Click Here (Amazon): Books for ICSE Class 10 Board Exams

Question 1: A line segment joining A(-1, \frac{5}{3}) \ and \ B(a, 5) is divided in the ratio 1 : 3 \ at \ P , the point where the line segment AB   intersects the y-axis .

(i) Calculate the value of  a

(ii) Calculate the co-ordinates of P .      [1994]

Answer:

31Therefore for  P (0,y)

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

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

Hence P(0, \frac{5}{2}) 

\\

Question 2: In what ratio is the line joining A(0, 3) \ and \ B (4, -1) divided by the  x-axis ? Write the co-ordinates of the point where AB intersects the x-axis .     [1993]

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 (-1) +3}{k+1}

\Rightarrow k=3

\Rightarrow  m_1:m_2 = 3:1

Therefore  x = \frac{3(4)+1(0)}{3+1} = 3

Therefore P(3,0)

\\

Question 3: The mid-point of the segment AB , as shown in diagram, is C(4, -3) . Write down the coordinates of A \ and \  B .     [1996]

Answer:

Given Midpoint of AB = (4,-3)

Therefore

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

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

Therefore A = (8, 0) \ and \  B(0, -6)

\\

Question 4: AB is a diameter of a circle with center C = (-2, 5) . If A = (3, -7) , find

(i) the length of radius AC 

(ii) the coordinates of B . [2013]

Answer:

Given Midpoint of AB = C(-2,5)

Therefore

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

5 = \frac{1 \times (y)+1 \times (-7)}{1+1}  \Rightarrow y = 17

Therefore B = (-7, 17)

AC = \sqrt{(-7-3)^2+(17-(-7))^2} = \sqrt{676} = 26

\\

Question 5: Find the co-ordinates of the centroid of a triangle ABC whose vertices are : A(-1, 3), B(1, -1) \ and \ C(5, 1) .     [2006]

Answer:

Let O(x, y) be the centroid of triangle ABC .

Therefore

x=\frac{-1+1+5}{3}= \frac{5}{3}

y = \frac{3-1+1}{3}=1

Hence the coordinates of the centroid are (\frac{5}{3}, 1)

\\

Question 6: The mid-point of the line segment joining (2a, 4) \ and \ (-2, 2b) \ is \ (1, 2a+1) . Find the values of a \ and \ b .      [2007]

Answer:

Given Midpoint of  = C(1, 2a+1)

Therefore

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

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

\\

Question 7: (i) Write down the co-ordinates of the point P that divides the line joining A(- 4, l) \ and \ B(17, 10) in the ratio 1 : 2 .

(ii) Calculate the distance OP , where O is the origin.

(iii) In what ratio does the y-axis divide the line AB ? [1995]

Answer:

i) For P When Ratio: m_1:m_2 = 1:2 A(- 4, l) \ and \ B(17, 10)

Therefore

x = \frac{1 \times (17)+2 \times (-4)}{1+2}= 3  

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

Therefore the point P= (3, 4)

ii) OP = \sqrt{(3-0)^2+(4-0)^2} = \sqrt{25} = 5

iii)  Let the required ratio be  k:1    and the point be Q(0,y)

Since  y = \frac{ky_2+y_1}{k+1}

\Rightarrow 0 = \frac{k \times (17) -4}{k+1}

\Rightarrow k=\frac{4}{17}

\Rightarrow  m_1:m_2 = 4:17

\\

Question 8: Prove that the points A(-5,4); B(-1, -2) \ and \ C(5, 2) are the vertices of an isosceles right-angled triangle. Find the co-ordinates of D so that ABCD is a square.      [1992]

Answer:

AC = \sqrt{(5-(-5))^2+(2-4)^2} = \sqrt{104}  

AB= \sqrt{(-1-(-5))^2+(-2-(-4))^2} = \sqrt{52}  

BC = \sqrt{(-1-(-5))^2+(-2-2)^2} = \sqrt{52}  

Since AB=BC (two sides are equal). Hence triangle ABC is a isosceles triangle.

\\

Question 9: Calculate the ratio in which the line joining A(-4, 2) \ and \ B(3, 6) is divided by point P(x, 3) . Also, find (i) x (ii) length of AP .   [2014]

Answer:

Let P(x,3) divide MO in the ratio  k:1  

Since  y = \frac{ky_2+y_1}{k+1}

\Rightarrow 3 = \frac{k \times (6) +2}{k+1}

\Rightarrow k=\frac{1}{3} 41

\Rightarrow  m_1:m_2=1:3

Since  x = \frac{kx_2+x_1}{k+1}

\Rightarrow x = \frac{1 \times (3) +3 \times (-4)}{1+3} = \frac{-9}{4}

AP = \sqrt{(-\frac{9}{4}-(-4))^2+(3-2)^2} = \sqrt{(\frac{7}{4})^2+1)} = \sqrt{\frac{65}{16}}  

\\

Question 10: 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)

\\

Question 11: 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

\\

Question 12: 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)

\\

Question 13: 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]21

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)  .

\\

Question 14: 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 .

\\

Question 15: KM is a straight line of 13 units. If  K has the coordinates  (2, 5)   and  M  has the coordinates  (x, -7) , find the value of  x .     [2004]

Answer:

K(2, 5)  and M(x, -7)  are the two points.

Distance between them is  13  units.

Therefore

\sqrt{x-2)^2+(-7-5)^2} = 13 

x^2+4-4x+144=169 

x^2-4x-21=0 

(x-7)(x+3)=0 \Rightarrow x = 7 \ or \ -3 

\\  

Question 16: The mid point of the line segment joining (3m, 6) and (-4, 3n) is (1, 2m-1). Find the values of m and n.     [2006]

Answer:

Given Midpoint of  = (1, 2m-1)

Therefore

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

2(m)+1 = \frac{1 \times (6)+1 \times (3n)}{1+1}  \Rightarrow n = 0

\\

Question 17: ABC is a triangle and  G(4,3)  is the centroid of the triangle. If  A(1,3), B(4,b) \ and \ C(a,1) , find  a \ and \ b . Find the length of the side  BC .    [2011]11

Answer:

Since G is the centroid

4 = \frac{1+4+a}{3} \Rightarrow  a=7 

3= \frac{3+b+1}{3} \Rightarrow b = 5 

Therefore B(4, 5) \ and \ C(7, 1) 

Therefore BC=\sqrt{(7-4)^2+(1-5)^2} = \sqrt{25} = 5   units.

\\

Click Here (Amazon): Books for ICSE Class 10 Board Exams

 

Advertisements