What is an equation of the line that is perpendicular to y plus 1-3(x-5) and passes through the point (4-6)?

What is an equation of the line that is perpendicular to y plus 1-3(x-5) and passes through the point (4-6)?
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What equation describes a line that passes through the point (-14) and is perpendicular to the line 4x-3y-9?

Description : What equation describes a line that passes through the point (-14) and is perpendicular to the line 4x-3y-9?

Last Answer : If you mean point (-1, 4) and equation of 4x-3y = -9 then y =4/3x+3Slope of equation: 4/3Perpendicular slope: -3/4Perpendicular equation: y-4 = -3/4(x--1) => 4y = -3x+13

What is the perpendicular bisector equation of the line y equals 5x plus 10 spanning the parabola y equals x squared plus 4?

Description : What is the perpendicular bisector equation of the line y equals 5x plus 10 spanning the parabola y equals x squared plus 4?

Last Answer : If: y = 5x +10 and y = x^2 +4Then: x^2 +4 = 5x +10Transposing terms: x^2 -5x -6 = 0Factorizing the above: (x-6)(X+1) = 0 meaning x = 6 or x =-1Therefore by substitution endpoints of the line are ... .5 = -1/5(x-2.25) => 5y= -x+114.75Perpendicular bisector equation in its general form: x+5y-114.75= 0

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 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°.

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.

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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.

The graph of the linear equation y = x passes through the point. -Maths 9th

Description : The graph of the linear equation y = x passes through the point. -Maths 9th

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The graph of the linear equation y = x passes through the point. -Maths 9th

Description : The graph of the linear equation y = x passes through the point. -Maths 9th

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Description : The slope of a line perpendicular to the line which passes through the points (–k, h) and (b, – f ) is -Maths 9th

Last Answer : (b) (2, 2)The line 3x + 4y - 24 = 0 cuts the axis at A. To obtain the co-ordinates of A put y = 0, as on x-axis, y = 0. ∴ A ≡ (8, 0) ...(i) Also, on y-axis, x = 0, therefore B ≡ (0, 6 ... 8+10},rac{6 imes0+8 imes6+10 imes0}{6+8+10}\bigg)\)= \(\bigg(rac{48}{24},rac{48}{24}\bigg)\) = (2, 2).

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.

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

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Description : What equation represents the circle whose center is (-5,3) and that passes through the point (-1,3)?

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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.

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As a 10 kilogram mass on the end of a spring passes through its equilibrium position, the kinetic energy of the mass is 20 joules. The speed of the mass is: w) 2.0 meters per second x) 4.0 meters per second y) 5.0 meters per second z) 6.3 meters per second

Description : As a 10 kilogram mass on the end of a spring passes through its equilibrium position, the kinetic energy of the mass is 20 joules. The speed of the mass is: w) 2.0 meters per second x) 4.0 meters per second y) 5.0 meters per second z) 6.3 meters per second

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Last Answer : answer:

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Description : What is the perpendicular bisector equation of a line segment with endpoints of p q and 7p 3q?

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Description : What is the perpendicular bisector equation of a line with endpoints of s 2s and 3s 8s?

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Description : What is the perpendicular bisector equation of a line with endpoints of s 2s and 3s 8s?

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What is the tangent line equation of the circle 2x2 plus 2y2 -8x -5y -1 equals 0 at the point 1 -1 on its circumference?

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Draw a circle with centre at point O and radius 5 cm. Draw its chord AB, draw the perpendicular bisector of line segment AB. Does it pass through the centre of the circle? -Maths 9th

Description : Draw a circle with centre at point O and radius 5 cm. Draw its chord AB, draw the perpendicular bisector of line segment AB. Does it pass through the centre of the circle? -Maths 9th

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Description : Prove that through a given point, we can draw only one perpendicular to a given line. -Maths 9th

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Description : what- A map is drawn on a coordinate grid, as shown. The line represents an incoming flight path. A plane is about to leave on a perpendicular flight path that runs through the point (0, 4).Which line represents the outgoing flight path?

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In step 4 of the construction of a perpendicular line through a point why must the compass point be placed on the points where the arc intersects with the original line How would the construction be d?

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What is the slope of y plus 2 3(x - 7) and a point on the line?

Description : What is the slope of y plus 2 3(x - 7) and a point on the line?

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What is the point of contact between the line y equals x plus 4 and the circle x2 plus y2 -8x plus 4y equals 30?

Description : What is the point of contact between the line y equals x plus 4 and the circle x2 plus y2 -8x plus 4y equals 30?

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Description : What is the point of contact when the tangent line y equals x plus c touches the circle x squared plus y squared equals 4?

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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

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Description : 0 Pick up the correct statement from the following: (A) The point through which the resultant of the shear stresses passes is known as shear centre (B) In the standard rolled channels, the shear centre ... horizontal plane and away from the C.G., outside of the leg projection (D) All the above

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Description : The graph of a linear equation in two variables always passes through three quadrants of the graph paper. True/false -Maths 9th

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