC-AB AB+BC=AB HENCE PROVED

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Description : If the coordinates of two points are P( -2,3) and Q ( -3, 5) then find (abscissa of P)–(abscissa of Q) -Maths 9th

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Description : If P, Q and R are three points on a line and Q is between P and R,then prove that PR - QR= PQ. -Maths 9th

Last Answer : Solution :-

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Description : If a point O lies between two points P and R such that PO=OR then prove that PO= 1/2PR. -Maths 9th

Last Answer : THINGS WHICH ARE COINCIDE WITH EACH OTHER ARE EQUAL TO ONE ANOTHER PO+OR=PR 2PO=PR PO=OR PO=1/2PR HENCE PROVED

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

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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

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Write the coordinates of each of the points P, Q, R, S, T and 0 from the figure . -Maths 9th

Description : Write the coordinates of each of the points P, Q, R, S, T and 0 from the figure . -Maths 9th

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Description : Plot the points (x, y) given by the following table. Use scale 1 cm= 0.25 unit. -Maths 9th

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Description : Taking 0.5 cm as 1 unit, plot the following points on the graph paper. -Maths 9th

Last Answer : Here, in point 4(1, 3) both x and y-coordinates are positive, so it lies in I quadrant. In point 8(-3, -1),both x and y-coordinates are negative, so it lies in III quadrant. In point C(1, -4), x- ... is zero, so it lies on Y-axis and in point F(1,0) y-coordinate is zero, so it lies on X-axis.

Plot the points P(1, 0), Q(4, 0) and 5(1, 3). Find the coordinates of the point R such that PQRS is a square. -Maths 9th

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Description : Points A(5, 3), B(-2, 3) and 0(5, – 4) are three vertices of a square ABCD. -Maths 9th

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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

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.

Which of the points P(0, 3), Q(l, 0), R(0, – 1), S(-5, 0) and T(1, 2) do not lie on the X-axis ? -Maths 9th

Description : Which of the points P(0, 3), Q(l, 0), R(0, – 1), S(-5, 0) and T(1, 2) do not lie on the X-axis ? -Maths 9th

Last Answer : (c) We know that, if a point is of the form (x, 0)i.e., its y-coordinate is zero, then it will lie on X-axis otherwise not. Here, y-coordinates of points P(0, 3), R (0, -1) and T (1,2) are not zero, so these points do not lie on the X-axis.

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Description : Write the coordinates of each of the points P, Q, R, S, T and 0 from the figure . -Maths 9th

Last Answer : Here, points P and S lie in I quadrant so their both coordinates will be positive. Now, perpendicular distance of P from both axes is 1, so coordinates of P are (1, 1). Also, perpendicular distance of S ... 0 is the intersection of both axes, so it is the origin and its coordinates are O (0,0).

Plot the points (x, y) given by the following table. Use scale 1 cm= 0.25 unit. -Maths 9th

Description : Plot the points (x, y) given by the following table. Use scale 1 cm= 0.25 unit. -Maths 9th

Last Answer : Let X’OX and X’ OX be the coordinate axes. Plot the given points (1.25, -0.5), (0.25, 1), (1.5,1.5) and (-1.75, – 0.25) on the graph paper.

Taking 0.5 cm as 1 unit, plot the following points on the graph paper. -Maths 9th

Description : Taking 0.5 cm as 1 unit, plot the following points on the graph paper. -Maths 9th

Last Answer : Here, in point 4(1, 3) both x and y-coordinates are positive, so it lies in I quadrant. In point 8(-3, -1),both x and y-coordinates are negative, so it lies in III quadrant. In point C(1, -4), x- ... is zero, so it lies on Y-axis and in point F(1,0) y-coordinate is zero, so it lies on X-axis.

Points A(5, 3), B(-2, 3) and 0(5, – 4) are three vertices of a square ABCD. -Maths 9th

Description : Points A(5, 3), B(-2, 3) and 0(5, – 4) are three vertices of a square ABCD. -Maths 9th

Last Answer : The graph obtained by plotting the points A, B and C and D is given below. Take a point C on the graph such that ABCD is a square i.e., all sides AB, BC, CD, and AD are equal. So, abscissa of C should be ... of C should be equal to ordinate of D i.e., -4. Hence, the coordinates of C are (-2, - 4).

Plot the points P(1, 0), Q(4, 0) and 5(1, 3). Find the coordinates of the point R such that PQRS is a square. -Maths 9th

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Description : If the points (2, 1) and (1, – 2) are equidistant from the point (x, y), show that x + 3y = 0. -Maths 9th

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

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.