Whether the pair of given lines are parallel or not give reason. -Maths 9th

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Where are the lines please tell first

Related questions

Description : If a transversal intersects two parallel lines, prove that the bisectors of any pair of corresponding angles so formed are parallel. -Maths 9th

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Description : AP and BQ are the bisectors of the two alternate interior angles formed by the intersection of a transversal t with parallel lines l and m (in the given figure). Show that AP || BQ. -Maths 9th

Last Answer : Given In the figure l || m, AP and BQ are the bisectors of ∠EAB and ∠ABH, respectively. To prove AP|| BQ Proof Since, l || m and t is transversal. Therefore, ∠EAB = ∠ABH [alternate interior ... ∠PAB and ∠ABQ are alternate interior angles with two lines AP and BQ and transversal AB. Hence, AP || BQ.

Description : AP and BQ are the bisectors of the two alternate interior angles formed by the intersection of a transversal t with parallel lines l and m (in the given figure). Show that AP || BQ. -Maths 9th

Last Answer : Given In the figure l || m, AP and BQ are the bisectors of ∠EAB and ∠ABH, respectively. To prove AP|| BQ Proof Since, l || m and t is transversal. Therefore, ∠EAB = ∠ABH [alternate interior ... ∠PAB and ∠ABQ are alternate interior angles with two lines AP and BQ and transversal AB. Hence, AP || BQ.

Description : EF is the transversal to two parallel lines AB and CD. GM and HL are the bisector of the corresponding angles EGB and EHD.Prove that GL parallel to HL. -Maths 9th

Last Answer : AB || CD and a transversal EF intersects them ∴ ∠EGB = ∠GHD ( Corresponding Angles) ⇒ 2 ∠EGM = 2 ∠GHL ∵ GM and HL are the bisectors of ∠EGB and ∠EHD respectively. ⇒ ∠EGM = ∠GHL But these angles form a pair of equal corresponding angles for lines GM and HL and transversal EF. ∴ GM || HL.

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

Description : Two lines are respectively perpendicular to two parallel lines. Show that they are parallel to each other. -Maths 9th

Last Answer : Given Two lines m and n are parallel and another two lines p and q are respectively perpendicular to m and n. i.e., p ⊥ m, p ⊥ n, q ⊥ m, q ⊥ n To prove p||g Proof Since, ... of interior angles on the same side of the transversal is supplementary, then the two lines are parallel. Hence, p||g.

Description : A transversal intersects two parallel lines. -Maths 9th

Last Answer : Given Two lines AB and CD are parallel and intersected by transversal t at P and 0, respectively. Also, EP and FQ are the bisectors of angles ∠APG and ∠CQP, respectively.

Description : If APB and CQD are two parallel lines, then the bisectors of the angles APQ, BPQ, CQP and PQD form -Maths 9th

Last Answer : According to question the bisectors of the angles APQ, BPQ, CQP and PQD form

Description : EF is the transversal to two parallel lines AB and CD. GM and HL are the bisector of the corresponding angles EGB and EHD.Prove that GL parallel to HL. -Maths 9th

Last Answer : AB || CD and a transversal EF intersects them ∴ ∠EGB = ∠GHD ( Corresponding Angles) ⇒ 2 ∠EGM = 2 ∠GHL ∵ GM and HL are the bisectors of ∠EGB and ∠EHD respectively. ⇒ ∠EGM = ∠GHL But these angles form a pair of equal corresponding angles for lines GM and HL and transversal EF. ∴ GM || HL.

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

Description : Two lines are respectively perpendicular to two parallel lines. Show that they are parallel to each other. -Maths 9th

Last Answer : Given Two lines m and n are parallel and another two lines p and q are respectively perpendicular to m and n. i.e., p ⊥ m, p ⊥ n, q ⊥ m, q ⊥ n To prove p||g Proof Since, ... of interior angles on the same side of the transversal is supplementary, then the two lines are parallel. Hence, p||g.

Description : A transversal intersects two parallel lines. -Maths 9th

Last Answer : Given Two lines AB and CD are parallel and intersected by transversal t at P and 0, respectively. Also, EP and FQ are the bisectors of angles ∠APG and ∠CQP, respectively.

Description : If APB and CQD are two parallel lines, then the bisectors of the angles APQ, BPQ, CQP and PQD form -Maths 9th

Last Answer : According to question the bisectors of the angles APQ, BPQ, CQP and PQD form

Description : Does Euclid's fifth postulate imply the existence of parallel lines?Explain. -Maths 9th

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Description : A transversal intersects two lines in such a way that the two interior angle on the same side of transversal are equal.Will the two lines always be parallel? -Maths 9th

Last Answer : Solution :- The two lines will not be always parallel as the sum of the two equal angles will not always be 180°. Lines will be parallel when each of the equal angles is equal to 90°.

Description : In Fig.6.5,which of the two lines are parallel? -Maths 9th

Last Answer : Solution :- l||m, because angles on the same side of the transversal are supplementary, i.e., 128° +52° = 180°. Therefore p is not parallel to q, because 105° + 74° = 179°.

Description : In Fig.6.6, find the value of x for which the lines l and m are parallel. -Maths 9th

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Description : For what value of x will the lines l and m be parallel to each other ? -Maths 9th

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Description : Two parallel lines l and m are intersected by a transversal p (see Fig. 8.46). Show that the quadrilateral formed by the bisectors of interior angles is a rectangle. -Maths 9th

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Description : l,m and n are three parallel lines intersected by transversal p and q such that l,m and n cut-off equal intersepts AB and BC on p (Fig.8.55). Show that l,m and n cut - off equal intercepts DE and EF on q also. -Maths 9th

Last Answer : Given:l∥m∥n l,m and n cut off equal intercepts AB and BC on p So,AB=BC To prove:l,m and n cut off equal intercepts DE and EF on q i.e.,DE=EF Proof:In △ACF, B is the mid-point of ... a triangle, parallel to another side, bisects the third side. Since E is the mid-point of DF DE=EF Hence proved.

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 : The relation “is parallel to” on a set S of all straight lines in a plane is : -Maths 9th

Last Answer : (d) An equivalence relationLet R = {(x, y) : line x is parallel to line y, x y ∈ set of coplanar straight lines}. Every line is parallel to itself. So, if x ∈S, then (x, x) ∈R ⇒ R is ... | z ⇒ (x, z) ∈R ⇒ R is transitive ∴ R being reflexive, symmetric and transitive, it is an equivalence relation.

Description : If m parallel lines in a plane are intersected by a family of n parallel lines, find the number of parallelograms formed? -Maths 9th

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Description : If two parallel lines are intersected by a transversal, then the bisectors of the interior angles form which one of the following? -Maths 9th

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Description : Show that the equation of the parallel line midway between the parallel lines -Maths 9th

Last Answer : ∵ Length of perpendicular from point (x1, y1) to line a\(x\) + by + c = 0 = \(rac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}\)∴ Length of perpendicular from (0, 0) to \(rac{x}{a}\) + \(rac{y}{b}\) = 1 ⇒ \(rac{\big|rac{1}{a} imes0+ ... {1}{b^2}+rac{1}{b^2}}\) ⇒ \(rac{1}{p^2}\) = \(rac{1}{a^2}\) + \(rac{1}{b^2}\)

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.

Description : The area of triangle ABC is 15 cm sq. If ΔABC and a parallelogram ABPD are on the same base and between the same parallel lines then what is the area of parallelogram ABPD. -Maths 9th

Last Answer : area of parallelogram=2× area of triangle ABC =2×15=30sq cm theorem on area

Description : In the following equations , verify whether the given value of the variable is a solution of the equation : -Maths 9th

Last Answer : (i) If we substitute x = 4, we get L.H.S = x + 4 = 4 + 4 = 8 and R.H.S = 2x = 2 x 4 = 8 ∴ L.H.S = R.H.S. Hence, 4 is a solution of x + 4 = 2x. (ii) If we substitute y = 3, we get L.H.S = y - 7 = 3 - ... S = 2u + 7 = 2(5) + 7 = 10 + 7 = 17 ∴ L.H.S = R.H.S. Hence , 5 is a solution of 3u + 2 = 2u + 7

Description : In the following equations , verify whether the given value of the variable is a solution of the equation : -Maths 9th

Last Answer : (i) f we substitute x = √2, we get L.H.S. = 2x - 3 = 2(√2) - 3 = 2√2 - 3 and R.H.S = x / 2 - 2 = √2 / 2 - 2 ∴ L.H.S. ≠ R.H.S . Hence, √2 is not a solution of 2x - 3 = x / 2 - 2 (ii) If we substitute ... = 7 ∴ L.H.S. ≠ R.H.S . Hence , -1 is not a solution of 24 - 3 (u - 2) = u + 8

Description : In the following equations , verify whether the given value of the variable is a solution of the equation : -Maths 9th

Last Answer : (i) If we substitute x = 4, we get L.H.S = x + 4 = 4 + 4 = 8 and R.H.S = 2x = 2 x 4 = 8 ∴ L.H.S = R.H.S. Hence, 4 is a solution of x + 4 = 2x. (ii) If we substitute y = 3, we get L.H.S = y - 7 = 3 - ... S = 2u + 7 = 2(5) + 7 = 10 + 7 = 17 ∴ L.H.S = R.H.S. Hence , 5 is a solution of 3u + 2 = 2u + 7

Description : In the following equations , verify whether the given value of the variable is a solution of the equation : -Maths 9th

Last Answer : (i) f we substitute x = √2, we get L.H.S. = 2x - 3 = 2(√2) - 3 = 2√2 - 3 and R.H.S = x / 2 - 2 = √2 / 2 - 2 ∴ L.H.S. ≠ R.H.S . Hence, √2 is not a solution of 2x - 3 = x / 2 - 2 (ii) If we substitute ... = 7 ∴ L.H.S. ≠ R.H.S . Hence , -1 is not a solution of 24 - 3 (u - 2) = u + 8

Description : A circle with centre O and diameter COB is given. If AB and CD are parallel, then show that chord AC is equal to chord BD. -Maths 9th

Last Answer : O Join AC and BD. Given, COB is the diameter of circle. ∠CAB = ∠BDC = 90° [angle in a semi-circle] Also, AB II CD ∠ABC = ∠DCB (alternate angles] Now, ∠ACB = 90° - ∠ABC and ∠DBC = 90° - ∠DCB = ... = ∠DBC BC = BC [common sides] ΔABC = ΔDCB [by ASA congruency] ∴ AC = BD [by CPCT] Hence Proved.

Description : In the given figure, bisectors AP and BQ of the alternate interior angles are parallel, then show that l || m. -Maths 9th

Last Answer : Given, In the figure AP|| BQ, AP and BQ are the bisectors of alternate interior angles ∠CAB and ∠ABF. To show l || m Proof Since, AP|| BQ and t is transversal, therefore ∠PAB = ∠ABQ [alternate interior angles] ⇒ 2 ∠PAB = 2 ∠ABQ [multiplying both sides by 2]

Description : A circle with centre O and diameter COB is given. If AB and CD are parallel, then show that chord AC is equal to chord BD. -Maths 9th

Last Answer : O Join AC and BD. Given, COB is the diameter of circle. ∠CAB = ∠BDC = 90° [angle in a semi-circle] Also, AB II CD ∠ABC = ∠DCB (alternate angles] Now, ∠ACB = 90° - ∠ABC and ∠DBC = 90° - ∠DCB = ... = ∠DBC BC = BC [common sides] ΔABC = ΔDCB [by ASA congruency] ∴ AC = BD [by CPCT] Hence Proved.

Description : In the given figure, bisectors AP and BQ of the alternate interior angles are parallel, then show that l || m. -Maths 9th

Last Answer : Given, In the figure AP|| BQ, AP and BQ are the bisectors of alternate interior angles ∠CAB and ∠ABF. To show l || m Proof Since, AP|| BQ and t is transversal, therefore ∠PAB = ∠ABQ [alternate interior angles] ⇒ 2 ∠PAB = 2 ∠ABQ [multiplying both sides by 2]

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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Description : In the given figure, ABC is an equilateral triangle of side length 30 cm. XY is parallel to BC, XP is parallel to AC and YQ is parallel to AB. -Maths 9th

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Description : In the given figure, (not drawn to scale), P is a point on AB such that AP : PB = 4 : 3. PQ is parallel to AC and QD -Maths 9th

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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 : Given triangle ABC with medians AE, BF, CD; FH parallel and equal in length to AE; BH and HE are drawn; -Maths 9th

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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°.

Description : Give the equations of two lines passing through (4, –2). How many more such lines are there, and why? -Maths 9th

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Description : Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases: (i) p(x) = 2x3+x2–2x–1, g(x) = x+1 -Maths 9th

Last Answer : Solution: p(x) = 2x3+x2–2x–1, g(x) = x+1 g(x) = 0 ⇒ x+1 = 0 ⇒ x = −1 ∴Zero of g(x) is -1. Now, p(−1) = 2(−1)3+(−1)2–2(−1)–1 = −2+1+2−1 = 0 ∴By factor theorem, g(x) is a factor of p(x

Description : 3. Check whether 7+3x is a factor of 3x3+7x. -Maths 9th

Last Answer : Solution: 7+3x = 0 ⇒ 3x = −7 ⇒ x = -7/3 ∴Remainder: 3(-7/3)3+7(-7/3) = -(343/9)+(-49/3) = (-343-(49)3)/9 = (-343-147)/9 = -490/9 ≠ 0 ∴7+3x is not a factor of 3x3+7x

Description : Examine , whether the following numbers are rational or irrational : -Maths 9th

Last Answer : ∴ It is an irrational number .

Description : Examine whether the following numbers are rational or irrational: -Maths 9th

Last Answer : 1 irrational no. 2 rational no. 3 irrational no.

Description : State whether the following statements are true or false ? Justify your answer. -Maths 9th

Last Answer : (i) False, here √2 is an irrational number and 3 is a rational number, we know that when we divide irrational number by non-zero rational number it will always give an irrational number. (ii) False, ... in the form p/q, q ≠0. p,q both are integers and these numbers are called irrational numbers.

Description : Write whether the following statements are true or false. Justify your answer. ’ -Maths 9th

Last Answer : (i) False, because a binomial has exactly two terms. (ii) False, because every polynomial is not a binomial . e.g., (a) 3x2 + 4x + 5 [polynomial but hot a binomial] (b) 3x2 + 5 [polynomial and also a binomial] (Hi ... = x4 + 2 + (-x4 + 4x3 + 2x) = 4x3 + 2x + 2 which is not a polynomial of degree 4.

Description : Verify whether the following are true or false. -Maths 9th

Last Answer : The following are true or false

Description : Check whether p(x) is a multiple of g(x) or not -Maths 9th

Last Answer : p(x) is a multiple of g(x) or not