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

In figure LM is a line parallel to the Y-axis at a distance of 3 units. -Maths 9th
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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 point 0 = (3, -1) [since, its perpendicular distance from X-axis is 1 in negative direction of Y-axis]. Coordinate of point R = (3, 0) [since its lies on X-axis, so its y-coordinate is zero]. (ii) Abscissa of point L = 3, abscissa of point M = 3 Difference between the abscissa of the points L and M = 3 -3 = 0
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In figure LM is a line parallel to the Y-axis at a distance of 3 units. -Maths 9th

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

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

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

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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)

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Description : 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

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)

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Description : In the given figure, YZ is parallel to MN, XY is parallel is LM and XZ is parallel to LN. Then MY is -Maths 9th

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

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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).

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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).

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The equation of a line parallel to the y- axis is of the form ................. -Maths 9th

Description : The equation of a line parallel to the y- axis is of the form ................. -Maths 9th

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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 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

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

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Write the equation of a line parallel to x-axis and passing through the point (–3, –4). -Maths 9th

Description : Write the equation of a line parallel to x-axis and passing through the point (–3, –4). -Maths 9th

Last Answer : Solution :- y = -4

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Description : Which axis is parallel to the line on which the two points with coordinates (4, 3) and (4,–2) lie? -Maths 9th

Last Answer : Solution :- As x-coordinate of both points is 4. So, both points lie on the line x = 4 which is parallel to y-axis.

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Description : Two points with coordinates (3, 4) and (–5, 4) lie on a line parallel to which axis? Justify your answer. -Maths 9th

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Write the equation of line parallel to x- axis. -Maths 9th

Description : Write the equation of line parallel to x- axis. -Maths 9th

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

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.

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

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.

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

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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

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

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Description : P is the mid - point of side AB of a parallelogram ABCD. A line through B parallel to PD meets DC at Q and AD produced at R (see figure). -Maths 9th

Last Answer : (i) In △ARB,P is the mid point of AB and PD || BR. ∴ D is a mid - point of AR [converse of mid - point theorem] ∴ AR = 2AD But BC = AD [opp sides of ||gm ABCD] Thus, AR = 2BC (ii) ∴ ABCD is a ... a mid - point of AR and DQ || AB ∴ Q is a mid point of BR [converse of mid - point theorem] ⇒ BR = 2BQ

The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q, then parallelogram PBQR is completed (see figure). -Maths 9th

Description : The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q, then parallelogram PBQR is completed (see figure). -Maths 9th

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P is the mid - point of side AB of a parallelogram ABCD. A line through B parallel to PD meets DC at Q and AD produced at R (see figure). -Maths 9th

Description : P is the mid - point of side AB of a parallelogram ABCD. A line through B parallel to PD meets DC at Q and AD produced at R (see figure). -Maths 9th

Last Answer : (i) In △ARB,P is the mid point of AB and PD || BR. ∴ D is a mid - point of AR [converse of mid - point theorem] ∴ AR = 2AD But BC = AD [opp sides of ||gm ABCD] Thus, AR = 2BC (ii) ∴ ABCD is a ... a mid - point of AR and DQ || AB ∴ Q is a mid point of BR [converse of mid - point theorem] ⇒ BR = 2BQ

The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q, then parallelogram PBQR is completed (see figure). -Maths 9th

Description : The side AB of a parallelogram ABCD is produced to any point P. A line through A and parallel to CP meets CB produced at Q, then parallelogram PBQR is completed (see figure). -Maths 9th

Last Answer : Join AC and QP, also it is given that AQ || CP ∴ △ACQ and △APQ are on the same base AQ and lie between the same parallels AQ || CP. ∴ ar(△ACQ) = ar(△APQ) or ar(△ABC) + ar(△ABQ) = ar(△BPQ) + ar(△ABQ) or ar(△ABC) = ar( △BPQ) or 1/2 ar(||gm ABCD) = 1/2 ar(||gm PBQR) or ar(||gm ABCD) = ar(||gm PBQR)

In the given figure, line DE is parallel to line AB. CD = 3 while DA = 6. Which of the following must be true? -Maths 9th

Description : In the given figure, line DE is parallel to line AB. CD = 3 while DA = 6. Which of the following must be true? -Maths 9th

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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

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

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The equation of the line bisecting the join of (3, – 4) and (5, 2) and having its intercepts on the x-axis and y-axis in the ratio 2 : 1 is -Maths 9th

Description : The equation of the line bisecting the join of (3, – 4) and (5, 2) and having its intercepts on the x-axis and y-axis in the ratio 2 : 1 is -Maths 9th

Last Answer : (d) \(\lambda\) = \(\mu\) Let the equation of the line in the intercept from be\(rac{x}{\lambda}\)+ \(rac{y}{\mu}\) = 1Since it passes through (4, 3) and (2, 5)\(rac{4}{\lambda}\) + \(rac{3}{\mu}\) = 1 ... ) = 1 - \(rac{3}{7}\) = \(rac{4}{7}\) = \(\lambda\) = 7∴ \(\lambda\) = \(\mu\) = 7.

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

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).

For what value of x + y in figure will ABC be a line? Justify your answer. -Maths 9th

Description : For what value of x + y in figure will ABC be a line? Justify your answer. -Maths 9th

Last Answer : For ABC to be a line, the sum of the two adjacent angles must be 180° i.e.,x + y = 180°.

For what value of x + y in figure will ABC be a line? Justify your answer. -Maths 9th

Description : For what value of x + y in figure will ABC be a line? Justify your answer. -Maths 9th

Last Answer : For ABC to be a line, the sum of the two adjacent angles must be 180° i.e.,x + y = 180°.

In a trapezoid ABCD, side BC is parallel to side AD. Also, the lengths of the sides AB, BC, CD and AD are 8, 2, 8 and 10 units respectively -Maths 9th

Description : In a trapezoid ABCD, side BC is parallel to side AD. Also, the lengths of the sides AB, BC, CD and AD are 8, 2, 8 and 10 units respectively -Maths 9th

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What is the equation of the line having the y-intercept –1 and parallel to the line y = 5x – 7 ? -Maths 9th

Description : What is the equation of the line having the y-intercept –1 and parallel to the line y = 5x – 7 ? -Maths 9th

Last Answer : Slope of AB = \(rac{2-4}{1-0}\) = -2, Slope of BC = \(rac{3-2}{3-1}\) = \(rac{1}{2}\)Slope of AC = \(rac{3-4}{3-0}\) = \(-rac{1}{3}\)Slope of AB × Slope of BC = -2 x \(rac{1}{2}\) = -1∴ AB ⊥ BC, i.e, ∠B = 90º ⇒ ΔABC is a right angled.

The line L is given by x/5 + y/b = 1 passes through the point (13, 32). The line K is parallel to L and has the equation -Maths 9th

Description : The line L is given by x/5 + y/b = 1 passes through the point (13, 32). The line K is parallel to L and has the equation -Maths 9th

Last Answer : (a) 45º The equations of the given lines are: A\(x\) + By = A + B ⇒ By = -A\(x\) + (A + B) ⇒ y = \(-rac{A}{B}x\) + \(rac{(A+B)}{B}\) ....(i)and A(\(x\) - y) + B(\(x\) ... (ii) = m2 = \(rac{(A+B)}{B-A}\)Let θ be the angle between both the lines, then∴ tan θ = 1 ⇒ θ = tan-1 (1) = 45°.

A straight line is parallel to the lines 3x – y – 3 = 0 and 3x – y + 5 = 0 and lies between them. -Maths 9th

Description : A straight line is parallel to the lines 3x – y – 3 = 0 and 3x – y + 5 = 0 and lies between them. -Maths 9th

Last Answer : (c) 3x - 4y + 15 = 0 Let (m > 0) be the gradient (slope) of the required line. Then, Equation of any line through (-5, 0) having slope = m is y - 0 = m(x - (-5)) or mx - y + 5m = 0 ...(i) Its ... (rac{3}{4}\) (∵ m is +ve)∴ Required equation: y = \(rac{3}{4}\) (x + 5)⇒ 3x - 4y + 15 = 0.

The area of the triangular region, in the following figure, bounded by the lines 2x-y=1 and x+2y=13 and Y-axis is (a) 11.25 sq.units (b) 9.75 sq.units (c) 16.25 sq,units (d) 18.75 sq.units

Description : The area of the triangular region, in the following figure, bounded by the lines 2x-y=1 and x+2y=13 and Y-axis is (a) 11.25 sq.units (b) 9.75 sq.units (c) 16.25 sq,units (d) 18.75 sq.units

Last Answer : (a) 11.25 sq.units

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

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

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ABC is an acute angled triangle. CD is the altitude through C. If AB = 8 units, CD = 6 units, find the distance -Maths 9th

Description : ABC is an acute angled triangle. CD is the altitude through C. If AB = 8 units, CD = 6 units, find the distance -Maths 9th

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l, m and n are three parallel lines intersected by transversals p and q such that l, m and n cut off equal intercepts AB and BC on p (see figure). -Maths 9th

Description : l, m and n are three parallel lines intersected by transversals p and q such that l, m and n cut off equal intercepts AB and BC on p (see figure). -Maths 9th

Last Answer : Though E, draw a line parallel to p intersecting L at G and n at H respectively. Since l | | m ⇒ AG | | BE and AB | | GE [by construction] ∴ Opposite sides of quadrilateral AGEB are ... ∠DGE = ∠FHE [alternate interior angles] By ASA congruence axiom, we have △DEG ≅ △FEH Hence, DE = EF