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

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

1 Answer

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 perpendicular distance from origin is 3⇒ \(rac{\pm|\,m.\,0-0+5m\,|}{\sqrt{m^2+(-1)^2}}\) = 3 ⇒ | 5m | = 3\(\sqrt{m^2+1}\)\(\bigg(\because ext{Distance of}\,(x_1,y_1)\, ext{from line}\,as+by+c=rac{|as_1+by_1+c|}{\sqrt{a^2+b^2}}\bigg)\)⇒ 25m2 = 9(m2 + 1) ⇒ 16m2 = 9 ⇒ m = \(rac{3}{4}\) (∵ m is +ve)∴ Required equation: y = \(rac{3}{4}\) (x + 5)⇒ 3x – 4y + 15 = 0.
by

Related questions

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

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

Last Answer : Solution :-

Find the equation of the straight line passing through the point (4, 5) and perpendicular to 3x – 2y + 5 = 0. -Maths 9th

Description : Find the equation of the straight line passing through the point (4, 5) and perpendicular to 3x – 2y + 5 = 0. -Maths 9th

Last Answer : There are two ways to prove it. 1st way: Area of triangle formed by the given points = 0 if they are collinear.∴ \(rac{1}{2}\) [\(x\)(2 - (y + 1) ) + 1((y + 1) - 1) + 0(1 - 2)] = 0⇒ \(rac{1}{2}\) [2\(x\) - \(x\)y - \( ... y - \(x\)y - 1 + \(x\) ⇒ x + y = \(x\)y ⇒ \(rac{1}{x}\) + \(rac{1}{y}\) = 1.

The point (2,3) lies on the graph of the linear equation 3x - (a -1)y =2a -1. If the same point also lies on the graph of the linear equation 5x + (1-2a)y = 3b, then find the value of b. -Maths 9th

Description : The point (2,3) lies on the graph of the linear equation 3x - (a -1)y =2a -1. If the same point also lies on the graph of the linear equation 5x + (1-2a)y = 3b, then find the value of b. -Maths 9th

Last Answer : Given, point (2,3) lies on the line. So, the point (2, 3) is the solution of 3x - (a -1) y = 2a - 1 On putting x = 2 and y = 3 in given solution. ∴ 3 2 - (a-1) 3 = 2a - 1 ⇒ 6 - 3a + 3 = 2a - 1 ⇒ - 3a ... 2 2) 3 = 3b ⇒ 10 - 9 = 3b ⇒ 1 = 3b ⇒ 1 / 3 = b Hence, the value of b is 1 / 3.

The point (2,3) lies on the graph of the linear equation 3x - (a -1)y =2a -1. If the same point also lies on the graph of the linear equation 5x + (1-2a)y = 3b, then find the value of b. -Maths 9th

Description : The point (2,3) lies on the graph of the linear equation 3x - (a -1)y =2a -1. If the same point also lies on the graph of the linear equation 5x + (1-2a)y = 3b, then find the value of b. -Maths 9th

Last Answer : Given, point (2,3) lies on the line. So, the point (2, 3) is the solution of 3x - (a -1) y = 2a - 1 On putting x = 2 and y = 3 in given solution. ∴ 3 2 - (a-1) 3 = 2a - 1 ⇒ 6 - 3a + 3 = 2a - 1 ⇒ - 3a ... 2 2) 3 = 3b ⇒ 10 - 9 = 3b ⇒ 1 = 3b ⇒ 1 / 3 = b Hence, the value of b is 1 / 3.

Draw a graph of the equation x + Y = 5 & 3x - 2y =0 on the same graph paper. Find the coordinates of the point whose two lines intersect. -Maths 9th

Description : Draw a graph of the equation x + Y = 5 & 3x - 2y =0 on the same graph paper. Find the coordinates of the point whose two lines intersect. -Maths 9th

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

Draw a graph of the equation x+ y=5 & 3x -2y=0 in the same graph paper find the coordinates of the point whose two two lines intersect. -Maths 9th

Description : Draw a graph of the equation x+ y=5 & 3x -2y=0 in the same graph paper find the coordinates of the point whose two two lines intersect. -Maths 9th

Last Answer : From x + y = 5, If x = 0 0 + y = 5 y = 5 Therefore (0,5) If x = 1 1 + y = 5 y =5 - 1 y = 4 Therefore (1,4) Draw a graph for this And From 3x - 2y = 0 If x = 0 3 (0) - 2y = 0 0 - ... 2y = 0 -2y = -6 y = -6/-2 y = 3 Therefore (2,3) Draw a graph for these points And the point of intersection is (2,3)

What is the angle between the lines whose equations are: 3x + y – 7 = 0 and x + 2y + 9 = 0. -Maths 9th

Description : What is the angle between the lines whose equations are: 3x + y – 7 = 0 and x + 2y + 9 = 0. -Maths 9th

Last Answer : (c) (8, 6)Let AB be the given line 4x + 3y = 25 Let O′(a, b) be the image of O in the given line AB. Let O O′ cut AB in point P. Also OP ⊥ AB and P is the mid-point of OO′. ∴ Co-ordinates of P are \(\bigg( ... 4 imes6}{3}\) = 8∴ The image of the point O(0, 0) in the line 4x + 3y - 25 = 0 is (8, 6).

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

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.

What is the equation of the line joining the origin with the point of intersection of the lines 4x + 3y = 12 and 3x + 4y = 12 ? -Maths 9th

Description : What is the equation of the line joining the origin with the point of intersection of the lines 4x + 3y = 12 and 3x + 4y = 12 ? -Maths 9th

Last Answer : (b) (5, 6)Let the foot of the perpendicular be M(x1, y1) Slope of line AB, i.e., y = -x + 11 = -1 Slope of line PM = \(rac{y_1-3}{x_1-2}\)Now, PM ⊥ AB⇒ \(\bigg(rac{y_1-3}{x_1-2}\bigg)\) x - ... get 2x1 = 10 ⇒ x1 = 5 Putting x1 in (ii), we get y1 = 6. ∴ Required foot of the perpendicular M is (5, 6).

What is the acute angle between the two straight lines y = ( ) 2– 3 x + 5 and y = ( ) 2 3 + x – 7 ? -Maths 9th

Description : What is the acute angle between the two straight lines y = ( ) 2– 3 x + 5 and y = ( ) 2 3 + x – 7 ? -Maths 9th

Last Answer : (a) 60ºThe two lines are: y = \((2-\sqrt3)\) \(x\) + 5 ...(i) y = \((2-\sqrt3)\) \(x\) - 7 ...(ii) Slope of line (i), m1 = \(2-\sqrt3\)Slope of line (ii), m2 = \(2 ... \sqrt3)}{1+(2-\sqrt3)(2+\sqrt3)}\bigg|\)= \(\bigg|rac{-2\sqrt3}{1+1}\bigg|=\sqrt3\)∴ θ = -1 tan ( 3) = 60º.

In figure X and Y are the mid-points of AC and AB respectively, QP || BC and CYQ and BXP are straight lines. -Maths 9th

Description : In figure X and Y are the mid-points of AC and AB respectively, QP || BC and CYQ and BXP are straight lines. -Maths 9th

Last Answer : Given X and Y are the mid-points of AC and AB respectively. Also, QP|| BC and CYQ, BXP are straight lines. To prove ar (ΔABP) = ar (ΔACQ) Proof Since, X and Y are the mid-points of AC and AB respectively. So, ... ar (ΔBYX) + ar (XYAP) = ar (ΔCXY) + ar (YXAQ) ⇒ ar (ΔABP) = ar (ΔACQ) Hence proved.

In figure X and Y are the mid-points of AC and AB respectively, QP || BC and CYQ and BXP are straight lines. -Maths 9th

Description : In figure X and Y are the mid-points of AC and AB respectively, QP || BC and CYQ and BXP are straight lines. -Maths 9th

Last Answer : Given X and Y are the mid-points of AC and AB respectively. Also, QP|| BC and CYQ, BXP are straight lines. To prove ar (ΔABP) = ar (ΔACQ) Proof Since, X and Y are the mid-points of AC and AB respectively. So, ... ar (ΔBYX) + ar (XYAP) = ar (ΔCXY) + ar (YXAQ) ⇒ ar (ΔABP) = ar (ΔACQ) Hence proved.

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

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.

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

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.

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

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.

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).

Find the point which lies on the line y x = -3 having abscissa 3. -Maths 9th

Description : Find the point which lies on the line y x = -3 having abscissa 3. -Maths 9th

Last Answer : Solution :- When x=3 then y= -9,thus the point is (3,-9)

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

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

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

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 ⇒ \(rac{\big|rac{1}{a} imes0+ ... {1}{b^2}+rac{1}{b^2}}\) ⇒ \(rac{1}{p^2}\) = \(rac{1}{a^2}\) + \(rac{1}{b^2}\)

Find the equation of the line which passes through the point of intersection of the lines 2x – y + 5 = 0 -Maths 9th

Description : Find the equation of the line which passes through the point of intersection of the lines 2x – y + 5 = 0 -Maths 9th

Last Answer : (a) 45º 3x + y - 7 = 0 ⇒ y = -3x + 7 ⇒ Slope (m1) = -3 x + 2y + 9 = 0 ⇒ y = \(rac{-x}{2}\) - \(rac{9}{2}\) ⇒ Slope (m2) = \(-rac{1}{2}\)If θ is the angle between the given lines, then tan θ = \(\ ... \bigg|rac{-rac{5}{2}}{1+rac{3}{2}}\bigg|\)= \(\bigg|rac{-rac{5}{2}}{rac{5}{2}}\bigg|\) = 1∴ θ = 45°.

What is the equation of the straight line which passes through (3, 4) and the sum of whose x-intercept and y-intercept is 14 ? -Maths 9th

Description : What is the equation of the straight line which passes through (3, 4) and the sum of whose x-intercept and y-intercept is 14 ? -Maths 9th

Last Answer : (a) 4x + 3y = 24 Let the x-intercept = a. Then, y-intercept = 14 - a ∴ Eqn of the straight line is \(rac{x}{a}\) + \(rac{y}{14-a}\) = 1Since it passes through (3, 4), so\(rac{3}{a}\) + \(rac{4}{14-a}\) = 1⇒ 3(14 - ... = 1 ⇒ x + y = 7or \(rac{x}{6}\) + \(rac{y}{8}\) = 1 ⇒ 8x + 6y = 48 ⇒ 4x + 3y = 24.

Check whether the graph of the equation y = 3x + 5 passes through the origin or not. -Maths 9th

Description : Check whether the graph of the equation y = 3x + 5 passes through the origin or not. -Maths 9th

Last Answer : Solution :-

Draw the graph of the linear equation 3x + 4y = 6. At what points, does the graph cut the x-axis and the y-axis? -Maths 9th

Description : Draw the graph of the linear equation 3x + 4y = 6. At what points, does the graph cut the x-axis and the y-axis? -Maths 9th

Last Answer : hope it helps

Draw the graph of the following equations (1) Y =3x+2 (b) Y=x -Maths 9th

Description : Draw the graph of the following equations (1) Y =3x+2 (b) Y=x -Maths 9th

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

If the equation `3x^(2)+6xy+my^(2)=0` represents a pair of coincident straight lines, then `(3m)/2` is equal to

Description : If the equation `3x^(2)+6xy+my^(2)=0` represents a pair of coincident straight lines, then `(3m)/2` is equal to

Last Answer : If the equation `3x^(2)+6xy+my^(2)=0` represents a pair of coincident straight lines, then `(3m)/2` is equal to

The ratio in which the line 3x + 4y = 7 divides the line joining the points (–2, 1) and (1, 2) is -Maths 9th

Description : The ratio in which the line 3x + 4y = 7 divides the line joining the points (–2, 1) and (1, 2) is -Maths 9th

Last Answer : (a) (–24, –2)Co-ordinates of the point of external division are\(\bigg(rac{m_1\,x_2-m_2\,x_1}{m_1-m_2},rac{m_1y_2-m_2y_1}{m_1-m_2}\bigg)\), i.e.,∴ Required point = \(\bigg(rac{3 imes-6-2 imes4}{3-2},rac{3 imes2-2 imes4}{3-2}\bigg)\)= \(\big(rac{-24}{1},rac{-2}{1}\big)\), i.e., (–24, –2).

Write the equation of a line parallel to y-axis and passing through the point (–4, –5). -Maths 9th

Description : Write the equation of a line parallel to y-axis and passing through the point (–4, –5). -Maths 9th

Last Answer : Solution :- x= -4

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°.

Find the equation of the straight line with a positive gradient which passes through the point (–5, 0) -Maths 9th

Description : Find the equation of the straight line with a positive gradient which passes through the point (–5, 0) -Maths 9th

Last Answer : (d) Both (a) and (c)Since the line passes through A(a, 0) and B(0, b), it makes intercepts a and b on x-axis and y-axis respectively. Let the equation of this line in the intercept from be \(rac{x}{a}\) + \(rac{y}{a}\) ... \(rac{x}{-12}\) + \(rac{y}{-5}\) = 1⇒ 5x + 12y = 60 and 5x + 12y + 60 = 0.

A straight line passes through the points (5, 0) and (0, 3). The length of the perpendicular from the point (4, 4) on the line is: -Maths 9th

Description : A straight line passes through the points (5, 0) and (0, 3). The length of the perpendicular from the point (4, 4) on the line is: -Maths 9th

Last Answer : (b) \(rac{\sqrt{17}}{2}\)Equation of the line through the points (5, 0) and (0, 3) y - 0 = \(rac{3-0}{0-5}\) (x - 5)⇒ y = \(rac{-3}{5}\)(x - 5)⇒ 5y + 3x - 15 = 0 ∴ Distance of perpendicular from ... (rac{|20+12-15|}{\sqrt{25+9}{}}\) = \(rac{17}{\sqrt{34}}\) units. = \(rac{\sqrt{17}}{2}\) units.

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

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).

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

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).

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

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)

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

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

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)

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

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

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

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

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

Last Answer : answer:

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.

What is the equation of the line passing through (2, –3) and parallel to the y-axis ? -Maths 9th

Description : What is the equation of the line passing through (2, –3) and parallel to the y-axis ? -Maths 9th

Last Answer : (c) x = 2 Slope of y-axis = tan 90º (∵ y-axis ⊥ x-axis) ∴ Equation of line passing through (2, –3) parallel to y-axis is (y + 3) = tan 90º (x – 2) ⇒ (y + 3) = ∞ (x – 2) ⇒ (x – 2) = \(rac{1}{\infty}\) (y + 3) = 0 ⇒ x = 2.

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

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°.

A straight line passes through the points (a, 0) and (0, b). The length of the line segment contained between the axes is 13 and the product of -Maths 9th

Description : A straight line passes through the points (a, 0) and (0, b). The length of the line segment contained between the axes is 13 and the product of -Maths 9th

Last Answer : (d) \(rac{23}{\sqrt{17}}\)The given lines are:L : \(rac{x}{5}+rac{y}{b}=1\) ....(i)K : \(rac{x}{c}+rac{y}{3}=1.\) ...(ii)Since line L passes through (13, 32),\(rac{13}{ ... between parallel lines ax + by + c1 = 0 and ax + by c2 = 0 is d = \(rac{|c_2-c_1|}{\sqrt{a^2+b^2}}\bigg)\)

The straight line ax + by + c = 0 and the co-ordinate axes form an isosceles triangle under which of the following conditions ? -Maths 9th

Description : The straight line ax + by + c = 0 and the co-ordinate axes form an isosceles triangle under which of the following conditions ? -Maths 9th

Last Answer : (a) | a | = | b | The equation of line AB, i.e., ax + by + c = 0 in intercept form is ax + by = - c⇒ \(rac{x}{\big(-rac{c}{a}\big)}\) + \(rac{x}{\big(-rac{c}{b}\big)}\) = 1Δ AOB is isosceles Δ if OA = OB, i.e., ... \(rac{-c}{a}\) = \(rac{-c}{a}\) ⇒ \(rac{1}{a}\) = \(rac{1}{a}\) ⇒ | a | = | b |.

The equation whose roots are the negatives of the roots of the equation x^7 + 3x^5 + x^3 – x^2 + 7x + 2 = 0 is : -Maths 9th

Description : The equation whose roots are the negatives of the roots of the equation x^7 + 3x^5 + x^3 – x^2 + 7x + 2 = 0 is : -Maths 9th

Last Answer : answer:

If y-coordinate of a point is zero, then this point always lies -Maths 9th

Description : If y-coordinate of a point is zero, then this point always lies -Maths 9th

Last Answer : (c) If y-coordinate of a point is zero, then this point always lies on X-axis. Because perpendicular distance of the point from X-axis measured along Y-axis is zero.

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

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.

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

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).

If y-coordinate of a point is zero, then this point always lies -Maths 9th

Description : If y-coordinate of a point is zero, then this point always lies -Maths 9th

Last Answer : (c) If y-coordinate of a point is zero, then this point always lies on X-axis. Because perpendicular distance of the point from X-axis measured along Y-axis is zero.

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

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.