The equation of line which passes through the point of intersection of 2x + 3y – 5 = 0 and x + y = 2 and also which is farthest from (2, 3) is

2y + 3x = 5

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2x + y = 3

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x + 2y = 3

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2x + 3y = 5

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Solution

The correct option is C
x + 2y = 3

Point of intersection (1, 1)

Required line will be perpendicular to line joining (1, 1) and (2, 3).

Slope of line $= m = - (\frac{2 - 1}{3 - 1}) = \frac{- 1}{2}$

Equation is $y - 1 = - \frac{1}{2} (x - 1)$

2y – 2 = –x + 1

2y + x = 3

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The equation of line which passes through the point of intersection of 2x + 3y – 5 = 0 and x + y = 2 and also which is farthest from (2, 3) is

The correct option is C x + 2y = 3 Point of intersection (1, 1) Required line will be perpendicular to line joining (1, 1) and (2, 3). Slope of line =m=−(2−13−1)=−12 Equation is y−1=−12(x−1) 2y – 2 = –x + 1 2y + x = 3

The correct option is C
x + 2y = 3

Point of intersection (1, 1)

Required line will be perpendicular to line joining (1, 1) and (2, 3).

Slope of line $= m = - (\frac{2 - 1}{3 - 1}) = \frac{- 1}{2}$

Equation is $y - 1 = - \frac{1}{2} (x - 1)$

2y – 2 = –x + 1

2y + x = 3