Class 10: Distance and Section Formula – ICSE Board Problems – ICSE / ISC / CBSE Mathematics Portal for K12 Students

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:

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

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

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

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

$\displaystyle \\$

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 $\displaystyle (x,0)$

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

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

$\displaystyle \Rightarrow k=3$

$\displaystyle \Rightarrow m_1:m_2 = 3:1$

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

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

$\displaystyle \\$

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

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

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

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

$\displaystyle \\$

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

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

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

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

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

$\displaystyle \\$

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

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

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

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

$\displaystyle \\$

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

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

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

$\displaystyle \\$

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 $\displaystyle Q(0,y)$

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

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

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

$\displaystyle \Rightarrow m_1:m_2 = 4:17$

$\displaystyle \\$

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.

$\displaystyle \\$

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 $\displaystyle k:1$

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

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

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

$\displaystyle \Rightarrow m_1:m_2=1:3$

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

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

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

$\displaystyle \\$

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 = $\displaystyle (4.25, 2)$

$\displaystyle \\$

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 $\displaystyle (x,0)$

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

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

$\displaystyle \Rightarrow 6k-3=0$

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

$\displaystyle \Rightarrow m_1:m_2 = 1:2$

$\displaystyle \\$

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 $\displaystyle K(x, 0)$

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

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

$\displaystyle \Rightarrow 5k-3=0$

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

$\displaystyle \Rightarrow m_1:m_2 = 3:5$

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

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

$\displaystyle \\$

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

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

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

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

$\displaystyle \\$

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

$\displaystyle \\$

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

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

$\displaystyle x^2+4-4x+144=169$

$\displaystyle x^2-4x-21=0$

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

$\displaystyle \\$

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

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

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

$\displaystyle \\$

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.

ICSE / ISC / CBSE Mathematics Portal for K12 Students

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Class 10: Distance and Section Formula – ICSE Board Problems

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