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Note: We will use the formula. If a point divides two points (x_1, y_1) and (x_2, y_2) in the ratio m_1:m_2 , then the coordinates of the point at

x = \frac{m_1x_2+m_2x_1}{m_1+m_2}  

y = \frac{m_1y_2+m_2y_1}{m_1+m_2}  

 

Question 1: Calculate the co-ordinates of the point P which divides the line segment joining:

(i) A (1, 3) and B (5, 9) in the ratio 1: 2

(ii) A (-4, 6) and B (3, -5) in the ratio 3: 2

Answer:

i)

Ratio: m_1:m_2 = 1:2

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

Therefore

x = \frac{1 \times 5+2 \times 1}{1+2} = \frac{7}{3} 

y = \frac{1 \times 9+2 \times 3}{1+2}  = \frac{15}{3} = 5

Therefore P = (\frac{7}{3}, 5)

ii)

Ratio: m_1:m_2 = 3:2

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

Therefore

x = \frac{2 \times (-4)+3 \times 3}{3+2} = \frac{1}{5} 

y = \frac{2 \times 6+3 \times (-5)}{3+2}  = -\frac{3}{5}

Therefore P = (\frac{1}{5}, -\frac{3}{5})

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Question 2: In what ratio is the line joining (2, -3) and (5, 6) divided by the x-axis .

Answer:

Let the required ratio be  k:1   and the point of  x-axis    be  (x,0)

Since  y = \frac{ky_2+y_1}{k+1}

\Rightarrow 0 = \frac{k \times 6 -3}{k+1}

\Rightarrow 6k-3=0

\Rightarrow k = \frac{1}{2}

\Rightarrow  m_1:m_2 = 1:2

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Question 3: In what ratio is the line joining (2, -4) and (-3, 6) divided by the y-axis .

Answer:

Let the required ratio be  k:1   and the point of  y-axis    be  (0,y)

Since  x = \frac{kx_2+x_1}{k+1}

\Rightarrow 0 = \frac{k \times (-3) +2}{k+1}

\Rightarrow 3k-2=0

\Rightarrow k = \frac{2}{3}

\Rightarrow  m_1:m_2 = 2:3

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Question 4: In what ratio does the point (1, a) divide the join of (-1, 4) and (4, -1) ? Also, find the value of a .

Answer:

Let the point (1, a) divide the join of (-1, 4) and (4, -1) in the ratio k: 1

Since x = \frac{kx_2+x_1}{k+1}

\Rightarrow 1 = \frac{k \times (4) -1}{k+1}

\Rightarrow k+1=4k-1

\Rightarrow k = \frac{2}{3}

\Rightarrow  m_1:m_2 = 2:3

Now calculate a

\Rightarrow a = \frac{\frac{2}{3} \times (-1) +4} {\frac{2}{3} +1}

\Rightarrow a = \frac{10}{5}=2

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Question 5: In what ratio does the point (a, 6) divide the join of (-4, 3) and (2, 8) ? Also, find the value of a .

Answer:

Let the point (a, 6) divide the join of (-4, 3) and (2, 8) in the ratio k: 1

Since y = \frac{ky_2+y_1}{k+1}

\Rightarrow 6 = \frac{k \times (8) +3}{k+1}

\Rightarrow 6k+6=8k+3

\Rightarrow k = \frac{3}{2}

\Rightarrow  m_1:m_2 = 3:2

Now calculate a

\Rightarrow a = \frac{3 \times (2) +2 \times (-4)} {3+2}

\Rightarrow 5a = 6-8=-2

\Rightarrow a = -\frac{2}{5}

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Question 6: In what ratio is the join of (4, 3) and (2, -6) divided by the x-axis . Also, find the co-ordinates of the point of intersection.

Answer:

Let the required ratio be  k:1   and the point of  x-axis    be  (x,0)

Since  y = \frac{ky_2+y_1}{k+1}

\Rightarrow 0 = \frac{k \times (-6) +3}{k+1}

\Rightarrow 6k-3=0

\Rightarrow k = \frac{1}{2}

\Rightarrow m_1:m_2 = 1:2

Now calculate the coordinate of the point of intersection

x = \frac{1 \times (2)+2 \times (4)}{1+2} = \frac{10}{3} 

y = \frac{1 \times (-6)+2 \times (3)}{1+2}  = 0

Co-ordinates of the point of intersection = (\frac{10}{3}, 0)

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Question 7: Find the ratio in which the join of (-4, 7) and (3,0) is divided by the y-axis . Also, find the co-ordinates of the point of intersection.

Answer:

Let the required ratio be  k:1   and the point of  x-axis    be  (0, y)

Since  x = \frac{kx_2+x_1}{k+1}

\Rightarrow 0 = \frac{k \times (3) -4}{k+1}

\Rightarrow 3k-4=0

\Rightarrow k = \frac{4}{3}

\Rightarrow m_1:m_2 = 4:3

Now calculate the coordinate of the point of intersection

y = \frac{4 \times (0)+3 \times (7)}{4+3}  = 3

Co-ordinates of the point of intersection = (0, 3)

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Question 8: Points A, B, C \ and \  D divide the line segment joining the point (5, -10) and the origin in five equal parts. Find the co-ordinates of B \ and \ D .

Answer:

PB: BQ=2:3

PD: DQ=4:1

For B

Ratio =2:3

x = \frac{2 \times (5) +3 \times (0)}{2+3} = 2   

y = \frac{2 \times (-10) +3 \times (0)}{2+3}  = -4  

Hence the coordinates of B = (2, -4)

For D

Ratio =4:1

x = \frac{4 \times (5) +1 \times (0)}{4+1} = 4   

y = \frac{4 \times (-10) +1 \times (0)}{4+1}  = -8  

Hence the coordinates of D = (4, -8)

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Question 9: The line joining the points A (-3, -10) and B (-2,6) is divided by the point P  such that \frac{PB}{AB}=\frac{1}{5} Find the co-ordinates of P .

Answer:

Given \frac{PB}{AB}=\frac{1}{5}

This implies that AP: PB=4:1

x = \frac{4 \times (-2) +1 \times (-3)}{4+1} = -\frac{11}{5}   

y = \frac{4 \times (6) +1 \times (-10)}{4+1}  = -\frac{14}{5}  

Hence the coordinates of the point = (-\frac{11}{5} , -\frac{14}{5} )

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Question 10: P is a point on the line joining A (4, 3) and B (-2, 6) such that 5AP = 2BP Find the coordinates of P .

Answer:

Given 5AP = 2BP

This implies \frac{AP}{BP}=\frac{2}{5}

This implies that AP: PB=2:5

x = \frac{2 \times (-2) +5 \times (4)}{2+5} = \frac{16}{7}   

y = \frac{2 \times (6) +5 \times (3)}{2+5}  = \frac{27}{7}  

Hence the coordinates of the point = (\frac{16}{7}, \frac{27}{7})  

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Question 11: Calculate the ratio in which the line joining the points (-3, -1) and (5, 7) is divided by the line x = 2 . Also, find the co-ordinates of the point of intersection.

Answer:

Let the required ratio be k: 1 and the point of x=2   be (2, y)

Since  x = \frac{kx_2+x_1}{k+1}

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

\Rightarrow 2k+2=5k-3

\Rightarrow k = \frac{5}{3}

\Rightarrow m_1:m_2 = 5:3

Now calculate the coordinate of the point of intersection

y = \frac{5 \times (7)+3 \times (-1)}{5+3}  = 4

Co-ordinates of the point of intersection = (2, 4)

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Question 12: Calculate the ratio in which the line joining A (6, 5) and B (4, -3) is divided by the line y=2 .      [2006]

Answer:

Let the required ratio be k: 1 and the point of y=2   be (x, 2)

Since  y = \frac{ky_2+y_1}{k+1}

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

\Rightarrow 2k+2=-3k+5

\Rightarrow k = \frac{3}{5}

\Rightarrow m_1:m_2 = 3:5

Now calculate the coordinate of the point of intersection

x = \frac{3 \times (4)+5 \times (6)}{3+5}  = 4.25

Co-ordinates of the point of intersection = (4.25, 2)

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Question 13: The point P (5, -4) divides the line segment AB , as shown in the figure, in the ratio 2: 5 . Find the co-ordinates of points A \ and \ B .

Answer:

Ratio: m_1:m_2 = 2:5

Therefore

5 = \frac{2 \times (0)+5 \times x}{2+5} \Rightarrow x = 7 

-4 = \frac{2 \times y+5 \times (0)}{2+5}  \Rightarrow y = -14

Therefore A = (7, 0) \ and \ B=(0, -14)

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Question 14: Find the co-ordinates of the points of trisection of the line joining the points (-3, 0) and (6, 6) .

Answer:

When Ratio: m_1:m_2 = 1:2

Therefore

x = \frac{1 \times (6)+2 \times (-6)}{1+2}= 0 

y = \frac{1 \times 6+2 \times (0)}{1+2} = 2

Therefore the point = (0, 2)

When Ratio: m_1:m_2 = 2:1

Therefore

x = \frac{2 \times (6) +1 \times (-3)}{2+1}=3 

y = \frac{2 \times 6+1 \times (0)}{2+1} = 4

Therefore the point = (3, 4)

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Question 15: Show that the Line segment joining the points (-5, 8) and (10, -4) is trisection by the co-ordinate axes.

Answer:

Let the two points trisecting the points (-5, 8) and (10, -4) are A (x, 0) \ and \ B (0, y) .

When Ratio for B: m_1:m_2 = 1:2

Therefore

y = \frac{1 \times (-4)+2 \times (8)}{1+2} = 4

Therefore the point = (0, 4)

When Ratio for A: m_1:m_2 = 2:1

Therefore

x = \frac{2 \times (10)+1 \times (-5)}{2+1} = \frac{5}{3}

Therefore the point = (\frac{5}{3}, 0)

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Question 16: Show that A (3, -2) is a point of trisection of the line-segment joining the points (2, 1) and (5, -8) . Also find the co-ordinates of the other point of trisection.

Answer:

Let the point A (3, -2)  divide the join of (2, 1) and (5, -8) in the ratio k: 1

Since x = \frac{kx_2+x_1}{k+1}

\Rightarrow 3 = \frac{k \times (5) +2}{k+1}

\Rightarrow 3k+3=5k+2

\Rightarrow k = \frac{1}{2}

\Rightarrow  m_1:m_2 = 1:2

For the other point

When Ratio: m_1:m_2 = 2:1

Therefore

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

y = \frac{2 \times (-8)+1 \times (1)}{2+1} = -5

Therefore the point = (4, -5)

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Question 17: lf A = (-4, 3) and B = (8, -6)

(i) find the length of AB

(ii) In what ratio is the line joining A \ and \  B , divided by the x-axis ?      [2008]

Answer:

AB = \sqrt{(8-(-4))^2+(-6-3)^2} = \sqrt{144+81} = \sqrt{225} = 15

Let the required ratio be k:1   and the point of  x-axis    be  (x,0)

Since y = \frac{ky_2+y_1}{k+1}

\Rightarrow 0 = \frac{k \times (-6)+3}{k+1}

\Rightarrow 6k-3=0

\Rightarrow k = \frac{1}{2}

\Rightarrow  m_1:m_2 = 1:2

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Question 18: The line segment joining the points M (5, 7) and N (-3, 2) is intersected by the y-axis at point L . Write down the abscissa of L . Hence, find the ratio in which L divides MN . Also, find the co-ordinates of L .

Answer:

Let the required ratio be k: 1 and the point of y-axis   be (0, y)

Since x = \frac{kx_2+x_1}{k+1}

\Rightarrow 0 = \frac{k \times (-3)+5}{k+1}

\Rightarrow 3k-5=0

\Rightarrow k = \frac{5}{3}

\Rightarrow  m_1:m_2 = 5:3

y = \frac{5 \times (2)+3 \times (7)}{5+3} = \frac{31}{8}

Therefore the point = (0, \frac{31}{8})

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Question 19: A (2, 5), B (-1, 2) \ and \ C (5, 8) are the co-ordinates of the vertices of the triangle ABC . Points P \ and \ Q lie on AB and AC respectively, such that: AP: PB = AQ: QC = 1 : 2

(i) Calculate the co-ordinates of P and Q .

(ii) Show that PQ = \frac{1}{3} BC

Answer:

For P When Ratio: m_1:m_2 = 1:2

Therefore

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

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

Therefore the point P= (1, 4)

For Q When Ratio: m_1:m_2 = 1:2

Therefore

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

y = \frac{1 \times 8+2 \times (5)}{1+2} = 6

Therefore the point = (3, 6)

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Question 20: A (-3,4), B (3, -1) \ and \  C (-2, 4) are the vertices of a triangle ABC . Find the length of line segment AP , where point P lies inside BC , such that BP: PC = 2 : 3

Answer:

For P When Ratio: m_1:m_2 = 2:3

Therefore

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

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

Therefore the point P= (1,1)

AP = \sqrt{(1-(-3))^2+(1-4)^2} = \sqrt{16+9} = \sqrt{25} = 5

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Question 21: The line segment joining A(2, 3) and 8(6, -5) is intercepted by x-axis at the point K . Write down the ordinate of the point K . Hence, find the ratio in which K divides AB . Also, find the co-ordinates of the point K .      [1990, 2006]

Answer:

Let the required ratio be k: 1 and the point of x-axis   be K(x, 0)

Since y = \frac{ky_2+y_1}{k+1}

\Rightarrow 0 = \frac{k \times (-5)+3}{k+1}

\Rightarrow 5k-3=0

\Rightarrow k = \frac{3}{5}

\Rightarrow  m_1:m_2 = 3:5

x = \frac{3 \times (6)+5 \times (2)}{3+5} = \frac{14}{4}

Therefore the point K= (\frac{14}{4}, 0)

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Question 22: The line segment joining A(4, 7) and B(-6, -2) is intercepted by the y-axis at the point K . Write down the abscissa of the point K . Hence, find the ratio in which K divides AB . Also, find the co-ordinates of the point K .

Answer:

Let the required ratio be k: 1 and the point of y-axis   be K(0, y)

Since x = \frac{kx_2+x_1}{k+1}

\Rightarrow 0 = \frac{k \times (-6)+4}{k+1}

\Rightarrow 6k-4=0

\Rightarrow k = \frac{2}{3}

\Rightarrow  m_1:m_2 = 2:3

y = \frac{2 \times (-2)+3 \times (7)}{2+3} = \frac{17}{5}

Therefore the point K= (0, \frac{17}{5})

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Question 23: The line joining P(-4, 5) and Q(3, 2) intersects the y-axis at point R . PM and QN are perpendiculars from P \ and \ Q on the x-axis . Find:

(i) The ratio PR: RQ .

(ii) The co-ordinates of R .

(iii) The area of the quadrilateral PMNQ .

Answer:

Let the required ratio be k: 1 and the point of y-axis   be R(0, y)

Since x = \frac{kx_2+x_1}{k+1}

\Rightarrow 0 = \frac{k \times (3)-4}{k+1}

\Rightarrow 3k-4=0

\Rightarrow k = \frac{4}{3}

\Rightarrow  m_1:m_2 = 4:3

y = \frac{\frac{4}{3} \times (2)+1 \times (5)}{ \frac{4}{3}+1} = \frac{23}{7}

Therefore the point R= (0, \frac{23}{7})

Area of quadrilateral PMNQ = 7 \times \frac{5+2}{2} = 24.5 sq. units

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Question 24: In the given figure, line APB meets the x-axis at point A and y-axis at point B . P is the point (-4,2) and AP: PB = 1 :2  Find the co-ordinates of A \ and \  B .     [1999, 2013]

Answer:

Given AP: PB = 1:2 

Therefore

-4 = \frac{1 \times (0)+2 \times x}{1+2} \Rightarrow x = -6 

2 = \frac{1 \times y+2 \times (0)}{1+2}  \Rightarrow y = 6

Therefore A (-6, 0) \ and \ B(0,6)  .

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Question 25: Given a line segment AB joining the points A(-4,6) and B(8,-3) . Find:

  1. i) The ratio in which AB is divided by y-axis .
  2. ii) Find the coordinates of point of intersection

iii) The length of AB      [2012]

Answer:

Let the required ratio be k: 1 and the point of intersection y-axis   be (0, y)

Since x = \frac{kx_2+x_1}{k+1}

\Rightarrow 0 = \frac{k \times (8)-4}{k+1}

\Rightarrow 8k-4=0

\Rightarrow k = \frac{1}{2}

\Rightarrow  m_1:m_2 = 1:2

y = \frac{1 \times (-3)+2 \times (6)}{ 1+2} = 3

Therefore the point intersection is = (0, 3)

Length of AB = \sqrt{(8-(-4))^2+(-3-6)^2} = \sqrt{144+81} = \sqrt{225} = 15 units .

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