Identify an equation in slope-intercept from for the line parallel to y=4x-9 that passes through (-5,3)?

Identify an equation in slope-intercept from for the line parallel to y=4x-9 that passes through (-5,3)?
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rewrite the equation 4x + 4y = 20 in slope-intercept form. -General Knowledge

Description : rewrite the equation 4x + 4y = 20 in slope-intercept form. -General Knowledge

Last Answer : 4x + 4y = 20 written in slope-intercept form is y = -x + 5. 4x + 4y = 20 4y = -4x + 20 (4y = -4x + 20)/(4) y = -x + 5.

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

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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.

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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

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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.

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Last Answer : If you mean y = 4x-1 then the slope is 4 and the y intercept is-1

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Description : What slope-intercept equation for a line the coefficient of the -term gives the slope?

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What slope-intercept equation for a line the coefficient of the -term gives the slope?

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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°.

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