Class 11: The Straight Line – Exercise 23.5 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Note: Equation of a like is also defined as $\displaystyle y - y_1 = m ( x-x_1)$

Question 1: Find the equation of the straight through the following pair of points:

i) $\displaystyle (0, 0) \text{ and } (2,-2)$ ii) $\displaystyle (a, b) \text{ and } (a + c \sin \alpha, b + c \cos \alpha)$

iii) $\displaystyle (0, -a) \text{ and } ( b, 0)$ iv) $\displaystyle (a, b) \text{ and } (a + b , a -b)$

v) $\displaystyle ( a t_1, a/t_1) \text{ and } ( a t_2, a/t_2)$ vi) $\displaystyle (a \cos \alpha, a\sin \alpha) \text{ and } (a \cos \beta, a\sin \beta)$

Answer:

i) Given points $\displaystyle A(0, 0)$ and $\displaystyle B(2,-2)$

$\displaystyle \text{ Slope of } AB = m = \frac{-2-0}{2-0} = \frac{-2}{2} = 1$

Substituting in $\displaystyle y - y_1 = m ( x-x_1)$ we get the equation of $\displaystyle AB$ as:

$\displaystyle y - 0 = -1(x-0) \hspace{0.5cm} \Rightarrow y + x = 0$

ii) Given points $\displaystyle A(a, b)$ and $\displaystyle B(a + c \sin \alpha, b + c \cos \alpha)$

$\displaystyle \text{ Slope of } AB = m = \frac{b+c \cos \alpha - b}{a+c \sin \alpha - a} = \frac{c \cos \alpha}{c \sin \alpha} = \cot \alpha$

Substituting in $\displaystyle y - y_1 = m ( x-x_1)$ we get the equation of $\displaystyle AB$ as:

$\displaystyle y - b = \cot \alpha (x-a) \hspace{0.5cm} \Rightarrow y - x \cot \alpha + a - b = 0$

iii) Given points $\displaystyle A(0, -a)$ and $\displaystyle B( b, 0)$

$\displaystyle \text{ Slope of } AB = m = \frac{0-(-a)}{b-0} = \frac{a}{b}$

Substituting in $\displaystyle y - y_1 = m ( x-x_1)$ we get the equation of $\displaystyle AB$ as:

$\displaystyle y - (-a) = \frac{a}{b} (x-0)$

$\displaystyle \Rightarrow by + ab = ax$

$\displaystyle \Rightarrow ax-by-ab=0$

iv) Given points $\displaystyle A(a, b)$ and $\displaystyle B(a + b , a -b)$

$\displaystyle \text{ Slope of } AB = m = \frac{a-b-b}{a+b-a} = \frac{a-2b}{b}$

Substituting in $\displaystyle y - y_1 = m ( x-x_1)$ we get the equation of $\displaystyle AB$ as:

$\displaystyle y - b = \frac{a-2b}{b} (x-a)$

$\displaystyle \Rightarrow yb-b^2=(a-2b)x-a(a-2b)$

$\displaystyle \Rightarrow yb-b^2 = (a-2b)x - a^2+2ab$

$\displaystyle \Rightarrow (a-2b)x - yb + b^2 + 2ab - a^2=0$

v) Given points $\displaystyle A( a t_1, a/t_1)$ and $\displaystyle B( a t_2, a/t_2)$

$\displaystyle \text{ Slope of } AB = m = \frac{\frac{a}{t_2} - \frac{a}{t_1}}{at_2 - at_1} = \frac{t_1-t_2}{t_1t_2(t_2-t_1)} = - \frac{1}{t_1t_2}$

Substituting in $\displaystyle y - y_1 = m ( x-x_1)$ we get the equation of $\displaystyle AB$ as:

$\displaystyle y - \frac{a}{t_1} = \frac{-1}{t_1t_2} (x-at_1)$

$\displaystyle \Rightarrow t_1t_2y- at_2 = -x + at_1$

$\displaystyle \Rightarrow x + t_1t_2y-a(t_1+t_2)=0$

vi) Given points $\displaystyle A(a \cos \alpha, a\sin \alpha)$ and $\displaystyle B(a \cos \beta, a\sin \beta)$

$\displaystyle \text{ Slope of } AB = m = \frac{a \sin \beta - a \sin \alpha}{a \cos \beta - a \cos \alpha} = \frac{\sin \beta - \sin \alpha}{ \cos \beta - \cos \alpha}$

Substituting in $\displaystyle y - y_1 = m ( x-x_1)$ we get the equation of $\displaystyle AB$ as:

$\displaystyle y - a \sin \alpha = \frac{\sin \beta - \sin \alpha}{ \cos \beta - \cos \alpha} (x-a \cos \alpha)$

$\displaystyle \Rightarrow y ( \cos \beta - \cos \alpha) - a \sin \alpha ( \cos \beta - \cos \alpha) = x ( \sin \beta - \sin \alpha) - a \cos \alpha ( \sin \beta - \sin \alpha)$

$\displaystyle \Rightarrow y ( \cos \beta - \cos \alpha) - x ( \sin \beta - \sin \alpha) = a \sin \alpha ( \cos \beta - \cos \alpha) - a \cos \alpha ( \sin \beta - \sin \alpha)$

$\displaystyle \Rightarrow y . 2 \sin \Big(\frac{\beta + \alpha}{2} \Big) \sin \Big(\frac{ \alpha - \beta}{2} \Big) - x . 2 \sin \Big(\frac{\beta - \alpha}{2} \Big) \cos \Big(\frac{ \beta + \alpha }{2} \Big) = a ( \sin \alpha \cos \beta - \cos \alpha \sin \beta)$

$\displaystyle \Rightarrow y . 2 \sin \Big(\frac{\beta + \alpha}{2} \Big) \sin \Big(\frac{ \alpha - \beta}{2} \Big) - x . 2 \sin \Big(\frac{\beta - \alpha}{2} \Big) \cos \Big(\frac{ \beta + \alpha }{2} \Big) = 2a \sin \Big(\frac{\alpha-\beta}{2} \Big) \cos \Big(\frac{ \alpha - \beta}{2} \Big)$

$\displaystyle \Rightarrow y \sin \Big(\frac{\beta + \alpha}{2} \Big)+ x \cos \Big(\frac{ \beta + \alpha }{2} \Big) = a \cos \Big(\frac{ \alpha - \beta}{2} \Big)$

$\displaystyle \\$

Question 2: Find the equation to the sides of the triangles the coordinates of whose angular points are respectively: (i) $\displaystyle (1, 4),(2, -3)$ and $\displaystyle (-1, -2)$ ii) $\displaystyle (0, 1), (2, 0)$ and $\displaystyle ( -1, -2)$

Answer:

i) Given points $\displaystyle A(1, 4), B(2, -3)$ and $\displaystyle C(-1, -2)$

$\displaystyle \text{ Slope of } AB = m_1 = \frac{-3-4}{2-1} = \frac{-7}{1} = -7$

Therefore equation of $\displaystyle AB$ :

$\displaystyle y - 4 = - 7 ( x-1) \hspace{0.5cm} \Rightarrow 7x + y - 11 = 0$

$\displaystyle \text{ Slope of } BC = m_2 = \frac{-2-(-3)}{-1-2} = \frac{1}{-3} = \frac{-1}{3}$

Therefore equation of $\displaystyle BC$ :

$\displaystyle y - (-3) = \frac{-1}{3} ( x-2) \hspace{0.5cm} \Rightarrow 3y + 9 = - x + 2 \hspace{0.5cm} \Rightarrow x + 3y + 7 = 0$

$\displaystyle \text{ Slope of } CA = m_3 = \frac{4-(-2)}{1-(-1)} = \frac{6}{2} = 3$

Therefore equation of $\displaystyle CA$ :

$\displaystyle y - 4 = 3 ( x-1) \hspace{0.5cm} \Rightarrow 3x-y+1 = 0$

ii) Given points $\displaystyle A(0,1), B(2,0)$ and $\displaystyle C(-1, -2)$

$\displaystyle \text{ Slope of } AB = m_1 = \frac{0-1}{2-0} = \frac{-1}{2}$

Therefore equation of $\displaystyle AB$ :

$\displaystyle y - 1 = \frac{-1}{2} ( x-0) \hspace{0.5cm} \Rightarrow x+2y-2 = 0$

$\displaystyle \text{ Slope of } BC = m_2 = \frac{-2-0}{-1-2} = \frac{-2}{-3} = \frac{2}{3}$

Therefore equation of $\displaystyle BC$ :

$\displaystyle y - 0 = \frac{2}{3} ( x-2) \hspace{0.5cm} \Rightarrow 2x-3y-4=0$

$\displaystyle \text{ Slope of } CA = m_3 = \frac{1-(-2)}{0-(-1)} = \frac{3}{1} = 3$

Therefore equation of $\displaystyle CA$ :

$\displaystyle y - 1 = 3 ( x-0) \hspace{0.5cm} \Rightarrow 3x-y+1 = 0$

$\displaystyle \\$

Question 3: Find the equations of the medians of a triangle, the the coordinates of the vertices are $\displaystyle (-1, 6), (- 3, - 9)$ and $\displaystyle (5, -8)$ .

Answer:

Given points $\displaystyle A(-1,6), B(-3,-9)$ and $\displaystyle C(5,-8)$ . Please refer to the figure shown.

Let $\displaystyle D, E \ \& \ F$ be the mid point of $\displaystyle AB, BC$ and $\displaystyle CA$ respectively. Therefore,

$\displaystyle D = \Big( \frac{-3-1}{2} , \frac{-9+6}{2} \Big ) = \Big( -2, \frac{-3}{2} \Big )$

$\displaystyle E = \Big( \frac{5-3}{2} , \frac{-8-9}{2} \Big ) = \Big( 1, \frac{-17}{2} \Big )$

$\displaystyle F = \Big( \frac{-1+5}{2} , \frac{6-8}{2} \Big ) = ( 2, -1)$

$\displaystyle \text{ Slope of } AE = m_1 = \frac{\frac{-17}{2}-6}{1-(-1)} = \frac{-29}{4}$

Therefore equation of $\displaystyle AE$ :

$\displaystyle y - 6 = \frac{-29}{4} ( x-(-1)) \hspace{0.5cm} \Rightarrow 4y-24=-29x-29 \hspace{0.5cm} \Rightarrow 29x+4y+5 = 0$

$\displaystyle \text{ Slope of } BF = m_2 = \frac{-1-(-9)}{2-(-3)} = \frac{8}{5}$

Therefore equation of $\displaystyle BF$ :

$\displaystyle y - (-9) = \frac{8}{5} ( x-(-3)) \hspace{0.5cm} \Rightarrow 5y+45 = 8x+ 24 \hspace{0.5cm} \Rightarrow 8x-5y-21=0$

$\displaystyle \text{ Slope of } CD = m_3 = \frac{\frac{-3}{2} - (-8)}{-2-5} = \frac{-13}{14}$

Therefore equation of $\displaystyle CD$ :

$\displaystyle y - (-8) = \frac{-13}{14} ( x-5) \hspace{0.5cm} \Rightarrow 14y + 112 = - 13x +65 \hspace{0.5cm} \Rightarrow 13x + 14y + 47 = 0$

$\displaystyle \\$

Question 4: Find the equations to the diagonals of the rectangle the equations of whose sides are $\displaystyle x = a, x = a', y = b$ and $\displaystyle y = b'$ .

Answer:

Please refer to the figure shown. Therefore the four points are $\displaystyle A(a, b), B ( a', b), C( a', b'), D( a , b')$

$\displaystyle \text{ Slope of } AC = m_3 = \frac{b'-b}{a'-a}$

Therefore equation of $\displaystyle AC$ :

$\displaystyle y - b= \frac{b'-b}{a'-a} ( x-a)$

$\displaystyle \Rightarrow (a'-a)y-a'b+ab=(b'-b)x-ab'+ab$

$\displaystyle \Rightarrow (b'-b)x-(a'-a)y - ab'+a'b = 0$

$\displaystyle \Rightarrow (a'-a)y - ( b'-b)x = ba'-ab'$

$\displaystyle \text{ Slope of } BD = m_3 = \frac{b'-b}{a-a'}$

Therefore equation of $\displaystyle BD$ :

$\displaystyle y - b= \frac{b'-b}{a-a'} ( x-a')$

$\displaystyle \Rightarrow (a-a')y - ab + ba' = ( b'-b) x - a'b' + a'b$

$\displaystyle \Rightarrow (b-b')x - ( a - a') y + ab - a'b'=0$

$\displaystyle \Rightarrow ( a - a') y + (b-b')x = a'b' - ab$

$\displaystyle \\$

Question 5: Find the equation of the side $\displaystyle BC$ of the $\displaystyle \triangle ABC$ whose vertices are $\displaystyle A(- 1, - 2), B (0, 1)$ and $\displaystyle C (2,0)$ respectively. Also, find the equation of the median through $\displaystyle A ( -1, -2)$ .

Answer:

Given points $\displaystyle A(-1,-2), B(0,1)$ and $\displaystyle C(2,0)$

$\displaystyle \text{ Slope of } BC = m_1 = \frac{0-1}{2-0} = \frac{-1}{2}$

Therefore equation of $\displaystyle BC$ :

$\displaystyle y - 1 = \frac{-1}{2} ( x-0) \hspace{0.5cm} \Rightarrow 2y-2=-x \hspace{0.5cm} \Rightarrow x+ 2y - 2 = 0$

$\displaystyle \text{ Mid point D of } BC = \Big( \frac{0+2}{2} , \frac{1+0}{2} \Big)= \Big( 1, \frac{1}{2} \Big)$

$\displaystyle \text{ Slope of } AD = m_2 = \frac{\frac{1}{2} -(-2)}{1-(-1)} = \frac{5}{4}$

Therefore equation of $\displaystyle AD$ :

$\displaystyle y - (-2) = \frac{5}{4} ( x-(-1))$

$\displaystyle \Rightarrow y+2 = \frac{5}{4} ( x+1) \Rightarrow 4y+ 8 = 5x + 5 \Rightarrow 5x - 4y - 3$

$\displaystyle \\$

Question 6: Using the concept of the equation of a line, prove that the three points $\displaystyle (-2, -2) (8, 2)$ and $\displaystyle ( 3, 0)$ are collinear.

Answer:

Given points $\displaystyle A(-2,-2), B(8,2)$ and $\displaystyle C(3,0)$

$\displaystyle \text{ Slope of } AC = m_1 = \frac{0-(-2)}{3-(-2)} = \frac{2}{5}$

Therefore equation of $\displaystyle AC$ :

$\displaystyle y - (-2) = \frac{2}{5} ( x-(-2))$

$\displaystyle \Rightarrow y+2 = \frac{2}{5} (x+2) \hspace{0.5cm} \Rightarrow 5y+10 = 2x + 4 \hspace{0.5cm} \Rightarrow 2x - 5y - 6 = 0$

Now check if $\displaystyle B( 8, 2)$ satisfy the equation

$\displaystyle \therefore 2 ( 8) - 5(2) - 6 = 0$

$\displaystyle \Rightarrow 16-10-6 = 0$

Therefore $\displaystyle A, B, C$ are collinear.

$\displaystyle \\$

Question 7: Prove that the line $\displaystyle y -x+2=0$ , divides the join of points $\displaystyle (3,-1)$ and $\displaystyle (8,9)$ in the ratio $\displaystyle 2:3$ .

Answer:

Let $\displaystyle y -x+2=0$ , divides the join of points $\displaystyle (3,-1)$ and $\displaystyle (8,9)$ at a point P in the ratio of $\displaystyle k:1$

$\displaystyle \therefore P = \Big( \frac{3+8k}{k+1} , \frac{-1+9k}{k+1} \Big)$

Since $\displaystyle P$ lies on $\displaystyle y - x+2 = 0$ , it should satisfy the equation.

$\displaystyle \therefore \frac{-1+9k}{k+1} - \frac{3+8k}{k+1} + 2 = 0$

$\displaystyle \Rightarrow -1 + 9k - 3 - 8 k + 2 k + 2 = 0$

$\displaystyle \Rightarrow 3k - 2 = 0$

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

Hence the line $\displaystyle y -x+2=0$ , divides the join of points $\displaystyle (3,-1)$ and $\displaystyle (8,9)$ in the ratio $\displaystyle 2:3$ .

$\displaystyle \\$

Question 8: Find the equation to the straight line which bisects the distance between the points $\displaystyle (a, b), (a', b')$ and also bisects the distance between the points $\displaystyle ( -a, b)$ and $\displaystyle (a', -b')$ .

Answer:

Given points $\displaystyle A(a, b), B(a', b')$ and $\displaystyle C( -a, b)$ and $\displaystyle D(a', -b')$

$\displaystyle \text{ Mid point P of } AB = \Big( \frac{a+a'}{2} , \frac{b+b'}{2} \Big)$

$\displaystyle \text{ Mid point Q of } CD = \Big( \frac{-a+a'}{2} , \frac{b-b'}{2} \Big)$

$\displaystyle \text{ Slope of } PQ = m_1 = \frac{\frac{b-b'}{2} - \frac{b+b'}{2} }{\frac{-a+a'}{2} - \frac{a+a'}{2}} = \frac{b-b'-b-b'}{-a+a'-a-a'} = \frac{-2b'}{-2a} = \frac{b'}{a}$

Therefore equation of $\displaystyle PQ$ :

$\displaystyle y - \frac{b+b'}{2} = \frac{b'}{a} \Big( x- \frac{a+a'}{2} \Big)$

$\displaystyle \Rightarrow 2y - b - b' = \frac{b'}{a} ( 2x - a - a')$

$\displaystyle \Rightarrow 2ay - ab - ab' = 2b'x-ab'-a'b'$

$\displaystyle \Rightarrow 2b'x-2ay+ab-a'b'=0$

$\displaystyle \Rightarrow 2ay - 2b'x = ab - a'b'$

$\displaystyle \\$

Question 9: In what ratio is the line joining the points $\displaystyle (2, 3)$ and $\displaystyle (4, -5)$ divided by the line passing through the points $\displaystyle (6,8)$ and $\displaystyle (-3, -2)$ .

Answer:

Given points: $\displaystyle (6,8)$ and $\displaystyle (-3, -2)$

$\displaystyle \text{ Slope of } AB = \frac{-2-8}{-3-6} = \frac{-10}{-9} = \frac{10}{9}$

Therefore equation of $\displaystyle AB$ :

$\displaystyle y - 8 = \frac{10}{9} ( x - 6) \hspace{0.5cm} \Rightarrow 9y - 72 = 10 x - 60 \hspace{0.5cm} \Rightarrow 10x - 9 y + 12 = 0$

Let $\displaystyle 10x - 9 y + 12 = 0$ divides the line joining $\displaystyle (2, 3)$ and $\displaystyle (4, -5)$ in the ratio of $\displaystyle k:1$

$\displaystyle \therefore$ point of intersection $\displaystyle = \Big( \frac{4k+2}{k+1} , \frac{-5k+3}{k+1} \Big)$

This point is on line $\displaystyle 10x - 9 y + 12 = 0$

$\displaystyle \therefore 10 \Big( \frac{4k+2}{k+1} \Big ) - 9 \Big( \frac{-5k+3}{k+1} \Big) + 12 = 0$

$\displaystyle \Rightarrow 40k+20+45k-27+12k+12=0$

$\displaystyle \Rightarrow 97k + 5 = 0$

$\displaystyle \Rightarrow k = - \frac{5}{97}$

Hence the required ratio is $\displaystyle 5:97$ externally.

$\displaystyle \\$

Question 10: The vertices of a quadrilateral are $\displaystyle A(-2,6),8(1,2),C(10, 4)$ and $\displaystyle D(7, 8)$ . Find the equations of its diagonals.

Answer:

Given points: $\displaystyle A(-2,6), B(1,2) , C(10,4), D(7,8)$

$\displaystyle \text{ Slope of } AC = \frac{4-6}{10-(-2)} = \frac{-2}{12} = \frac{-1}{6}$

Therefore equation of $\displaystyle AC$ :

$\displaystyle y - 6 = \frac{-1}{6} ( x - (-2)) \hspace{0.5cm} \Rightarrow 6y-36=-x-2 \hspace{0.5cm} \Rightarrow x+6y-34=0$

$\displaystyle \text{ Slope of } BD = \frac{8-2}{7-1} = \frac{6}{6} = 1$

Therefore equation of $\displaystyle BD$ :

$\displaystyle y - 2 =$ $\displaystyle 1 ( x - 1) \hspace{0.5cm} \Rightarrow x-y+1=0$

$\displaystyle \\$

Question 11: The length $\displaystyle L$ (in centimeters) of a copper rod is a linear function of its Celsius temperature $\displaystyle C$ . In an experiment, if $\displaystyle L = 124.942$ when $\displaystyle C = 20$ , and $\displaystyle L = 125.134$ when $\displaystyle C = 110$ , express $\displaystyle L$ in terms of $\displaystyle C$ .

Answer:

Given two points $\displaystyle ( 124.942, 20)$ and $\displaystyle (125.134, 110)$

$\displaystyle \text{ Slope of line} = \frac{110-20}{125.134-124.942} = \frac{90}{0.192}$

Therefore the equation of line:

$\displaystyle y - 20 = \frac{90}{0.192} ( x - 124.942)$

$\displaystyle \Rightarrow 0.192 y - 3.84 = 90x - 11244.78$

$\displaystyle \Rightarrow 90x = 0.192y + 11240.78$

$\displaystyle \Rightarrow x = \frac{0.192}{90} y + \frac{11240.78}{90}$

$\displaystyle \Rightarrow x = \frac{4}{1875} y + 124.898$

$\displaystyle \\$

Question 12: The owner of a milk store finds that he can sell $\displaystyle 980$ liters milk each week at ( $\displaystyle 14$ per liter and $\displaystyle 1220$ liters of milk each week at Rs. $\displaystyle 16$ per liter. Assuming a linear relationship between selling price and demand, how many liters can he sell weekly at Rs. $\displaystyle 17$ per liter.

Answer:

Given two points $\displaystyle ( 14, 980)$ and $\displaystyle (16, 1220)$

$\displaystyle \text{ Slope of line} = \frac{1220-980}{16-14} = \frac{240}{2} = 120$

Therefore the equation of line:

$\displaystyle y - 980 = 120(x-14) \hspace{0.5cm} \Rightarrow y = 120x - 700$

When $\displaystyle x = 17, y = 120 \times 17 - 700 = 1340$

Hence the owner of the milk store would be able to sell $\displaystyle 1340$ liters of milk at Rs. $\displaystyle 17$ / liter.

$\displaystyle \\$

Question 13: Find the equation of the bisector of $\displaystyle \angle A$ of the triangle whose vertices are $\displaystyle A (4,3), B(0,0)$ and $\displaystyle C( 2, 3)$ .

Answer:

Let $\displaystyle AD$ be the bisector of $\displaystyle \angle BAC$ .

We know, $\displaystyle \frac{AB}{AC} = \frac{BD}{DC}$

Now $\displaystyle AB = \sqrt{(4-0)^2 + ( 3-0)^2} = \sqrt{25} = 5$

$\displaystyle AC = \sqrt{(4-2)^2 + ( 3-3)^2} = \sqrt{4} = 2$

$\displaystyle \therefore \frac{BD}{DC} = \frac{5}{2}$

Therefore coordinates of $\displaystyle D = \Big( \frac{2 \times 5 + 0 \times 2}{5+2} , \frac{3 \times 5 + 0 \times 2}{5+2} \Big) = \Big( \frac{10}{7} , \frac{15}{7} \Big)$

$\displaystyle \text{ Slope of } AD = \frac{\frac{15}{7}-3}{\frac{10}{7}-4} = \frac{15-21}{10-28} = \frac{-6}{-18} = \frac{1}{3}$

Therefore equation of $\displaystyle AD$ :

$\displaystyle y - 3 = \frac{1}{3} ( x - 4) \hspace{0.5cm} \Rightarrow 3y-9=x-4 \hspace{0.5cm} \Rightarrow x-3y+5=0$

$\displaystyle \\$

Question 14: Find the equations to the straight lines which go through the origin and trisect the portion of the straight line $\displaystyle 3x+ y = 12$ which is intercepted between the axes of coordinates.

Answer:

Given line $\displaystyle 3x+y =12$

x-intercept $\displaystyle = A ( 4, 0)$

y-intercept $\displaystyle = B ( 0, 12$

$\displaystyle C$ divides $\displaystyle BC$ in the ratio of $\displaystyle 2:1$

Coordinates of $\displaystyle C = \Big( \frac{2 \times 4 + 1 \times 0}{2+1} , \frac{2 \times 0 + 1 \times 12}{2+1} \Big) = \Big( \frac{8}{3} , 4 \Big)$

$\displaystyle D$ divides $\displaystyle BC$ in the ratio of $\displaystyle 1:2$

Coordinates of $\displaystyle C = \Big( \frac{2 \times 4 + 1 \times 0}{1+2} , \frac{2 \times 0 + 1 \times 12}{1+2} \Big) = \Big( \frac{8}{3} , 4 \Big)$