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

\displaystyle 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 \displaystyle P which divides the line segment joining:

\displaystyle \text{i) } A (1, 3) \text{ and } B (5, 9) \text{ in the ratio } 1: 2  

\displaystyle \text{ii) } A (-4, 6) \text{ and } B (3, -5) \text{ in the ratio } 3: 2  

Answer:

\displaystyle \text{i) } \text{Ratio: } m_1:m_2 = 1:2  

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

Therefore

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

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

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

\displaystyle \text{ii) } \text{Ratio: } m_1:m_2 = 3:2  

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

Therefore

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

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

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

\displaystyle \\

Question 2: In what ratio is the line joining \displaystyle (2, -3) \text{ and } (5, 6) divided by the \displaystyle x-axis

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 6 -3}{k+1}  

\displaystyle \Rightarrow 6k-3=0  

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

\displaystyle \Rightarrow m_1:m_2 = 1:2  

\displaystyle \\

Question 3: In what ratio is the line joining \displaystyle (2, -4) \text{ and } (-3, 6) divided by the \displaystyle y-axis

Answer:

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

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

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

\displaystyle \Rightarrow 3k-2=0  

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

\displaystyle \Rightarrow m_1:m_2 = 2:3  

\displaystyle \\

Question 4: In what ratio does the point \displaystyle (1, a) divide the join of \displaystyle (-1, 4) \text{ and } (4, -1) ? Also, find the value of \displaystyle a

Answer:

Let the point \displaystyle (1, a) divide the join of \displaystyle (-1, 4) \text{ and } (4, -1) \text{ in the ratio } k: 1  

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

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

\displaystyle \Rightarrow k+1=4k-1  

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

\displaystyle \Rightarrow m_1:m_2 = 2:3  

Now calculate \displaystyle a  

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

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

\displaystyle \\

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

Answer:

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

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

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

\displaystyle \Rightarrow 6k+6=8k+3  

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

\displaystyle \Rightarrow m_1:m_2 = 3:2  

Now calculate \displaystyle a  

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

\displaystyle \Rightarrow 5a = 6-8=-2  

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

\displaystyle \\

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

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 (-6) +3}{k+1}  

\displaystyle \Rightarrow 6k-3=0  

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

\displaystyle \Rightarrow m_1:m_2 = 1:2  

Now calculate the coordinate of the point of intersection

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

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

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

\displaystyle \\

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

Answer:

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

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

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

\displaystyle \Rightarrow 3k-4=0  

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

\displaystyle \Rightarrow m_1:m_2 = 4:3  

Now calculate the coordinate of the point of intersection

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

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

\displaystyle \\

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

Answer:

\displaystyle PB: BQ=2:3  

\displaystyle PD: DQ=4:1  

For \displaystyle B  

Ratio \displaystyle =2:3  

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

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

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

For \displaystyle D  

Ratio \displaystyle =4:1  

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

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

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

\displaystyle \\

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

Answer:

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

This implies that \displaystyle AP: PB=4:1  

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

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

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

\displaystyle \\

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

Answer:

\displaystyle \text{ Given }  5AP = 2BP  

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

This implies that \displaystyle AP: PB=2:5  

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

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

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

\displaystyle \\

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

Answer:

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

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

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

\displaystyle \Rightarrow 2k+2=5k-3  

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

\displaystyle \Rightarrow m_1:m_2 = 5:3  

Now calculate the coordinate of the point of intersection

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

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

\displaystyle \\

Question 12: 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)  

\displaystyle \\

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

Answer:

\displaystyle \text{Ratio: } m_1:m_2 = 2:5  

Therefore

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

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

\displaystyle \text{Therefore } A = (7, 0) \ and \ B=(0, -14)  

\displaystyle \\

Question 14: Find the co-ordinates of the points of trisection of the line joining the points \displaystyle (-3, 0) \text{ and } (6, 6)

Answer:

When \displaystyle \text{Ratio: } m_1:m_2 = 1:2  

Therefore

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

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

\displaystyle \text{Therefore the point } = (0, 2)  

When \displaystyle \text{Ratio: } m_1:m_2 = 2:1  

Therefore

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

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

\displaystyle \text{Therefore the point } = (3, 4)  

\displaystyle \\

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

Answer:

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

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

Therefore

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

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

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

Therefore

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

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

\displaystyle \\

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

Answer:

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

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

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

\displaystyle \Rightarrow 3k+3=5k+2  

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

\displaystyle \Rightarrow m_1:m_2 = 1:2  

For the other point

When \displaystyle \text{Ratio: } m_1:m_2 = 2:1  

Therefore

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

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

\displaystyle \text{Therefore the point } = (4, -5)  

\displaystyle \\

Question 17: 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 \ 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  

\displaystyle \\

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

Answer:

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

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

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

\displaystyle \Rightarrow 3k-5=0  

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

\displaystyle \Rightarrow m_1:m_2 = 5:3  

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

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

\displaystyle \\

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

\displaystyle \text{(i) Calculate the co-ordinates of } P \text{ and } Q .

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

Answer:

\displaystyle \text{For P When Ratio: } m_1:m_2 = 1:2  

Therefore

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

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

\displaystyle \text{Therefore the point } P= (1, 4)  

\displaystyle \text{For Q When Ratio: } m_1:m_2 = 1:2  

Therefore

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

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

\displaystyle \text{Therefore the point } = (3, 6)  

\displaystyle \\

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

Answer:

For P When \displaystyle \text{Ratio: } m_1:m_2 = 2:3  

Therefore

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

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

\displaystyle \text{Therefore the point } P= (1,1)  

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

\displaystyle \\

Question 21: 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)  

\displaystyle \\

Question 22: The line segment joining \displaystyle A(4, 7) \text{ and } B(-6, -2) is intercepted by the \displaystyle y-axis at the point \displaystyle K . Write down the abscissa 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

Answer:

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

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

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

\displaystyle \Rightarrow 6k-4=0  

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

\displaystyle \Rightarrow m_1:m_2 = 2:3  

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

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

\displaystyle \\

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

\displaystyle \text{(i) The ratio  } PR: RQ

\displaystyle \text{(ii) The co-ordinates of  } R

\displaystyle \text{(iii) The area of the quadrilateral  } PMNQ

Answer:

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

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

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

\displaystyle \Rightarrow 3k-4=0  

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

\displaystyle \Rightarrow m_1:m_2 = 4:3  

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

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

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

\displaystyle \\

Question 24: 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 \ 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) \ and \ B(0,6) .

\displaystyle \\

Question 25: 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  

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

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