If the perpendicular distance of a point P from the X-axis is 5 units and the foot of the perpendicular lies on the negative direction of X-axis, then the point P has -Maths 9th

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Answer :

(d) We know that, the perpendicular distance of a point from the X-axis gives y-coordinate of that point. Here, foot of perpendicular lies on the negative direction of X-axis, so perpendicular distance can be measure in II quadrant or III quadrant. Hence, the point P has y-coordinate = 5 or -5.

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Description : If the perpendicular distance of a point P from the X-axis is 5 units and the foot of the perpendicular lies on the negative direction of X-axis, then the point P has -Maths 9th

Last Answer : (d) We know that, the perpendicular distance of a point from the X-axis gives y-coordinate of that point. Here, foot of perpendicular lies on the negative direction of X-axis, so perpendicular distance can be measure in II quadrant or III quadrant. Hence, the point P has y-coordinate = 5 or -5.

Description : The point which lies on Y-axis at a distance of 5 units in the negative direction of Y-axis is -Maths 9th

Last Answer : (C) Given the point lies on X-axis this shows that its ^-coordinate is zero. Also, it is at a distance of 5 units in negative direction of X-axis, so its y-coordinate” is negative.Hence, the required point is (0, – 5).

Description : The point which lies on Y-axis at a distance of 5 units in the negative direction of Y-axis is -Maths 9th

Last Answer : (C) Given the point lies on X-axis this shows that its ^-coordinate is zero. Also, it is at a distance of 5 units in negative direction of X-axis, so its y-coordinate” is negative.Hence, the required point is (0, – 5).

Description : A point lies on positive direction of X-axis at a distance of 7 units from the Y-axis. What are its coordinates ? -Maths 9th

Last Answer : Given, point lies on the positive direction of X-axis, so its y-coordinate will be zero and it is at a distance of 7 units from the X-axis, so its coordinates are (7, 0). If it lies on negative ... x-coordinate will be zero and its distance from X-axis is 7 units, so its coordinates are (0, -7).

Description : A point lies on positive direction of X-axis at a distance of 7 units from the Y-axis. What are its coordinates ? -Maths 9th

Last Answer : Given, point lies on the positive direction of X-axis, so its y-coordinate will be zero and it is at a distance of 7 units from the X-axis, so its coordinates are (7, 0). If it lies on negative ... x-coordinate will be zero and its distance from X-axis is 7 units, so its coordinates are (0, -7).

Description : Write the coordinates of a point on x-axis at a distance of 6 units from the origin in the positive direction of x-axis and then justify your answer. -Maths 9th

Last Answer : Solution :- As, any point on x-axis has coordinates (,)x0 where x is the distance from origin, so required coordinates are (6, 0).

Description : Find the perpendicular distance of the point P(5, 7) from the y-axis. -Maths 9th

Last Answer : Solution :- 5

Description : The perpendicular distance of the point P(3, 4) from the Y-axis is -Maths 9th

Last Answer : (a) We know that, abscissa or the x-coordinate of a point is its perpendicular distance from the Y-axis. So, perpendicular distance of the point P(3, 4)from Y-axis = Abscissa = 3.

Description : The perpendicular distance of the point P(3, 4) from the Y-axis is -Maths 9th

Last Answer : (a) We know that, abscissa or the x-coordinate of a point is its perpendicular distance from the Y-axis. So, perpendicular distance of the point P(3, 4)from Y-axis = Abscissa = 3.

Description : The position of a boy on the coordinate plane is given by the point (4,6) . What is the perpendicular distance from the x-axis and the y-axis ? -Maths 9th

Last Answer : answer:

Description : Write the coordinates of the vertices of a rectangle whose lenght and breadth are 7 and 4 units respectively,one vertex atthe the origin,the longer side lies on the x-axis and one of the vertices lies in the third quadrant. -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 : Write the coordinates of the vertices of a rectangle whose length and breadth are 6 and 3 units respectively, one vertex at the origin, the longer side lies on the y-axis and one of the vertices lies in the second quadrant. -Maths 9th

Last Answer : Solution :-

Description : At what point does the graph of the linear equation 2x + 3y = 9 meet a line which is parallel to the y-axis, at a distance of 4 units from the origin and on the right of the y-axis? -Maths 9th

Last Answer : hope its clear

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 : Draw the graph of the equation represented by a straight Line which is parallel to the X-axis and at a distance 3 units below it. -Maths 9th

Last Answer : Any straight line parallel to X-axis in negative direction of Y-axis is given by y = - k, where k is the distance of the line from the X-axis. Here, k = 3. Therefore, the equation of the line is y = -3. To ... , plot the points (1,-3), (2, -3) and (3, -3) and join them. This is the required graph.

Description : Draw the graph of the equation represented by a straight Line which is parallel to the X-axis and at a distance 3 units below it. -Maths 9th

Last Answer : Any straight line parallel to X-axis in negative direction of Y-axis is given by y = - k, where k is the distance of the line from the X-axis. Here, k = 3. Therefore, the equation of the line is y = -3. To ... , plot the points (1,-3), (2, -3) and (3, -3) and join them. This is the required graph.

Description : Write the equation of the line parallel to the x-axis at distance 3 units above x-axis. -Maths 9th

Last Answer : Solution :-

Description : In figure LM is a line parallel to the Y-axis at a distance of 3 units. -Maths 9th

Last Answer : Given, LM is a line parallel to the Y-axis and its perpendicular distance from Y-axis is 3 units. (i) Coordinate of point P = (3, 2) [since, its perpendicular distance from X-axis is 2] Coordinate of ... 3, abscissa of point M = 3 Difference between the abscissa of the points L and M = 3 -3 = 0

Description : In figure LM is a line parallel to the Y-axis at a distance of 3 units. -Maths 9th

Last Answer : Given, LM is a line parallel to the Y-axis and its perpendicular distance from Y-axis is 3 units. (i) Coordinate of point P = (3, 2) [since, its perpendicular distance from X-axis is 2] Coordinate of ... 3, abscissa of point M = 3 Difference between the abscissa of the points L and M = 3 -3 = 0

Description : In fig.4.7, LM is a line parallel to the y - axis at a distance of 2 units. -Maths 9th

Last Answer : Given , LM is a line parallel to the Y-axis and its perpendicular distance from Y-axis is 3 units. (i) Coordinates of point P=(3,2) [since,its perpendicular distance from X-axis is 2] Coordinate of point ... L=3 , abscissa of point M=3 ∴∴ Difference between the abscissa of the points L and M =3-3=0

Description : The point whose ordinate is 4 and which lies on K-axis is -Maths 9th

Last Answer : (b) Given ordinate of the point is 4 arid the point lies on Y-axis, so its abscissa is zero. Hence, the required point is (0, 4).

Description : The point whose ordinate is 4 and which lies on K-axis is -Maths 9th

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Description : Find the point whose ordinate is –3 and which lies on y-axis. -Maths 9th

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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

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Description : The acute angle which the perpendicular from the origin on the line 7x –3y = 4 makes with the x-axis is: -Maths 9th

Last Answer : (c) negativeAs the line from the origin is perpendicular to the line 7x - 3y = 4, so its slope = \(rac{-1}{ ext{slope of }\,7x-3y=4}\)Slope of 7x - 3y - 4 = \(rac{7}{3}\)∴ Slope of line from origin = \(rac{-1} ... of x-axis⇒ θ = tan-1 \(\big(rac{-3}{7}\big)\) = - tan-1 \(\big(rac{3}{7}\big)\)

Description : The work done by a body on application of a constant force is the product of the constant force and distance travelled by the body in the direction of force. Express this in the form of a linear equation in two variables and draw its graph by taking the constant force as 3 units. -Maths 9th

Last Answer : Solution :-

Description : The two vertices of a triangle are (2, –1), (3, 2) and the third vertex lies on the line x + y = 5. The area of the triangle is 4 units. -Maths 9th

Last Answer : (c) (5, 0) or (1, 4) Let the third vertex of the triangle be P(a, b). Since it lies on the line x + y = 5, a + b = 5 ...(i) Also, given area of triangle formed by the points (2, -1), (3, 2) and (a, b) = 4 ... b) - (-3a + b) = 5 + 15⇒ 4a = 20 ⇒ a = 5 ⇒ b = 0. ∴ The points are (1, 4) and (5, 0).

Description : In a right-angled triangle ABC, D is the foot of the perpendicular from B on the hypotenuse AC -Maths 9th

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Description : WXYZ is a square of side length 30. V is a point on XY and P is a point inside the square with PV perpendicular to XY. PW = PZ = PV – 5. Find PV. -Maths 9th

Last Answer : answer:

Description : If P (5,1), Q (8, 0), R(0, 4), S(0, 5) and O(0, 0) are plotted on the graph paper, then the points on the X-axis is/are -Maths 9th

Last Answer : (d) We know that, a point lies on X-axis, if its y-coordinate is zero. So, on plotting the given points on graph paper, we get Q and O lie on the X-axis.

Description : If P (5,1), Q (8, 0), R(0, 4), S(0, 5) and O(0, 0) are plotted on the graph paper, then the points on the X-axis is/are -Maths 9th

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Description : If P, Q and R are the mid-points of the sides, BC, CA and AB of a triangle and AD is the perpendicular from A on BC, then prove that P, Q, R and D are concyclic. -Maths 9th

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Description : If P, Q and R are the mid-points of the sides, BC, CA and AB of a triangle and AD is the perpendicular from A on BC, then prove that P, Q, R and D are concyclic. -Maths 9th

Last Answer : According to question prove that P, Q, R and D are concyclic.

Description : If the perpendicular bisector of a chord AB of a circle PXAQBY intersects the circle at P and Q, then prove that arc PXA ≅ arc PYB. -Maths 9th

Last Answer : Solution :- Let AB be a chord of a circle having centre at O. Let PQ be the perpendicular bisector of the chord AB intersect it say at M. Perpendicular bisector of the chord passes through the centre of the circle,i. ... = PM (Common) ∴ △APM ≅ △BPM (SAS) PA = PB (CPCT) Hence, arc PXA ≅ arc PYB

Description : Perpendiculars are drawn from the vertex of the obtuse angles of a rhombus to its sides. The length of each perpendicular is equal to a units. -Maths 9th

Last Answer : answer:

Description : Draw the graph of the equation represented by the straight line which is parallel to the x-axis and 3 units above it. -Maths 9th

Last Answer : Solution :- Any straight line parallel to x-axis is given by y = a, where a is the distance of the line from the x-axis. Here a = 3. Therefore the equation of the line is y = 3. To draw the graph of this equation plot the points (0, 3) and (3, 3) and join them.

Description : A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree. -Maths 9th

Last Answer : The broken part bends so that the top of the tree touches the ground making an angle 30 °. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Original height of ... After rationalising Now we have , A`BC = AB + BC Take LCM So , original height of tree is

Description : Which of the following points lies on Y-axis ? -Maths 9th

Last Answer : We know that, a point lies on the Y-axis, if its x-coordinate is zero. Here, x-coordinate of points C(0, 1), D(0, 0), E(0,-1) and G(0, 5) are zero. So, these points lie on Y-axis. Also ... 0) is the intersection point of both.the axes, so we can consider that it lies on Y-axis as well as on X-axis.

Description : Which of the following points lies on Y-axis ? -Maths 9th

Last Answer : We know that, a point lies on the Y-axis, if its x-coordinate is zero. Here, x-coordinate of points C(0, 1), D(0, 0), E(0,-1) and G(0, 5) are zero. So, these points lie on Y-axis. Also ... 0) is the intersection point of both.the axes, so we can consider that it lies on Y-axis as well as on X-axis.

Description : For what value of p the point (p, 2) lies on the line 3x + y = 11? -Maths 9th

Last Answer : Solution :-

Description : Draw a circle of diameter 6.4 cm. Then draw two tangents to the circle from a point P at a distance 6.4 cm from the centre of the circle. -Maths 9th

Last Answer : clear .

Description : Pick up the correct statement from the following:  (A) A level surface is perpendicular at all points to the direction of gravity  (B) A level line lies in level surface  (C) A horizontal surface is normal to the direction of gravity at only one point  (D) All the above 

Last Answer : (D) All the above 

Description : Find the perpendicular distance between the lines 9x + 40y – 20 = 0 and 9x + 40y + 21 = 0. -Maths 9th

Last Answer : Given line: x - 2y = 3 ⇒ y = \(rac{x}{2}\) - \(rac{3}{2}\) ....(i)∴ Its slope = m1 = \(rac{1}{2}\)Let m2 be the slope of line through (3, 2). Since this line is inclined at 45º to line (i),tan 45º = \(\bigg ... (x -3)⇒ 3y - 6 = - x + 3 or y - 2 = 3x - 9 ⇒ 3y + x - 9 = 0 or y - 3x + 7 = 0.

Description : If y-coordinate of a point is zero, then where will this point lie in the coordinate plane? On the x-axis. -Maths 9th

Last Answer : Solution :- On the x-axis.

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.

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.

Description : If the perpendicular bisector of a chord AB of a circle PXAQBY intersects the circle at P and Q, prove that arc PXA = arc PYB. -Maths 9th

Last Answer : Let AB be a chord of a circle having centre at OPQ be the perpendicular bisector of the chord AB, which intersects at M and it always passes through O. To prove arc PXA ≅ arc PYB Construction Join AP and BP. Proof In ... ΔBPM, AM = MB ∠PMA = ∠PMB PM = PM ∴ ΔAPM s ΔBPM ∴PA = PB ⇒arc PXA ≅ arc PYB

Description : If the perpendicular bisector of a chord AB of a circle PXAQBY intersects the circle at P and Q, prove that arc PXA = arc PYB. -Maths 9th

Last Answer : Let AB be a chord of a circle having centre at OPQ be the perpendicular bisector of the chord AB, which intersects at M and it always passes through O. To prove arc PXA ≅ arc PYB Construction Join AP and BP. Proof In ... ΔBPM, AM = MB ∠PMA = ∠PMB PM = PM ∴ ΔAPM s ΔBPM ∴PA = PB ⇒arc PXA ≅ arc PYB

Description : If p is the length of the perpendicular drawn from the origin to the line -Maths 9th

Last Answer : Let the x-intercept = a. Then y-intercept = -1 - a The equation of the required line is \(rac{x}{a}\) + \(rac{y}{-1-a}\) = 1Given, it passes through (4, 3), so,\(rac{4}{a}\) + \(rac{3}{-1-a}\) = 1⇒ - 4 - 4a ... 2}\) - \(rac{y}{3}\) = 1When a = -2, required line is \(rac{x}{-2}\) + \(rac{y}{3}\) = 1