161. Equation of a line is taken as 3x – 4y + 5 = 0. Its slope and intercept on y-axis

1. 1. 	3	,	−5
      	4		4

   
   
   

   2. 	−3	,	−5
      	4		4

   
   
   

   3. 	3	,	5
      	4		4

   
   
   

   4. 	5	,	3
      	4		4

   
   
   

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1. View Hint View Answer Discuss in Forum

   Here,
   3x – 4y + 5 = 0
   ⇒  –4y = – 3x – 5
   

   y =	−3	x	−5
   	–4		−4

   
   

   y =	3	x +	5
   	4		4

   
   Compare it with, y = mx + c we get
   

   m =	3	and c =	5
   	4		4

   Correct Option: C

   Here,
   3x – 4y + 5 = 0
   ⇒  –4y = – 3x – 5
   

   y =	−3	x	−5
   	–4		−4

   
   

   y =	3	x +	5
   	4		4

   
   Compare it with, y = mx + c we get
   

   m =	3	and c =	5
   	4		4

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162. What will be the point on the xaxis, which is eduidistant from the points (7, 6) and (–3, 4)

1. 1. (6, 0)

   
   
   

   2. (–2, 0)

   
   
   

   3. (4, 0)

   
   
   

   4. (3, 0)

   
   
   

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1. View Hint View Answer Discuss in Forum

   Let the point on x-axis be P(x, 0)
   
   A.T.Q
   PA = PB
   ⇒  PA² = PB²
   (x + 3)² + (4 – 0)² = (x – 7)² + (6 – 0)²
   x² + 9 + 6x + 16 = x² + 49 – 14x + 36
   x² + 6x + 25 = x² – 14x + 85
   20x = 60
   x = 3
   ∴  Point P is (3, 0)

   Correct Option: D

   Let the point on x-axis be P(x, 0)
   
   A.T.Q
   PA = PB
   ⇒  PA² = PB²
   (x + 3)² + (4 – 0)² = (x – 7)² + (6 – 0)²
   x² + 9 + 6x + 16 = x² + 49 – 14x + 36
   x² + 6x + 25 = x² – 14x + 85
   20x = 60
   x = 3
   ∴  Point P is (3, 0)

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163. What is the equation of line which has y-intercept 2 and is inclined at 60° to the x-axis.

1. 1. y = − √3x + 2

   
   
   

   2. y = − √3

   
   
   

   3. y = x + 2√3

   
   
   

   4. y = √3x + 2

   
   
   

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1. View Hint View Answer Discuss in Forum

   Here,
   c = 2
   m = tan60° = √3
   Equation of line be,
   y = mx + c
   ⇒  y = √3x + 2.

   Correct Option: D

   Here,
   c = 2
   m = tan60° = √3
   Equation of line be,
   y = mx + c
   ⇒  y = √3x + 2.

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164. What is the equation of a line, which passes through the points (–1, 1) and (2, –4).

1. 1. 5x + 3y + 2 = 0

   
   
   

   2. –5x + 3y + 4 = 0

   
   
   

   3. 3x + 5y + 6 = 0

   
   
   

   4. 5x + 3y + 3 = 0

   
   
   

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1. View Hint View Answer Discuss in Forum

   Equation of required line be
   

   y − y1	=	x − x1
   y2 − y1		x2 − x1

   
   

   y − 1	=	x + 1
   –4 − 1		2 + 1

   
   
   3y – 3 = –5x – 5
   5x + 3y + 2 = 0

   Correct Option: A

   Equation of required line be
   

   y − y1	=	x − x1
   y2 − y1		x2 − x1

   
   

   y − 1	=	x + 1
   –4 − 1		2 + 1

   
   
   3y – 3 = –5x – 5
   5x + 3y + 2 = 0

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165. The distance of the point (3, – 1) from the line12x – 5y – 7 = 0 will be ?

1. 1. 1	units
      13

   
   
   

   2. 43	units
      13

   
   
   

   3. 34	units
      13

   
   
   

   4. 1	units
      5

   
   
   

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1. View Hint View Answer Discuss in Forum

   Let required distance = d
   
   

   =	34	units
   	13

   Correct Option: C

   Let required distance = d
   
   

   =	34	units
   	13

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