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

(i) Calculate the value of \displaystyle a  

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

Answer:

\displaystyle \text{Therefore for } P (0,y)  

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

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

\displaystyle \text{Hence } P(0, \frac{5}{2} )  

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

Answer:

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

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

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

\displaystyle \Rightarrow k=3  

\displaystyle \Rightarrow m_1:m_2 = 3:1  

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

\displaystyle \text{Therefore } P(3,0)  

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

Answer:

\displaystyle \text{Given Midpoint of } AB = (4,-3)  

Therefore

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

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

\displaystyle \text{Therefore } A = (8, 0) \text{ and } B(0, -6)  

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

(i) the length of radius \displaystyle AC  

(ii) the coordinates of \displaystyle B [2013]

Answer:

\displaystyle \text{Given Midpoint of } AB = C(-2,5)  

Therefore

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

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

\displaystyle \text{Therefore } B = (-7, 17)  

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

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

Answer:

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

Therefore

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

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

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

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

Answer:

\displaystyle \text{Given Midpoint of } = C(1, 2a+1)  

Therefore

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

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

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

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

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

Answer:

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

Therefore

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

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

Therefore the point \displaystyle P= (3, 4)  

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

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

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

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

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

\displaystyle \Rightarrow m_1:m_2 = 4:17  

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

Answer:

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

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

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

\displaystyle \text{Since } AB=BC (two sides are equal). Hence triangle \displaystyle ABC is a isosceles triangle.

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

Answer:

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

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

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

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

\displaystyle \Rightarrow m_1:m_2=1:3  

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

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

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

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Question 10: Calculate the ratio in which the line joining \displaystyle A (6, 5) \text{ and } B (4, -3) is divided by the line \displaystyle y=2 [2006]

Answer:

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

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

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

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

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

\displaystyle \Rightarrow m_1:m_2 = 3:5  

Now calculate the coordinate of the point of intersection

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

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

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Question 11: lf \displaystyle A = (-4, 3) \text{ and } B = (8, -6)  

(i) find the length of \displaystyle AB  

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

Answer:

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

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

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

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

\displaystyle \Rightarrow 6k-3=0  

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

\displaystyle \Rightarrow m_1:m_2 = 1:2  

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Question 12: The line segment joining \displaystyle A(2, 3) \text{ and } 8(6, -5) is intercepted by \displaystyle x-axis at the point \displaystyle K Write down the ordinate of the point \displaystyle K Hence, find the ratio in which \displaystyle K divides \displaystyle AB Also, find the co-ordinates of the point \displaystyle K [1990, 2006]

Answer:

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

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

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

\displaystyle \Rightarrow 5k-3=0  

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

\displaystyle \Rightarrow m_1:m_2 = 3:5  

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

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

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Question 13: In the given figure, line \displaystyle APB meets the \displaystyle x-axis at point \displaystyle A \text{ and } y-axis at point \displaystyle B \displaystyle P is the point \displaystyle (-4,2) \text{ and } AP: PB = 1 :2 Find the co-ordinates of \displaystyle A \text{ and } B [1999, 2013]

Answer:

\displaystyle \text{Given } AP: PB = 1:2  

Therefore

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

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

\displaystyle \text{Therefore } A (-6, 0) \text{ and } B(0,6)  

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Question 14: Given a line segment \displaystyle AB joining the points \displaystyle A(-4,6) \text{ and } B(8,-3) Find:

i) The ratio in which \displaystyle AB is divided by \displaystyle y-axis  

ii) Find the coordinates of point of intersection

iii) The length of \displaystyle AB [2012]

Answer:

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

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

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

\displaystyle \Rightarrow 8k-4=0  

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

\displaystyle \Rightarrow m_1:m_2 = 1:2  

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

\displaystyle \text{Therefore the point intersection is } = (0, 3)  

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

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

Answer:

\displaystyle K(2, 5) \text{ and } M(x, -7) are the two points.

Distance between them is \displaystyle 13 units.

Therefore

\displaystyle \sqrt{x-2)^2+(-7-5)^2} = 13  

\displaystyle x^2+4-4x+144=169  

\displaystyle x^2-4x-21=0  

\displaystyle (x-7)(x+3)=0 \Rightarrow x = 7 \text{ or }  -3  

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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:

\displaystyle \text{Given Midpoint of } = (1, 2m-1)  

Therefore

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

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

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

Answer:

\displaystyle \text{Since } G is the centroid

\displaystyle 4 = \frac{1+4+a}{3} \Rightarrow a=7  

\displaystyle 3= \frac{3+b+1}{3} \Rightarrow b = 5  

\displaystyle \text{Therefore } B(4, 5) \text{ and } C(7, 1)  

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