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

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
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(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 -1 = -1⇒ y1 –3 = x1 – 2 ⇒ x1 – y1 = –1             ...(i) Also, (x1, y1) lies on AB ⇒ x1 + y1 – 11 = 0 ⇒ x1 + y1 = 11                           ...(ii) Adding (i) and (ii), we get 2x1 = 10 ⇒ x1 = 5 Putting x1 in (ii), we get y1 = 6. ∴ Required foot of the perpendicular M is (5, 6).
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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.

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