Class 10: Distance and Section Formulae – Miscellaneous Problems – Set 1 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: Find the the distance between the points (3, 6) and (0,2).

Answer:

$\text{Let } (3, 6) = ( x_1, y_1) \text{ and } (0, 2) = ( x_2, y_2)$

Therefore the distance between the given points

$= \sqrt{ (x_2-x_1)^2 +(y_2-y_1)^2 }$

$= \sqrt{ (0-3)^2 +(2-6)^2 }$

$= \sqrt{ 9 +16 } = \sqrt{ 25 } = 5 \text{ units }$

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Question 2: Find the distance between the origin and the point : (i) (-12, 5) (ii) (15, -8).

Answer:

$\displaystyle \text{(i) Since, distance between origin and } (x, y) = \sqrt{x^2+y^2}$

$\displaystyle \text{Distance between origin and the point } (-12, 5) \\ \\ = \sqrt{(-12)^2+(5)^2} = \sqrt{169} = 13 \text{ units }$

$\displaystyle \text{(ii) Distance between origin and the point } ( 15, -8) \\ \\ = \sqrt{(15)^2+(-8)^2} = \sqrt{289} = 17 \text{ units }$

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Question 3: Find the co-ordinates of points on the x-axis which are at a distance of 5 units from the point (6, -3).

Answer:

Let the co-ordinates of the point on the x-axis be $(x, 0)$

$\text{Distance } = \sqrt{ (x_2-x_1)^2 +(y_2-y_1)^2 }$

$\Rightarrow 5 = \sqrt{ (x-6)^2 +(0+3)^2 }$

Squaring both sides

$25 = x^2 - 12 x + 36 + 9$

$\Rightarrow x^2 - 12 x + 20 = 0$

$\Rightarrow ( x-2)(x-10) = 0$

$\Rightarrow x = 2, or x = 10$

Therefore the required points on the x-axis are $( 2, 0)$ and $(10, 0)$

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Question 4: KM is a straight line of 13 units. If K has the coordinate ( 2, 5) and M has the coordinates $( x, -7)$ , find the value of $x$ . [ICSE Board 2004]

Answer:

$\text{Let } K (2, 5) = (x_1, y_1) \text{ and } M (x, -7) = (x_2, y_2)$

$KM = 13$ unit

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

Squaring both sides

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

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

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

$\Rightarrow x = 7 \text{ or } x = - 3$

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Question 5: Which point on the y-axis is equidistant from the points (12,3) and (-5, 10).

Answer:

Let the required point on the y-axis be $(0, y)$ .

Given $(0, y)$ is equidistant from $(12, 3)$ and $(-5, 10)$

i.e. distance between $(0, y)$ and $(12,3) =$ distance between $(0, y)$ and $(-5, 10 )$

$\Rightarrow \sqrt{ (12-0)^2 +(3-y)^2 } = \sqrt{ (-5-0)^2 +(10-y)^2}$

$\Rightarrow 144 + 9 + y^2 - 6y = 25 + 100 + y^2 -20$

$\Rightarrow 14y = - 28$

$\Rightarrow y = -2$

Therefore the required point on the axis is $(0, -2)$

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Question 6: Use the distance formula to show that the points A(1, 1), B(6, 4) and C(4,2) are collinear.

Answer:

$AB = \sqrt{ (6-1)^2+(4+1)^2 } = \sqrt{ 25 + 25 } = \sqrt{50} = 5\sqrt{2}$

$BC = \sqrt{ (4-6)^2+(2-4)^2 } = \sqrt{ 4 + 4 } = \sqrt{8} = 2\sqrt{2}$

$AC = \sqrt{ (4-1)^2+(2+1)^2 } = \sqrt{ 9 + 9 } = \sqrt{18} = 3\sqrt{2}$

$\Rightarrow BC + AC = 2 \sqrt{2}+3\sqrt{3} = 5\sqrt{2} = AB$

Therefore given points are collinear.

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Question 7: Show that the points A (8, 3), B (0, 9) and C (14, 11) are the vertices of an isosceles right-angled triangle.

Answer:

$AB = \sqrt{ (0-8)^2+(9-3)^2 } = \sqrt{ 64 + 36 } = \sqrt{100} = 10$

$BC = \sqrt{ (14-0)^2+(11-9)^2 } = \sqrt{ 196 + 4 } = \sqrt{200} = 10\sqrt{2}$

$CA = \sqrt{ (8-14)^2+(3-11)^2 } = \sqrt{ 36 + 64 } = \sqrt{100} = 10$

$AB^2 + CA^2 = 100 + 100 = 200 = BC^2$

$BC^2 = AB^2 + CA^2 \Rightarrow$ the triangle is right-angled triangle.

and, $AB = CA \Rightarrow$ the triangle is isosceles.

Hence, the $\triangle ABC$ is an isosceles right-angled triangle.

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Question 8: Show that the quadrilateral ABCD with A (3, 1), B (0, -2), C (1, 1) and D (4, 4) is a parallelogram.

Answer:

$AB = \sqrt{ (0-3)^2+(-2-1)^2 } = \sqrt{ 9 + 9 } = \sqrt{18} = 3\sqrt{2}$

$BC = \sqrt{ (1-0)^2+(1+2)^2 } = \sqrt{ 1 + 9 } = \sqrt{10}$

$CD = \sqrt{ (4-1)^2+(4-1)^2 } = \sqrt{ 9 + 9 } = \sqrt{18} = 3\sqrt{2}$

$DA = \sqrt{ (3-4)^2+(1-4)^2 } = \sqrt{ 1 + 9 } = \sqrt{10}$

Since the opposite sides of the quadrilateral ABCD are equal, it is a parallelogram.

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Question 9: Find the area of a circle, whose center is (5, -3) and which passes through the point ( -7, 2). Take $\pi = 3.14$

Answer:

The radius $(r)$ of the circle = distance between the points $(5, -3)$ and $(-7, 2)$

$=\sqrt{ (-7-5)^2+(2+3)^2 } = \sqrt{ 144 + 25 } = \sqrt{169} = 13$

Therefore Area of the circle $= \pi r^2 = 3.14 \times 13^2 = 520.66$ sq. units

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Question 10: Find the points on the x-axis whose distances from the points A(7, 6) and B(-3, 4) are in the ratio 1 : 2.

Answer:

Let the required point $P$ on x-axis $= ( x, 0)$

$\displaystyle \text{Given } \frac{PA}{OB} = \frac{1}{2} \Rightarrow 2PA = PB$

$\displaystyle \Rightarrow 2\sqrt{ (x-7)^2+(0-6)^2 } = \sqrt{ (x+3)^2+(0-4)^2 }$

$\displaystyle \Rightarrow 4( x^2 - 14x + 49 + 36) = x^2 + 6x + 9 + 16$

$\displaystyle \Rightarrow 4x^2 - 56x + 196 + 144 = x^2 + 6x + 2$

$\displaystyle \Rightarrow 3x^2 - 62x + 315 = 0$

$\displaystyle \Rightarrow 3x^2 - 27x - 35x + 315 = 0$

$\displaystyle \Rightarrow (x-9)(3x-35) = 0$

$\displaystyle \Rightarrow x = 9 \text{ or } x=\frac{35}{3}$

$\displaystyle \text{Therefore required points on x-axis are } (9,0) \text{ and } (\frac{35}{3}, 0)$

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Question 11: Point $P(x, y)$ is equidistant from the points A(-2, 0) and B(3, – 4). prove that : $10x-8y=21$ .

Answer:

$\displaystyle \Rightarrow \text{Given: } PA = PB$

$\displaystyle \Rightarrow \sqrt{ (x+2)^2+(y-0)^2 } = \sqrt{ (x-3)^2+(y+4)^2 }$

$\displaystyle \Rightarrow x^2 + 4x + 4 + y^2 = x^2 - 6x + 9 + y^2 + 8y + 1$

$\displaystyle \Rightarrow 10x - 8y = 2$

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Question 12: Find the co-ordinates of the circumcenter of the triangle ABC; whose vertices A, B and C are (4,6), (0,4) and (6, 2) respectively.

Answer:

Let the circumcenter be $P (x, y)$

Therefore $PA = PB$

$\Rightarrow \sqrt{ (x-4)^2+(y-6)^2 } = \sqrt{ (x-0)^2+(y-4)^2 }$

$\Rightarrow x^2 - 8x + 16 + y^2 - 12 y + 36 = x^2 + y^2 - 8y + 16$

$\Rightarrow - 8x - 4y = - 36$

$\Rightarrow 2x+y = 9 \text{ ... ... ... ... ... i) }$

and $PA = PC$

$\Rightarrow \sqrt{ (x-4)^2+(y-6)^2 } = \sqrt{ (x-6)^2+(y-2)^2 }$

$\Rightarrow x^2 - 8x + 16 + y^2 - 12y + 36 = x^2 - 12 x + 36 + y^2 - 4y + 4$

$\Rightarrow 4x-8y = - 12$

$\Rightarrow x-2y = - 3\text{ ... ... ... ... ... ii) }$

On solving i) and ii), we get : $x = 3$ and $y - 3$

Therefore the circumcenter of the given triangle $= (3, 3)$

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Question 13: Find the co-ordinates of point P which divides the join of A (4, -5) and B(6, 3) in the ratio 2 : 5.

Answer: x13 1

Let the co-ordinates of $P$ be $(x, y)$

$\displaystyle \therefore x = \frac{m_1x_2+m_2x_1}{m_1+m_2} = \frac{2 \times 6 + 5 \times 4 }{2+5} = \frac{32}{7}$

$\displaystyle \text{and } y = \frac{m_1y_2+m_2y_1}{m_1+m_2} = \frac{2 \times 3 + 5 \times (-5) }{2+5} = \frac{-19}{7}$

$\displaystyle \therefore P = \Big(\frac{32}{7}, \frac{-19}{7} \Big)$

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Question 14: Find the ratio in which the point (5, 4) divides the line joining points (2, 1) and (7, 6)

Answer:

Let the required ratio be $m_1: m_2$

Take $(2, 1) = (x_1, y_1); (7, 6) = (x_2, y_2)$ and $(5, 4) = (x, y)$

$\displaystyle x = \frac{m_1x_2+m_2x_1}{m_1+m_2}$

$\displaystyle \Rightarrow 5 = \frac{m_1 \times 7+m_2 \times 2}{m_1+m_2}$

$\displaystyle \Rightarrow 5 m_1 + 5m_2 = 7m_1 + 2 m_2$

$\displaystyle \Rightarrow 2m_1 = 3m_2$

$\displaystyle \Rightarrow \frac{m_1}{m_2} = \frac{3}{2}$

Therefore the required ratio is $3:2$

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Question 15: In what ratio is the line joining the points (4, 2) and (3, -5) 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 on the x-axis be $( x, 0)$

$\displaystyle \text{Taking } (4,2) = (x_1, y_1) \text{ and } ( 3, -5) = (x_2, y_2)$

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

$\displaystyle \Rightarrow 0 = \frac{k \times (-5) + 2}{k+1}$

$\displaystyle \Rightarrow 0 = -5k+ 2$

$\displaystyle \Rightarrow k = \frac{2}{5}$

$\displaystyle \Rightarrow m_1 : m+2 = 2: 5$

$\displaystyle \text{Therefore, } x = \frac{2 \times 3 + 5 \times 4}{2 + 5} = \frac{26}{7}$

$\displaystyle \text{Therefore the ratio } = 2 : 5 \text{ and the required point of intersection } = \Big( \frac{26}{7}, 0 \Big)$

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Question 16: Calculate the ratio in which the line joining the points (4, 6) and (-5, -4) is divided by the line y = 3. Also, find the co-ordinates of the point of intersection.

Answer:

$\displaystyle \text{The co-ordinates of every point on the line } y = 3 \text{ will be of the type } (x, 3).$

$\displaystyle \text{Now, } y = \frac{m_1y_2+m_2y_1}{m_1+m_2}$

$\displaystyle \Rightarrow 3 = \frac{m_1(-4)+m_2(6)}{m_1+m_2}$

$\displaystyle \Rightarrow 3m_1 + 3m_2 = - 4 m_1 + 6m_2$

$\displaystyle \Rightarrow 7m_1 = 3m_2$

$\displaystyle \Rightarrow \frac{m_1}{m_2} = \frac{3}{7}$

$\displaystyle \text{Therefore the required ratio is } 3: 7$

$\displaystyle \text{Now, } x = \frac{m_1x_2+m_2x_1}{m_1+m_2}$

$\displaystyle \Rightarrow x = \frac{3 \times ( -5) + 7 \times 4}{3+7} = \frac{13}{10}$

$\displaystyle \text{Therefore the required point of intersection } = \Big( \frac{13}{10}, 3 \Big)$

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Question 17: The origin O, B (-6, 9) and C (12, -3) are vertices of triangle OBC. Point P divides OB in the ratio 1 : 2 and point Q divides OC in the ratio 1 : 2. Find the co-ordinates of points P and Q. Also, show that $PQ = \frac{1}{3} BC.$

Answer:

$\displaystyle \text{For point } P: m_1 : m_2 = 1:2, (x_1, y_1) = ( 0, 0) \text{ and } ( x_2, y_2) = ( -6, 9)$

$\displaystyle \therefore P = \Big( \frac{m_1x_2+m_2x_1}{m_1+m_2}, \frac{m_1y_2+m_2y_1}{m_1+m_2} \Big)$

$\displaystyle = \Big( \frac{1 \times ( -6) + 2 \times 0}{1+2}, \frac{1\times 9 + 2 \times 0}{1+2} \Big)$

$\displaystyle = ( -2, 3)$

$\displaystyle \text{For point } Q: m_1 : m_2 = 1:2, (x_1, y_1) = ( 0, 0) \text{ and } ( x_2, y_2) = ( 12, -3)$

$\displaystyle \therefore Q = \Big( \frac{m_1x_2+m_2x_1}{m_1+m_2}, \frac{m_1y_2+m_2y_1}{m_1+m_2} \Big)$

$\displaystyle = \Big( \frac{1 \times12 + 2 \times 0}{1+2}, \frac{1\times (-3) + 2 \times 0}{1+2} \Big)$

$\displaystyle = ( 4, -1)$

$\displaystyle \text{Now } PQ = \text{ Distance between } P (-2,3) \text{ and } Q (4, -1)$

$\displaystyle = \sqrt{(4+2)^2 + ( -1 - 3)^2} = \sqrt{36+16} = \sqrt{52} = 2\sqrt{13}$

$\displaystyle \text{and } BC = \sqrt{(12+6)^2 + ( -3-9)^2} = \sqrt{324+144} = \sqrt{468} = 6\sqrt{13}$

$\displaystyle PQ = 2\sqrt{13} \text{ and } BC = 6\sqrt{13}$

$\displaystyle \Rightarrow PQ = \frac{1}{3} BC$

$\\$

Question 18: Find the coordinates of the points of trisection of the line segment joining the points A (6, -2) and B (-8, 10).

Answer:

$\displaystyle \text{Let } P \text{ and } Q \text{ be the points of trisection so that } AP = PQ = QB.$

For P:

$\displaystyle m_1:m_2 = AP : PB = 1 : 2; (x_1, y_1) = ( 6, -2) \text{ and } (x_2, y_2) = (-8, 10)$

$\displaystyle \therefore x = \frac{m_1x_2+m_2x_1}{m_1+m_2} = \frac{1 \times (-8) + 2 \times 6 }{1+2} = \frac{4}{3}$

$\displaystyle \text{and } y = \frac{m_1y_2+m_2y_1}{m_1+m_2} = \frac{1 \times 10 + 2 \times (-2) }{1+2} = 2$

$\displaystyle \therefore \text{Point } P = \Big( \frac{4}{3}, 2 \Big)$

For Q:

$\displaystyle m_1:m_2 = AQ:QB = 2 : 21; (x_1, y_1) = ( 6, -2) \text{ and } (x_2, y_2) = (-8, 10)$

$\displaystyle \therefore x = \frac{m_1x_2+m_2x_1}{m_1+m_2} = \frac{2 \times (-8) + 1 \times 6 }{2+1} = \frac{-10}{3}$

$\displaystyle \text{and } y = \frac{m_1y_2+m_2y_1}{m_1+m_2} = \frac{2 \times 10 + 1 \times (-2) }{2+1} = 6$

$\displaystyle \therefore \text{Point } P = \Big( \frac{-10}{3}, 6 \Big)$

$\\$

Question 19: Show that $P (3, m -5)$ is a point of trisection of the line segment joining points A (4, -2) and B (1, 4). Hence, find the value of $m$ .

Answer:

$P$ will be a point of trisection of $AB$ if it divides $AB$ in the ratio $1:2$ or $2:1$

$\displaystyle \text{Since } x = \frac{m_1x_2+m_2x_1}{m_1+m_2}$

$\displaystyle \Rightarrow 3 = \frac{m_1 \times 1+m_2 \times 4}{m_1+m_2}$

$\displaystyle \Rightarrow 3m_1+3m_2 = m_1 + 4m_2$

$\displaystyle \Rightarrow 2m_1 = m_2$

$\displaystyle \Rightarrow \frac{m_1}{m_2} = \frac{1}{2}$

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

Hence $P$ is a point of trisection of $AB.$

$\displaystyle \text{Now, } y = \frac{m_1y_2+m_2y_1}{m_1+m_2}$

$\displaystyle \Rightarrow m-5 = \frac{1 \times 4 + 2 \times (-2) }{1+2}$

$\displaystyle \Rightarrow m = 5$

$\\$

Question 20: Find the co-ordinates of the mid-point of the line segment joining the points P(4, -6) and Q(-2, 4).

Answer:

$\displaystyle \text{Mid point }= \Big( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \Big) = \Big( \frac{4-2}{2}, \frac{-6+4}{2} \Big) = ( 1, -1)$

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Question 21: The mid-point of line segment AB (shown in the diagram) is (-3, 5). Find the co-ordinates of A and B. x21

Answer:

Since, point $A$ lies on the x-axis; let $A = (x, 0)$

Since, point $B$ lies on the y-axis; let $B = (0, y)$

$\displaystyle \text{Mid point of AB }= \Big( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \Big) = \Big( \frac{x+0}{2}, \frac{0+y}{2} \Big) = ( -3, 5)$

$\displaystyle \Rightarrow \frac{x}{2} = -3 \Rightarrow x = -6$

$\displaystyle \Rightarrow \frac{y}{2} = 5 \Rightarrow y = 10$

Co-ordinates of $A = (-6, 0 )$ and co-ordinates of $B = (0, 10)$

$\\$

Question 22: A(14, -2), B(6, -2) and D (8, 2) are the three vertices of a parallelogram ABCD. Find the coordinates of the fourth vertex C.

Answer: x22

Let $C=(x, y)$

Since the diagonals of a parallelogram bisect each other;

Mid-point of AC = mid-point of BD

$\displaystyle \Rightarrow \Big( \frac{14+x}{2}, \frac{-2+y}{2} \Big) = \Big( \frac{8+6}{2}, \frac{2+(-2)}{2} \Big) = ( 7, 0)$

$\displaystyle \Rightarrow \frac{14+x}{x} = 7 \Rightarrow x = 0$

$\displaystyle \Rightarrow \frac{-2+y}{2} = 0 \Rightarrow y = 2$

Therefore the vertex is $C = ( 0, 2)$

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Question 23: In triangle ABC, P (-2,5) is mid-point of AB, Q (2, 4) is mid-point of BC and R (-1, 2) is mid-point of AC. Calculate the co-ordinates of vertices A, B and C.

Answer:

$\displaystyle \text{Let } A = ( x_1, y_1), B = ( x_2, y_2), \text{ and } C = ( x_3, y_3)$

$\displaystyle \text{Since P is mid-point of AB}$

$\displaystyle \Rightarrow \frac{x_1+x_2}{2} = -2 \text{ and } \frac{y_1+y_2}{2} = 5$

$\displaystyle \Rightarrow x_1+x_2 = - 4 \text{ ... ... ... i)} \text{ and } y_1+y_2 = 10\text{ ... ... ... ii)}$

$\displaystyle \text{Since Q is mid-point of BC}$

$\displaystyle \Rightarrow \frac{x_2+x_3}{2} = 2 \text{ and } \frac{y_2+y_3}{2} =4$

$\displaystyle \Rightarrow x_2+x_3 = 4 \text{ ... ... ... iii)} \text{ and } y_2+y_3 = 8 \text{ ... ... ... iv}$

$\displaystyle \text{Since R is mid-point of AC}$

$\displaystyle \Rightarrow \frac{x_1+x_3}{2} = -1 \text{ ... ... ... v)} \text{ and } \frac{y_1+y_3}{2} = 2 \text{ ... ... ... vi)}$

$\displaystyle \Rightarrow x_1+x_3 = - 2 \text{ and } y_1+y_3 = 4$

Adding equations i) , iii) and v) we get

$\displaystyle x_1 + x_2 + x_2 + x_ 3+ x_1 + x_ 3 = - 4 +4-2$

$\displaystyle \Rightarrow 2( x_1 + x_ 2 + x_ 3) = - 2$

$\displaystyle \Rightarrow x_1 + x_ 2 + x_ 3 = -1\text{ ... ... ... vii)}$

Subtracting equation i) from vii) we get $\displaystyle : x_3 = -1 + 4 = 3$

Subtracting equation iii) from vii) we get $\displaystyle : x_1 = -1 - 4 = -5$

Subtracting equation v) from vii) we get $\displaystyle : x_2 = -1 + 2 = 1$

Adding equations ii) , iv) and vi) we get

$\displaystyle y_1 + y_2 + y_2 + y_ 3+ y_1 + y_ 3 = 10+8+4$

$\displaystyle \Rightarrow 2( y_1 + y_ 2 + y_ 3) = 22$

$\displaystyle \Rightarrow y_1 + y_ 2 + y_ 3 = 11\text{ ... ... ... viii)}$

Subtracting equation ii) from viii) we get $\displaystyle : y_3 =11-10 = 1$

Subtracting equation iv) from viii) we get $\displaystyle : y_1 = 11-8 = 3$

Subtracting equation vi) from viii) we get $\displaystyle : y_2 = 11-4=7$

$\displaystyle \therefore A = ( x_1, y_1) = ( -5, 3), B = ( x_2, y_2) = (1, 7) ,\text{ and } C = ( x_3, y_3) = ( 3, 1)$

$\\$

Question 24: The mid-point of the line segment joining (3m, 6) and (- 4,3n) is (1,2m -1). Find the values of m and n. [ICSE Board 2006]

Answer: x24

According to the adjoining figure, we have :

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

$\displaystyle \frac{6+3n}{2} = 2m - 1 \\ \\ \Rightarrow 6 + 3n = 4m - 2 \Rightarrow 3n = 4m - 8 \Rightarrow 3n = 8-8 \Rightarrow n =0$

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Question 25: Find the co-ordinates of the point of intersection of the medians of triangle ABC; given A = (-2,3), B = (6, 7) and C = (4, 1). x25

Answer:

Let $D$ be the mid-point of $BC.$

$\displaystyle \therefore D = \Big( \frac{6+4}{2}, \frac{7+1}{2} \Big) = ( 5, 4)$

If $G$ is the point of intersection of medians (centroid), it divides the median $AD$ in the ratio $2 : 1$

$\displaystyle \therefore G = \Big( \frac{2 \times 5 + 1 \times ( -2)}{2+1}, \frac{2 \times 4 + 1 \times 3}{2+1} \Big) = \Big( \frac{8}{3}, \frac{11}{3} \Big)$

Alternate way:

$\displaystyle \text{For vertices } A (x_1, y_1), B(x_2, y_2) \text{ and } C( x_3, y_3) \text{ of } \triangle ABC, \\ \\ \text{ its centroid } = \Big( \frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3} \Big)$

$\displaystyle \text{Therefore the coordinates of the centroid will be } \\ \\ = \Big( \frac{-2+6+4}{3}, \frac{3+7+1}{3} \Big) = \Big( \frac{8}{3}, \frac{11}{3} \Big)$

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Question 26: 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 side BC. [ICSE Board 2011]

Answer:

Since, $G$ is the centroid of $\triangle ABC$