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(a) The equation of a straight line, 4, is given as y=-²x = ³

(i) State the gradient of the line, l₁.
(ii)The point A (b, 4) lies on the line l₁. Determine the value of b.
(iii) A line, 12, passing through the point (0,-5), is perpendicular to l₁. Find the equation of the straight line l2. ​

User MrGreg
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(i) The gradient of the line l₁ is -2.
(ii) Substituting the coordinates of point A (b, 4) into the equation of the line l₁, 4 = -2b + 3. Solving for b, we have b = -1/2.
(iii) Since the line 12 is perpendicular to l₁, its gradient is the negative reciprocal of the gradient of l₁. Thus, the gradient of the line 12 is 1/2. Using the point-slope form of the equation of a straight line, the equation of the line 12 is y + 5 = 1/2(x - 0) or y = 1/2x - 5.
User Jovan MSFT
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