Question

to find the gradient and equation of the line
(a) passing through point (2,3) and acrosses the y-axis at -5
(b) having y-itercept at 2 and x-intercept at 3

Answer 1

Step-by-step explanation:

. Given that a straight line in the xy-plane passes through the point (3,48) in the first quadrant and intersects the positive x- and y-axes at points P and Q respectively.

2. Let O designate the origin.

Asked: Find the minimum value of OP×OQ.

Let the equation of the line l be

x/a + y/b = 1

where a, b > 0

3/a + 48/b = 1

3b + 48a = ab

Minimum value of OP * OQ = min(ab) = 1/ Max (1/a*1/b)

3/a + 48/b = 1

Sum is fixed; product is maximum when terms are equal

3/a = 48/b

48a = 3b

b = 16a

48a + 48a = 16a^2 = 96a

a = 96/16 = 6

b = 96

Minimum value of OP * OQ = 6*96 = 576

Answer 2

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i sure will solve it