Question 1: Point P divides the line segment joining the points A (8, 0) \ and \ B (16, -8) in the ratio 3 : 5 . Find the co-ordinates of point P . Also, find the equation of the line through P and parallel to 3x + 5y = 7 .

Answer:

Given P divides A (8, 0) \ and \ B (16, -8) in the ratio 3 : 5

Ratio: m_1:m_2 = 3 : 5

Let the coordinates of the point P \ be \ (x, y) . Therefore

x =  \frac{3 \times 16+5 \times 8}{3+5} = \frac{88}{8}  = 11 

y =  \frac{3 \times (-8)+5 \times 0}{3+5}  = \frac{-24}{8}   =  -3

Therefore P = (11, -3)

Equation of the line given : 3x+5y=7 \Rightarrow slope = m = \frac{-3}{5}

Therefore the required equation is

y - y_1 = m(x-x_1)

y -(-3) = \frac{-3}{5}(x-11) or 5y +3x=18

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Question 2: The line segment joining the points A (3, -4) \ and \ B (-2, 1) is divided in the ratio 1 : 3 at point P in it. Find the co-ordinates of P . Also, find the equation of the line through P and perpendicular to the line 5x - 3y = 4 .

Answer:

Given P divides A (3,-4) \ and \ B (-2,1) in the ratio 1:3

Ratio: m_1:m_2 = 1:3

Let the coordinates of the point P \ be \ (x, y) . Therefore

x = \frac{1 \times (-2)+3 \times 3}{1+3} = \frac{7}{4}   

y = \frac{1 \times 1+3 \times (-4)}{1+3}  = \frac{-11}{4} 

Therefore P = (\frac{7}{4} , \frac{-11}{4})

Equation of the line given : 5x - 3y = 4 \Rightarrow slope = m_1 = \frac{5}{3}

Therefore the slope of the line perpendicular to the above line = \frac{-3}{5}

Therefore the required equation is

y - y_1 = m(x-x_1)

y -\frac{-11}{4} =\frac{-3}{5}(x-\frac{7}{4})   or  10y +6x+17 = 0

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Question 3: A line 5x + 3y + 15 = 0 meets y-axis at point P . Find the co-ordinates of point P . Find the equation of a line through P and perpendicular to x - 3y + 4 = 0 .

Answer:

At y-axis,  \ x = 0 \Rightarrow y = -5

Therefore the coordinate of P = (0, -5)

x - 3y + 4 = 0 \Rightarrow slope = \frac{1}{3}

Therefore the slope of a like perpendicular to this line = -3

Hence the line passing through (0, -5) with a slope of -3 is

y-(-5) = -3(x-0) \Rightarrow y+3x+5 = 0 

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Question 4: Find the value of k for which the lines kx - 5y + 4 = 0 and 5x - 2y + 5 = 0 are perpendicular to each other. [2003]

Answer:

Slope of  kx - 5y + 4 = 0   \Rightarrow m_1 = \frac{k}{5} 

Slope of 5x - 2y + 5 = 0   \Rightarrow m_2 = \frac{5}{2} 

Since the two lines are perpendicular, m_1. m_2 = -1 

\Rightarrow \frac{k}{5} . \frac{5}{2} = -1  

\Rightarrow k = -2 

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Question 5: A straight line passes through the points P (-1, 4) \ and \ Q (5, -2) . It intersects the co-ordinate axes at points A \ and \ B . M is the mid-point of the line segment AB . Find:

  • The equation of line
  • The co-ordinates of A \ and \ B
  • The co-ordinates of M [2003]

Answer:

P (-1, 4) \ and \ Q (5, -2)

The equation of the line:

y - 4 = \frac{-2-4}{5-(-1)} (x+1)

\Rightarrow y-4 = -1 (x+1)

\Rightarrow y+x=3

The x-intercept A = (3,0) and the y-intercept B = (0,3)

The coordinate of M = (\frac{3+0}{2}, \frac{0+3}{2}) = (\frac{3}{2}, \frac{3}{2})

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Question 6: (1, 5) \ and \ (-3, -1) are the co-ordinates of vertices A \ and \  C respectively of rhombus ABCD . Find the equations of the diagonals AC \ and \  BD .

Answer:

Midpoint M = (\frac{-3+1}{2}, \frac{-1+5}{2}) = (-1, 2)

Slope of AC = \frac{-1-5}{-3-1} = \frac{3}{2}

Equation of  AC :

y - 5 =  \frac{3}{2} (x-1) \Rightarrow 3x-2y+7=0

Slope of BD = -\frac{2}{3}

Therefore equation of BD :

y - 2 = -\frac{2}{3}(x+1) \Rightarrow 3y+2x=4

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Question 7: Show that A (3, 2), B (6, -2) \ and \ C (2, -5) can be vertices of a square.

  • Find the co-ordinates of its fourth vertex D , if ABCD is a square
  • Without using the co-ordinates of vertex D , find the equation of side AD of the square and the equation of diagonal BD .

Answer:

Mid point of AC = (\frac{3+2}{2}, \frac{-5+2}{2}) = (\frac{5}{2}, \frac{-3}{2})

Let the coordinate of D be (x, y)

In a square, the diagonals bisect each other. Therefore

\frac{x+6}{2}= \frac{5}{2} \Rightarrow x = -1

\frac{y-2}{2}= \frac{-3}{2} \Rightarrow y = -1

Hence D is (-1, -1)

Slope  of AB = \frac{-2-2}{6-3} = \frac{-4}{3}

Since AB \perp AD , slope of AD = \frac{3}{4}

Hence the equation of AD :

y - 2 = \frac{3}{4} (x-3) \Rightarrow 4y-3x+1 = 0

Slope of BD = \frac{-2+1}{6+1} = -\frac{1}{7}

Hence the equation of BD :

y +2 = -\frac{1}{7}(x-6) \Rightarrow 7y+x+8=0

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Question 8: A line through origin meets the line x = 3y + 2 at right angles at point X . find the co-ordinates of point X .

Answer:

Given x = 3y + 2 … … … … (i)

Slope of line x = 3y + 2 is \frac{1}{3}

Slope of perpendicular = -3

The equation of a line passing through (0,0) and having slope 3 is

y - 0 = -3 (x-0) \Rightarrow y = -3x   … … … … (i)

Solving equations (i) and (ii)

x= 3 (-3x) + 2 \Rightarrow x = \frac{1}{5}

Hence y = -\frac{3}{5}

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Question 9: A straight line passes through the point (3, 2) and the portion of this line, intercepted between the positive axes, is bisected at this point. Find the equation of the line.

Answer:

Let y-intercept be (0, y) and x-intercept be (x, 0)

Given (3,2) is the mid point of (0,y) and (x,0) . Therefore:

y-intercept = (0, 4)

x-intercept = (6, 0)

Slope of line = \frac{0-4}{6-0} = -\frac{2}{3}

Equation of line:

y-0 = -\frac{2}{3} (x-6) \Rightarrow 3y+2x=12

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Question 10: Find the equation of the line passing through the point of intersection of 7x + 6y = 71 \ and \ 5x - 8y = -23 ; and perpendicular to the line 4x -2y = 1 .

Answer:

Solve equations

7x + 6y = 71     … … … … (i)

5x - 8y = -23    … … … … (ii)

Multiply (i) by 4 and (ii) by 3 and then add the equations, we get

43x = 215 \Rightarrow x = 5 

Substituting x = 5  in (i) we get

7(5)+6y=71 \Rightarrow y = 6 

Therefore the intercept is (5, 6) 

Sloe of line 4x -2y = 1 is 2 

Therefore the slope of perpendicular = -\frac{1}{2} 

Hence the equation of the perpendicular:

y - 6 = -\frac{1}{2} (x-5) 

2y -12 = -x+5 

2y+x=17 

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