If (–2, 6) is the image of the point (4, 2) with respect to the line L = 0, then L is equal to -Maths 9th

If (–2, 6) is the image of the point (4, 2) with respect to the line L = 0, then L is equal to -Maths 9th
by

1 Answer

Answer :

(b) 3x – 2y + 5 Let the equation of the L be y = mx + c Since O′ (–2, 6) is the image of the point O (4, 2) in line L = 0, the mid-point of OO′, i.e., \(\bigg(rac{-2+4}{2},rac{6+2}{2}\bigg)\), i.e.,(1, 4) will lie on the given line. Also, L ⊥ OO′, so Slope of L x Slope of OO′ = –1⇒ m x \(\bigg(rac{6-2}{-2-4}\bigg)\)= -1 ⇒ m = \(rac{-1}{rac{-2}{3}}\) = \(rac{3}{2}\)∴ Equation of L is y = \(rac{3}{2}\,x+c\)∵ It passes through (1, 4)4 = \(rac{3}{2}\) x 1 + c ⇒ c = 4 - \(rac{3}{2}\) = \(rac{5}{2}.\)∴ Required equation is y = \(rac{3}{2}x\) + \(rac{5}{2}\)⇒ 3x – 2y + 5 = 0 ∴ L = 3x – 2y + 5.
by

Related questions

A spherical ball is divided into two equal halves. If the curved surface area of each half is 56.57 cm?, find the volume of the spherical ball.11531/cylinder-radius-halved-and-height-doubled-then-find-volume-with-respect-original-volume -Maths 9th

Description : A spherical ball is divided into two equal halves. If the curved surface area of each half is 56.57 cm?, find the volume of the spherical ball.11531/cylinder-radius-halved-and-height-doubled-then-find-volume-with-respect-original-volume -Maths 9th

Last Answer : since curved surface of half of the spherical ball = 56.57 cm2 ∴ 2πr2 = 56.57 ⇒ r2 = 56.57 / 2 × 3.14 = 9 ⇒ r = 3 cm Now, volume of spherical ball = 4 / 3 πr3 = 4 / 3 × 3.14 × 3 × 3 × 3 = 113.04 cm3

A spherical ball is divided into two equal halves. If the curved surface area of each half is 56.57 cm?, find the volume of the spherical ball.11531/cylinder-radius-halved-and-height-doubled-then-find-volume-with-respect-original-volume -Maths 9th

Description : A spherical ball is divided into two equal halves. If the curved surface area of each half is 56.57 cm?, find the volume of the spherical ball.11531/cylinder-radius-halved-and-height-doubled-then-find-volume-with-respect-original-volume -Maths 9th

Last Answer : since curved surface of half of the spherical ball = 56.57 cm2 ∴ 2πr2 = 56.57 ⇒ r2 = 56.57 / 2 × 3.14 = 9 ⇒ r = 3 cm Now, volume of spherical ball = 4 / 3 πr3 = 4 / 3 × 3.14 × 3 × 3 × 3 = 113.04 cm3

If (log x)/(l + m - 2n) = (log y)/(m + n - 2l) = (log z)/(n + l - 2m), then xyz is equal to : -Maths 9th

Description : If (log x)/(l + m - 2n) = (log y)/(m + n - 2l) = (log z)/(n + l - 2m), then xyz is equal to : -Maths 9th

Last Answer : Let l+m−2nlogx​=m+n−2llogy​=n+l−2mlogz​=k(say) So, we get logx=k(l+m−2n) ....... (i) logy=k(m+n−2l) ....... (ii) logz=k(n+l−2m) ....... (iii) ∴logx+logy+logz=k(l+m−2n)+k(m+n−2l)+k(n+l−2m) ⇒logx+logy+logz=kl+km−2kn+km+kn−2kl+kn+kl−2km ⇒log(xyz)=0 ⇒logxyz=log1 ⇒xyz=1

If (log x)/(l + m - 2n) = (log y)/(m + n - 2l) = (log z)/(n + l - 2m), then xyz is equal to : -Maths 9th

Description : If (log x)/(l + m - 2n) = (log y)/(m + n - 2l) = (log z)/(n + l - 2m), then xyz is equal to : -Maths 9th

Last Answer : (b) 1Let \(rac{ ext{log}\,x}{l+m-2n}\) = \(rac{ ext{log}\,y}{m+n-2l}\) = \(rac{ ext{log}\,z}{n+l-2m}\) = k. Thenlog x = k(l + m – 2n), log y = k(m + n – 2l); log z = k(n + l – 2m) ⇒ log x + log y + log z = k(l + m – 2n) + k(m + n – 2l) + k(n + l – 2m)⇒ log(xyz) = 0 ⇒ log(xyz) = log 1 ⇒ xyz = 1.

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

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

Last Answer : if pqrs is a trapezium so pq and RS are parallel is you draw a diagonal you will divide the trapezium into two parts such that two equal triangle so

In given figure l || m and M is the mid-point of a line segment AB. -Maths 9th

Description : In given figure l || m and M is the mid-point of a line segment AB. -Maths 9th

Last Answer : Solution of this question

In trapezium ABCD, AB || DC and L is the mid-point of BC. Through L, a line PQ || AD has been drawn which meets AB in P and DC produced in Q. -Maths 9th

Description : In trapezium ABCD, AB || DC and L is the mid-point of BC. Through L, a line PQ || AD has been drawn which meets AB in P and DC produced in Q. -Maths 9th

Last Answer : According to question prove that ar (ABCD) = ar (APQD).

In given figure l || m and M is the mid-point of a line segment AB. -Maths 9th

Description : In given figure l || m and M is the mid-point of a line segment AB. -Maths 9th

Last Answer : Solution of this question

In trapezium ABCD, AB || DC and L is the mid-point of BC. Through L, a line PQ || AD has been drawn which meets AB in P and DC produced in Q. -Maths 9th

Description : In trapezium ABCD, AB || DC and L is the mid-point of BC. Through L, a line PQ || AD has been drawn which meets AB in P and DC produced in Q. -Maths 9th

Last Answer : According to question prove that ar (ABCD) = ar (APQD).

In Fig. 8.53,ABCD is a parallelogram and E is the mid - point of AD. A line through D, drawn parallel to EB, meets AB produced at F and BC at L.Prove that (i) AF = 2DC (ii) DF = 2DL -Maths 9th

Description : In Fig. 8.53,ABCD is a parallelogram and E is the mid - point of AD. A line through D, drawn parallel to EB, meets AB produced at F and BC at L.Prove that (i) AF = 2DC (ii) DF = 2DL -Maths 9th

Last Answer : Given, E is mid point of AD Also EB∥DF ⇒ B is mid point of AF [mid--point theorem] so, AF=2AB (1) Since, ABCD is a parallelogram, CD=AB ⇒AF=2CD AD∥BC⇒LB∥AD In ΔFDA ⇒LB∥AD ⇒LDLF​=ABFB​=1 from (1) ⇒LF=LD so, DF=2DL

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°.

The image of the origin with reference to the line 4x + 3y – 25 = 0 is -Maths 9th

Description : The image of the origin with reference to the line 4x + 3y – 25 = 0 is -Maths 9th

Last Answer : (c) 25x + 25y - 4 = 0 The point of intersection of the given lines can be obtained by solving the equations of the two lines simultaneously. 100x + 50y = 1 ...(i) 75x + 25y = -3 ... = \(rac{4}{25}\)∴ Eqn of required line: x + y = \(rac{4}{25}\) ⇒ 25 + 25y - 4 = 0.

If P(-l, 1), Q(3, -4), R(1, -1), S(-2, -3) and T(-4, 4) are plotted on the graph paper, then the point(s) in the fourth quadrant is/are -Maths 9th

Description : If P(-l, 1), Q(3, -4), R(1, -1), S(-2, -3) and T(-4, 4) are plotted on the graph paper, then the point(s) in the fourth quadrant is/are -Maths 9th

Last Answer : (b) In point P (-1, 1), x-coordinate is -1 unit and y-coordinate is 1 unit, so it lies in llnd quadrant. Similarly, we can plot all the points Q (3, -4), R (1, -1), S (-2, -3) and T (-4, 4), It is clear from the graph that points R and Q lie in fourth quadrant.

If P(-l, 1), Q(3, -4), R(1, -1), S(-2, -3) and T(-4, 4) are plotted on the graph paper, then the point(s) in the fourth quadrant is/are -Maths 9th

Description : If P(-l, 1), Q(3, -4), R(1, -1), S(-2, -3) and T(-4, 4) are plotted on the graph paper, then the point(s) in the fourth quadrant is/are -Maths 9th

Last Answer : (b) In point P (-1, 1), x-coordinate is -1 unit and y-coordinate is 1 unit, so it lies in llnd quadrant. Similarly, we can plot all the points Q (3, -4), R (1, -1), S (-2, -3) and T (-4, 4), It is clear from the graph that points R and Q lie in fourth quadrant.

If m is the mid-point and l is the upper class limit of a class in a continuous frequency distribution, then lower class limit of the class is -Maths 9th

Description : If m is the mid-point and l is the upper class limit of a class in a continuous frequency distribution, then lower class limit of the class is -Maths 9th

Last Answer : NEED ANSWER

If m is the mid-point and l is the upper class limit of a class in a continuous frequency distribution, then lower class limit of the class is -Maths 9th

Description : If m is the mid-point and l is the upper class limit of a class in a continuous frequency distribution, then lower class limit of the class is -Maths 9th

Last Answer : (b) Let x and y be the lower and upper class limit of a continuous frequency distribution. Now, mid-point of a class = (x + y)/2 = m [given] ⇒ x + y = 2 m =x + l = 2m [∴ y = l = upper class limit (given)] ⇒ x = 2 m-l Hence, the lower class limit of the class is 2m – l.

A line passes through the point of intersection of the lines 100x + 50y – 1 = 0 and 75x + 25y + 3 = 0 and makes equal intercepts on the axes. -Maths 9th

Description : A line passes through the point of intersection of the lines 100x + 50y – 1 = 0 and 75x + 25y + 3 = 0 and makes equal intercepts on the axes. -Maths 9th

Last Answer : (d) x + 2y = 2Let the required equation make intercept on x-axis = 2a ⇒ intercept made on y-axis = a ∴ Eqn of the given line in the intercept from:\(rac{x}{2a}+rac{y}{a}=1\) ...(i)Since the line ... 1 ⇒ a = 1.∴ Required equation of line : \(rac{x}{2 imes1}+rac{y}{1}=1\) ⇒ x + 2y = 2.

In Fig. 6.7, if l||m, then find the value of x. -Maths 9th

Description : In Fig. 6.7, if l||m, then find the value of x. -Maths 9th

Last Answer : Solution :-

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

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

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

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.

Two lines l and m are perpendicular to the same line n.Are l and m perpendicular to each other ? -Maths 9th

Description : Two lines l and m are perpendicular to the same line n.Are l and m perpendicular to each other ? -Maths 9th

Last Answer : Solution :- No, they are parallel.

In the given figure, D is the mid-point of BC and L mid-is the point of AD. -Maths 9th

Description : In the given figure, D is the mid-point of BC and L mid-is the point of AD. -Maths 9th

Last Answer : In △ABC, AD is the median ∴ ar(△ABD) = 1/2 ar(△ABC) Again, △ABD BL is the median ∴ ar(△ABL) = 1/2 ar(△ABD) = 1/2 × 1/2 ar((△ABC) = 1/4 ar((△ABC) Hence, value of x is 1/4.

In the given figure, ABCD is a parallelogram and L is the mid - point of DC. -Maths 9th

Description : In the given figure, ABCD is a parallelogram and L is the mid - point of DC. -Maths 9th

Last Answer : In ||gm ABCD, AC is the diagonal ∴ ar(△ABC) = ar(△ADC) = 1/2 ar ||gm ABCD) In△ADC, AL is the median ∴ ar(△ADL) = ar(△ACL)= 1/2 ar(△ADC) = 1/4 ar (||gm ABCD) Now, ar(quad.ABCL) = ar(△ABC) + ar(△ACL) = 3/4 ar ... ar(||gm ABCD) = 96 cm2 ∴ ar(△ADC) = 1/2 ar(||gm ABCD) = 1/2 96 = 48 cm2

In the given figure, D is the mid-point of BC and L mid-is the point of AD. -Maths 9th

Description : In the given figure, D is the mid-point of BC and L mid-is the point of AD. -Maths 9th

Last Answer : In △ABC, AD is the median ∴ ar(△ABD) = 1/2 ar(△ABC) Again, △ABD BL is the median ∴ ar(△ABL) = 1/2 ar(△ABD) = 1/2 × 1/2 ar((△ABC) = 1/4 ar((△ABC) Hence, value of x is 1/4.

In the given figure, ABCD is a parallelogram and L is the mid - point of DC. -Maths 9th

Description : In the given figure, ABCD is a parallelogram and L is the mid - point of DC. -Maths 9th

Last Answer : In ||gm ABCD, AC is the diagonal ∴ ar(△ABC) = ar(△ADC) = 1/2 ar ||gm ABCD) In△ADC, AL is the median ∴ ar(△ADL) = ar(△ACL)= 1/2 ar(△ADC) = 1/4 ar (||gm ABCD) Now, ar(quad.ABCL) = ar(△ABC) + ar(△ACL) = 3/4 ar ... ar(||gm ABCD) = 96 cm2 ∴ ar(△ADC) = 1/2 ar(||gm ABCD) = 1/2 96 = 48 cm2

Let O be any point inside a triangle ABC. Let L, M and N be the points on AB, BC and CA respectively, -Maths 9th

Description : Let O be any point inside a triangle ABC. Let L, M and N be the points on AB, BC and CA respectively, -Maths 9th

Last Answer : answer:

If two chords of a circle with a common end-point are inclined equally to the diameter through this common end-point, then prove that chords are equal. -Maths 9th

Description : If two chords of a circle with a common end-point are inclined equally to the diameter through this common end-point, then prove that chords are equal. -Maths 9th

Last Answer : Let AB and AC be two chords and AD be the diameter of the circle . Draw OL ⊥ AB and OM ⊥ AC . In ΔOLA and ΔOMA ∠ OLA = ∠ OMA = 90° OA = OA [common sides] ∠ OAL = ∠ OAM [given] . OLA ≅ OMA [by ASA congruency] OL =OM (by CPCT) Chords AM and AC are equidistant from O. ∴ AB = AC Hence proved.

If two chords of a circle with a common end-point are inclined equally to the diameter through this common end-point, then prove that chords are equal. -Maths 9th

Description : If two chords of a circle with a common end-point are inclined equally to the diameter through this common end-point, then prove that chords are equal. -Maths 9th

Last Answer : Let AB and AC be two chords and AD be the diameter of the circle . Draw OL ⊥ AB and OM ⊥ AC . In ΔOLA and ΔOMA ∠ OLA = ∠ OMA = 90° OA = OA [common sides] ∠ OAL = ∠ OAM [given] . OLA ≅ OMA [by ASA congruency] OL =OM (by CPCT) Chords AM and AC are equidistant from O. ∴ AB = AC Hence proved.

The point P is equidistant from A(1, 3), B(–3, 5) and C(5, –1). Then PB is equal to : -Maths 9th

Description : The point P is equidistant from A(1, 3), B(–3, 5) and C(5, –1). Then PB is equal to : -Maths 9th

Last Answer : (b) (2, 2)Let the point be P whose abscissa = ordinate = a. ∴ P ≡ (a, a) Given, PA = PB ⇒ (a + 1)2 + a2 = a2 + (a – 5)2 ⇒ 2a2 + 2a + 1 = 2a2 – 10a + 25 ⇒ 12a = 24 ⇒ a = 2. ∴ The point is (2, 2).

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

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]

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

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]

In the figure if l parallel m, then find the value of x -Maths 9th

Description : In the figure if l parallel m, then find the value of x -Maths 9th

Last Answer : as L ll m Step-by-step explanation: :. 30 + 40 + y = 180 --------------------------------------(let's take the third angle as y) (because of angle sum property of triangle) 70 + y = 180 y = 110 ... + x = 180 ----------------------(co interior angles) :. x = 180 - 110 = 70 hence solved!!!!

In the figure if l parallel m, then find the value of x -Maths 9th

Description : In the figure if l parallel m, then find the value of x -Maths 9th

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

Points P, Q, R and S divide a line segment joining A (2, 6) and B (7, -4) in five equal parts. Find the coordinates of P and R. -Maths 9th

Description : Points P, Q, R and S divide a line segment joining A (2, 6) and B (7, -4) in five equal parts. Find the coordinates of P and R. -Maths 9th

Last Answer : this is the ans hope its clear

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)

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

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

Last Answer : Solution :-

If 3sin x + 4cos x = 5, then 6 tan x/2 - 9 tan^2 x/2 is equal to -Maths 9th

Description : If 3sin x + 4cos x = 5, then 6 tan x/2 - 9 tan^2 x/2 is equal to -Maths 9th

Last Answer : answer:

The graph of the linear equation 2x+ 3y = 6 is a line which meets the X-axis at the point. -Maths 9th

Description : The graph of the linear equation 2x+ 3y = 6 is a line which meets the X-axis at the point. -Maths 9th

Last Answer : (c) Since, the graph of linear equation 2x + 3y = 6 meets the X-axis. So, we put y = 0 in 2x + 3y = 6 ⇒ 2x + 3(0) = 6 = 2x + 0 = 6 ⇒ x = 6/2 ⇒ x = 3 Hence, the coordinate on X-axis is (3, 0).

The graph of the linear equation 2x+ 3y = 6 is a line which meets the X-axis at the point. -Maths 9th

Description : The graph of the linear equation 2x+ 3y = 6 is a line which meets the X-axis at the point. -Maths 9th

Last Answer : (c) Since, the graph of linear equation 2x + 3y = 6 meets the X-axis. So, we put y = 0 in 2x + 3y = 6 ⇒ 2x + 3(0) = 6 = 2x + 0 = 6 ⇒ x = 6/2 ⇒ x = 3 Hence, the coordinate on X-axis is (3, 0).

The line segment joining P(5, –2) and Q(9, 6) is divided in the ratio 3 : 1 by a point A -Maths 9th

Description : The line segment joining P(5, –2) and Q(9, 6) is divided in the ratio 3 : 1 by a point A -Maths 9th

Last Answer : Comparing y = 5\(x\) –7 with y = m\(x\) + c, the slope of given line = m = 5 ∴ Equation of a line parallel to y = 5\(x\) – 7 having y-intercept = –1 is y = 5\(x\) – 1.

The mid-point of the line joining the points (–10, 8) and (–6, 12) divides the line joining the points (4, –2) and (– 2, 4) in the ratio -Maths 9th

Description : The mid-point of the line joining the points (–10, 8) and (–6, 12) divides the line joining the points (4, –2) and (– 2, 4) in the ratio -Maths 9th

Last Answer : (d) 2 : 1 externallyThe mid-point of the line joining the points (-10, 8) and (- 6, 12) is\(\bigg(rac{-10+(-6)}{2},rac{8+12}{2}\bigg)\), i.e., (-8, 10).Let (-8, 10) divide the join of (4 ... 6k = 12 ⇒ k = -2 Since the value of k is negative, it is a case of external division and the ratio is 2 : 1.

The point whose abscissa is equal to its ordinate and which is equidistant from A(–1, 0) and B(0, 5) is -Maths 9th

Description : The point whose abscissa is equal to its ordinate and which is equidistant from A(–1, 0) and B(0, 5) is -Maths 9th

Last Answer : Putting \(x\) = 0 in equation of one of the lines say 9\(x\) + 40y -20 = 0, we get y = \(rac{1}{2}\)∴ A point on 9\(x\) + 40y - 20 = 0 is \(\big(0,rac{1}{2}\big)\)∴ Distance of \(\big(0,rac{1}{2}\big) ... imesrac{1}{2}+21\big|}{\sqrt{9^2+40^2}}\) = \(rac{|41|}{\sqrt{1681}}\) = \(rac{41}{41}\) = 1.

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.

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.

Prove that sum of any two sides of a triangle is greater than twice the median with respect to the third side -Maths 9th

Description : Prove that sum of any two sides of a triangle is greater than twice the median with respect to the third side -Maths 9th

Last Answer : Solution :-

In Fig. 6.14, if x+y = w+z,then then prove that AOB is a line. -Maths 9th

Description : In Fig. 6.14, if x+y = w+z,then then prove that AOB is a line. -Maths 9th

Last Answer : Solution :-

If a + b + c =0, then a3+b3 + c3 is equal to -Maths 9th

Description : If a + b + c =0, then a3+b3 + c3 is equal to -Maths 9th

Last Answer : (d) Now, a3+b3 + c3 = (a+ b + c) (a2 + b2 + c2 – ab – be – ca) + 3abc [using identity, a3+b3 + c3 – 3 abc = (a + b + c)(a2+b2+c2 –ab–bc-ca)] = 0 + 3abc [∴ a + b + c = 0] a3+b3 + c3 = 3abc.

If a + b + c =0, then a3+b3 + c3 is equal to -Maths 9th

Description : If a + b + c =0, then a3+b3 + c3 is equal to -Maths 9th

Last Answer : (d) Now, a3+b3 + c3 = (a+ b + c) (a2 + b2 + c2 – ab – be – ca) + 3abc [using identity, a3+b3 + c3 – 3 abc = (a + b + c)(a2+b2+c2 –ab–bc-ca)] = 0 + 3abc [∴ a + b + c = 0] a3+b3 + c3 = 3abc.