CBSE Class 10 Mathematics Coordinate Geometry Worksheet Set 07

Very Short Answer Questions

Question. Find the distance of a point \( p(x, y) \) from the origin. 
Answer: Distance between \( (x, y) \) and \( (0, 0) \).
\( \implies \) \( \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2} \)
\( = \sqrt{(x - 0)^2 + (y - 0)^2} \)
\( = \sqrt{x^2 + y^2} \). The distance is \( \sqrt{x^2 + y^2} \). 

Question. If the distance between the points \( (4, k) \) and \( (1, 0) \) is 5, then what can be the possible values of \( k \)? 
Answer: Using distance formula \( \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \), we have
\( \implies \) \( \sqrt{(4 - 1)^2 + (k - 0)^2} = 5 \)
\( \implies \) \( 9 + k^2 = 25 \)
\( \implies \) \( k^2 = 16 \)
\( \implies \) \( k = \pm 4 \)

Question. Find the coordinates of a point \( A \), where \( AB \) is a diameter of the circle with centre \( (- 2, 2) \) and \( B \) is the point with coordinates \( (3, 4) \). 
Answer: Let co-ordinate of point \( A \) be \( (x, y) \).
\( \therefore -2 = \frac{x + 3}{2} \)
\( \implies \) \( x + 3 = -4 \)
\( \implies \) \( x = -7 \) and \( 2 = \frac{y + 4}{2} \)
\( \implies \) \( y + 4 = 4 \)
\( \implies \) \( y = 0 \) \( \therefore \) Co-ordinate of \( A \) are \( (-7, 0) \).

Question. Find the positive value of \( m \), if the distance between the points \( A(5, -3) \) and \( B(13, m) \) is 10 units. 
Answer: Given, \( A = 5, -3 \), \( B = 13, m \) \( AB = 10 \) units. Using distance formula; \( \sqrt{(13 - 5)^2 + (m + 3)^2} = 10 \) on squaring; \( 8^2 + (m + 3)^2 = 100 \)
\( \implies \) \( (m + 3)^2 = 100 - 64 \)
\( \implies \) \( \sqrt{(m + 3)^2} = \sqrt{36} \)
\( \implies \) \( (m + 3) = \pm 6 \) Considering only positive value; \( m = 6 - 3 \)
\( \implies m = 3 \). [Topper’s Answer 2019]

Question. If the centroid of triangle formed by points \( P(a, b) \), \( Q(b, c) \) and \( R(c, a) \) is at the origin, what is the value of \( a + b + c \)?
Answer: Centroid of \( \Delta PQR = \left( \frac{a + b + c}{3}, \frac{b + c + a}{3} \right) \) Given \( \left( \frac{a + b + c}{3}, \frac{b + c + a}{3} \right) = (0, 0) \)
\( \implies a + b + c = 0 \)

Question. The coordinates of the points \( P \) and \( Q \) are respectively \( (4, -3) \) and \( (-1, 7) \). Find the abscissa of a point \( R \) on the line segment \( PQ \) such that \( \frac{PR}{PQ} = \frac{3}{5} \). 
Answer: \( \frac{PQ}{PR} = \frac{5}{3} \)
\( \implies \) \( \frac{PR + RQ}{PR} = \frac{5}{3} \)
\( \implies \) \( 1 + \frac{RQ}{PR} = \frac{5}{3} \)
\( \implies \) \( \frac{RQ}{PR} = \frac{2}{3} \) i.e., \( R \) divides \( PQ \) in the ratio \( 3 : 2 \). Abscissa of \( R = \frac{3 \times (-1) + 2 \times 4}{3 + 2} = \frac{-3 + 8}{5} = 1 \)

Short Answer Questions-I

Question. Prove that the points \( (3, 0) \), \( (6, 4) \) and \( (-1, 3) \) are the vertices of a right isosceles triangle.
Answer: Let \( A = (3, 0) \); \( B = (6, 4) \) and \( C = (-1, 3) \). Applying distance formula – \( AB = \sqrt{(6 - 3)^2 + (4 - 0)^2} = \sqrt{3^2 + 4^2} = \sqrt{25} = 5 \) unit \( BC = \sqrt{(6 + 1)^2 + (4 - 3)^2} = \sqrt{7^2 + 1^2} = \sqrt{49 + 1} = \sqrt{50} = 5\sqrt{2} \) unit \( AC = \sqrt{(3 + 1)^2 + (0 - 3)^2} = \sqrt{25} = 5 \) unit. Since, \( AB = AC = 5 \) unit \( \Delta ABC \) is isosceles triangle. Also, \( AB^2 + AC^2 \)
\( \implies \) \( 5^2 + 5^2 \) \( 25 + 25 = 50 = BC^2 = (5\sqrt{2})^2 \)
\( \implies AB^2 + AC^2 = BC^2 \) Hence, by converse of Pythagoras Theorem,
\( \implies \Delta ABC \) is right angled triangle. [Topper’s Answer 2016]

Question. If the point \( A(0, 2) \) is equidistant from the points \( B(3, p) \) and \( C(p, 5) \), find \( p \). Also find the length of \( AB \). 
Answer: Given that \( A(0, 2) \) is equidistant from \( B(3, p) \) and \( C(p, 5) \). \( \therefore AB = AC \)
\( \implies AB^2 = AC^2 \)
\( \implies (3 - 0)^2 + (p - 2)^2 = (p - 0)^2 + (5 - 2)^2 \)
\( \implies 3^2 + p^2 + 4 - 4p = p^2 + 9 \)
\( \implies 4 - 4p = 0 \)
\( \implies 4p = 4 \)
\( \implies p = 1 \) Length of \( AB = \sqrt{(3 - 0)^2 + (1 - 2)^2} = \sqrt{3^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10} \) units

Question. If the point \( P(x, y) \) is equidistant from the points \( A(a + b, b - a) \) and \( B(a - b, a + b) \). Prove that \( bx = ay \). 
Answer: Given, \( PA = PB \) or \( (PA)^2 = (PB)^2 \) \( (a + b - x)^2 + (b - a - y)^2 = (a - b - x)^2 + (a + b - y)^2 \)
\( \implies (a + b)^2 + x^2 - 2ax - 2bx + (b - a)^2 + y^2 - 2by + 2ay = (a - b)^2 + x^2 - 2ax + 2bx + (a + b)^2 + y^2 - 2ay - 2by \)
\( \implies 4ay = 4bx \) or \( bx = ay \) Hence proved.

Question. If the distances of \( P(x, y) \) from \( A(5, 1) \) and \( B(-1, 5) \) are equal, then prove that \( 3x = 2y \). 
Answer: \( PA = PB \) \( \therefore PA^2 = PB^2 \) by distance formula, \( (5 - x)^2 + (1 - y)^2 = (-1 - x)^2 + (5 - y)^2 \)
\( \implies (5 - x)^2 + (1 - y)^2 = (1 + x)^2 + (5 - y)^2 \) \( 25 - 10x + x^2 + 1 - 2y + y^2 = 1 + 2x + x^2 + 25 - 10y + y^2 \) \( -10x - 2y = 2x - 10y \) \( 8y = 12x \) \( 4(2y) = 4(3x) \) \( 3x = 2y \) Hence, proved.

Question. Point \( P \) and \( Q \) trisect the line segment joining the points \( A(-2, 0) \) and \( B(0, 8) \), such that \( P \) is nearer to \( A \). Find the co-ordinates of points \( P \) and \( Q \).  
Answer: \( P \) divides \( AB \) in the ratio \( 1 : 2 \) \( \therefore \) Coordinates of \( P \) are \( \left( \frac{0 - 4}{3}, \frac{8 + 0}{3} \right) = \left( \frac{-4}{3}, \frac{8}{3} \right) \) \( Q \) divides \( AB \) in the ratio \( 2 : 1 \) \( \therefore \) Coordinates of \( Q \) are \( \left( \frac{0 - 2}{3}, \frac{16 + 0}{3} \right) = \left( \frac{-2}{3}, \frac{16}{3} \right) \) [CBSE Marking Scheme 2019 (30/3/1)]

Question. If the point \( C(-1, 2) \) divides internally the line segment joining the points \( A(2, 5) \) and \( B(x, y) \) in the ratio of \( 3 : 4 \), find the value of \( x^2 + y^2 \).
Answer: \( C(-1, 2) \) divides \( AB \) in the ratio \( 3 : 4 \). \( \frac{3x + 4(2)}{7} = -1 \)
\( \implies x = -5 \) and \( \frac{3y + 4(5)}{7} = 2 \)
\( \implies y = -2 \) \( \therefore x^2 + y^2 = (-5)^2 + (-2)^2 = 29 \)

Question. Let \( P \) and \( Q \) be the points of trisection of the line segment joining the points \( A(2, -2) \) and \( B(-7, 4) \) such that \( P \) is nearer to \( A \). Find the coordinates of \( P \) and \( Q \). 
Answer: We have – Line \( AB \), joining points \( A(2, -2) \) and \( B(-7, 4) \) \( P \) & \( Q \) are points of trisection, then, \( P \) divides \( AB \) in ratio \( 1 : 2 \) & \( Q \) in ratio \( 2 : 1 \) By section formula – \( x = \frac{mx_2 + nx_1}{m + n} \) & \( y = \frac{my_2 + ny_1}{m + n} \) \( P(x, y) = \frac{1 \times -7 + 2 \times 2}{3} \) and \( y = \frac{1 \times 4 + 2 \times -2}{3} \) \( P(x, y) = \frac{-7 + 4}{3} \) and \( \frac{4 - 4}{3} \) \( P(x, y) = (-1, 0) \) \( Q(x', y') = \frac{2 \times -7 + 1 \times 2}{3} \) and \( y = \frac{2 \times 4 + 1 \times -2}{3} \) \( Q(x', y') = \frac{-14 + 2}{3} \) and \( \frac{8 - 2}{3} \) \( Q(x', y') = (-4, 2) \). [Topper’s Answer 2016]

Question. Find the ratio in which \( P(4, m) \) divides the line segment joining the points \( A(2, 3) \) and \( B(6, -3) \). Hence find \( m \). 
Answer: Let the ratio be \( k : 1 \). By section formula, \( P(4, m) = \left( \frac{6k + 2}{k + 1}, \frac{-3k + 3}{k + 1} \right) \)
\( \implies \frac{6k + 2}{k + 1} = 4 \) \( 6k + 2 = 4k + 4 \) \( 2k = 2 \)
\( k = 1 \). The ratio is \( 1 : 1 \). now, \( m = \frac{-3k + 3}{k + 1} \) \( m = \frac{-3 + 3}{1 + 1} \) \( m = 0 \). Value of \( m \) is 0, the point is \( P(4, 0) \). 

Question. A line intersects the Y-axis and X-axis at the points \( P \) and \( Q \) respectively. If \( (2, -5) \) is the mid-point of \( PQ \), then find the co-ordinates of \( P \) and \( Q \). 
Answer: Let coordinates of \( P \) be \( (0, y) \) and of \( Q \) be \( (x, 0) \). \( (2, -5) \) is mid point of \( PQ \). by section formula, \( (2, -5) = \left( \frac{0 + x}{2}, \frac{y + 0}{2} \right) \) \( 2 = \frac{x}{2} \) and \( -5 = \frac{y}{2} \) \( x = 4 \) and \( y = -10 \). \( \therefore P \) is \( (0, -10) \) and \( Q \) is \( (4, 0) \) [Topper’s Answer 2017]

Short Answer Questions-II

Question. Find the coordinates of a point on the \( x \)-axis which is equidistant from the points \( A(2, -5) \) and \( B(-2, 9) \). 
Answer: Let \( P(x, 0) \) be any point on \( x \)-axis. Now, \( P(x, 0) \) is equidistant from point \( A(2, -5) \) and \( B(-2, 9) \). \( \therefore AP = BP \)
\( \implies \sqrt{(x - 2)^2 + (0 + 5)^2} = \sqrt{(x + 2)^2 + (0 - 9)^2} \) Squaring both sides, we have \( (x - 2)^2 + 25 = (x + 2)^2 + 81 \)
\( \implies x^2 + 4 - 4x + 25 = x^2 + 4 + 4x + 81 \)
\( \implies -8x = 56 \)
\( \therefore x = \frac{56}{-8} = -7 \) The point on the \( x \)-axis equidistant from given points is \( (-7, 0) \).

Question. If the point \( P(k - 1, 2) \) is equidistant from the points \( A(3, k) \) and \( B(k, 5) \), find the values of \( k \).
Answer: Since \( P \) is equidistant from \( A \) and \( B \), \( AP = BP \) or \( AP^2 = PB^2 \)
\( \implies [3 - (k - 1)]^2 + (k - 2)^2 = [k - (k - 1)]^2 + (5 - 2)^2 \)
\( \implies (3 - k + 1)^2 + (k - 2)^2 = (k - k + 1)^2 + (3)^2 \)
\( \implies (4 - k)^2 + (k - 2)^2 = (1)^2 + (3)^2 \)
\( \implies 16 + k^2 - 8k + k^2 + 4 - 4k = 1 + 9 \)
\( \implies 2k^2 - 12k + 20 = 10 \)
\( \implies k^2 - 6k + 10 = 5 \)
\( \implies k^2 - 6k + 5 = 0 \)
\( \implies k^2 - 5k - k + 5 = 0 \)
\( \implies k(k - 5) - 1(k - 5) = 0 \)
\( \implies k = 1 \) or \( k = 5 \)

Question. If the mid-point of the line segment joining the points \( A(3, 4) \) and \( B(k, 6) \) is \( P(x, y) \) and \( x + y - 10 = 0 \), find the value \( k \). 
Answer: \( x = \frac{3 + k}{2} \), \( y = 5 \) \( x + y - 10 = 0 \)
\( \implies \frac{3 + k}{2} + 5 - 10 = 0 \)
\( \implies k = 7 \) 

Question. In what ratio does the point \( P(-4, y) \) divide the line segment joining the points \( A(-6, 10) \) and \( B(3, -8) \) if it lies on \( AB \)? Hence find the value of \( y \). 
Answer: Let \( k : 1 \) be the required ratio. \( \therefore \) Using section formula \( x = \frac{kx_2 + x_1}{k + 1} \)
\( \implies -4 = \frac{k \times 3 + 1 \times (-6)}{k + 1} \)
\( \implies -4k - 4 = 3k - 6 \)
\( \implies -4 + 6 = 3k + 4k \)
\( \implies 2 = 7k \)
\( \implies k = \frac{2}{7} \) Ratio be \( 2 : 7 \). Now, \( y = \frac{k \times (-8) + 1 \times 10}{k + 1} = \frac{\frac{2}{7} \times (-8) + 10}{\frac{2}{7} + 1} \)
\( = \frac{-16 + 70}{9} = \frac{54}{9} = 6 \)
\( \therefore y = 6 \)

Question. In what ratio does the point \( \left( \frac{24}{11}, y \right) \) divide the line segment joining the point \( P(2, -2) \) and \( Q(3, 7) \)? Also find the value of \( y \). 
Answer: Using section formula, \( \left( \frac{24}{11}, y \right) = \left( \frac{3m + 2n}{m + n}, \frac{7m - 2n}{m + n} \right) \) — ①
\( \implies \frac{24}{11} = \frac{3m + 2n}{m + n} \) \( 24m + 24n = 33m + 22n \) \( 2n = 9m \) \( \frac{2}{9} = \frac{m}{n} \) \( \therefore \) The given point divides the line segment in ratio \( 2 : 9 \). Taking \( m = 2 \) and \( n = 9 \), \( y = \frac{7m - 2n}{m + n} \) (from ①) \( y = \frac{7(2) - 2(9)}{2 + 9} \) \( y = \frac{14 - 18}{11} \) \( y = -\frac{4}{11} \). 

Question. Find the coordinates of the points of trisection of the line segment joining the points \( (3, -1) \) and \( (6, 8) \). 
Answer: Case I: If \( C \) and \( D \) trisect \( AB \) then \( C \) divides \( AB \) in the ratio \( 1 : 2 \). Co-ordinates of \( C : x = \frac{1 \times 6 + 2 \times 3}{3} = 4 \) and \( y = \frac{1 \times 8 + 2 \times (-1)}{3} = 2 \) \( \therefore \) Co-ordinates of point \( C \) be \( (4, 2) \). Case II: If \( D \) divides \( AB \) in the ratio \( 2 : 1 \), then Co-ordinates of \( D : x' = \frac{2 \times 6 + 1 \times 3}{3} = 5 \) and \( y' = \frac{2 \times 8 + 1 \times (-1)}{3} = 5 \) \( \therefore \) Co-ordinates of \( D(5, 5) \). 

Question. Find the ratio in which the line \( x - 3y = 0 \) divides the line segment joining the points \( (- 2, - 5) \) and \( (6, 3) \). Find the coordinates of the point of intersection. 
Answer: Let the line \( x - 3y = 0 \) intersect the segment joining \( A(-2, -5) \) and \( B(6, 3) \) in the ratio \( k : 1 \)
\( \therefore \) Coordinates of \( P \) are \( \left( \frac{6k-2}{k+1}, \frac{3k-5}{k+1} \right) \)
\( P \) lies on \( x - 3y = 0 \)
\( \implies \) \( \frac{6k-2}{k+1} = 3 \left( \frac{3k-5}{k+1} \right) \)
\( \implies \) \( k = \frac{13}{3} \)
\( \therefore \) Ratio is \( 13 : 3 \)
\( \implies \) Coordinates of \( P \) are \( \left( \frac{9}{2}, \frac{3}{2} \right) \)
[CBSE Marking Scheme 2019 (30/2/1)]

Question. (i) Find the ratio in which the y-axis divides the line segment joining the points \( (6, - 4) \) and \( (- 2, - 7) \). Also find the point of intersection.
(ii) Show that the points \( (7, 10), (- 2, 5) \) and \( (3, - 4) \) are vertices of an isosceles right triangle. 
Answer: (i) Let the point \( P(0, y) \) on y-axis divides the line segment AB in \( k : 1 \)
\( \therefore \) \( 0 = \frac{-2k + 6}{k + 1} \)
\( \implies \) \( k = 3 \). Ratio is \( 3 : 1 \)
Also, \( y = \frac{3(-7) + 1(-4)}{3 + 1} = \frac{-25}{4} \)
\( \therefore \) Point of intersection is \( \left( 0, -\frac{25}{4} \right) \)
(ii) Let the points be \( A(7, 10), B(-2, 5) \) and \( C(3, -4) \)
\( AB = \sqrt{(-2 - 7)^2 + (5 - 10)^2} = \sqrt{106} \)
\( BC = \sqrt{(3 + 2)^2 + (-4 - 5)^2} = \sqrt{106} \)
\( AC = \sqrt{(3 - 7)^2 + (-4 - 10)^2} = \sqrt{212} \)
\( AB = BC \) and \( AC^2 = AB^2 + BC^2 \)
Hence ABC is isosceles right triangle.
[CBSE Marking Scheme 2020 (30/5/1)]

Question. Point \( P \) divides the line segment joining the points \( A(2,1) \) and \( B(5,-8) \) such that \( \frac{AP}{AB} = \frac{1}{3} \). If \( P \) lies on the line \( 2x - y + k = 0 \), find the value of \( k \). 
Answer: Let the point \( P \) be \( (x, y) \) which divides \( AB \) such that
\( \frac{AP}{AB} = \frac{1}{3} \)
\( \implies \) \( \frac{AB}{AP} = \frac{3}{1} \)
\( \implies \) \( \frac{AB}{AP} - 1 = \frac{3}{1} - 1 \)
\( \implies \) \( \frac{AB - AP}{AP} = \frac{2}{1} \)
\( \implies \) \( \frac{PB}{AP} = \frac{2}{1} \)
\( \implies \) \( \frac{AP}{BP} = \frac{1}{2} \)
\( \implies \) \( AP : BP = 1 : 2 \)
\( \therefore \) Using section formula, we have
\( x = \frac{1 \times 5 + 2 \times 2}{1 + 2} = \frac{9}{3} = 3 \)
\( y = \frac{1 \times (-8) + 2 \times 1}{1 + 2} = \frac{-8 + 2}{3} = \frac{-6}{3} = - 2 \)
\( \therefore \) Co-ordinate of \( P \) be \( (3, -2) \).
Now, \( P(3, - 2) \) lies on the line \( 2x - y + k = 0 \).
\( \therefore 2 \times 3 - (-2) + k = 0 \)
\( \implies 6 + 2 + k = 0 \)
\( \implies k = -8 \)

Very Short Answer Questions: 

Question. Find the values of \( x \) for which the distance between the points \( A(x, 2) \) and \( B(9, 8) \) is 10 units. 
Answer: \( x = 17, x = 1 \)

Question. Find the value(s) of \( x \), if the distance between the points \( A(0, 0) \) and \( B(x, - 4) \) is 5 units. 
Answer: \( x = \pm 3 \)

Question. \( A(5, 1); B(1, 5) \) and \( C(- 3, -1) \) are the vertices of \( \Delta ABC \). Find the length of median \( AD \). 
Answer: \( \sqrt{37} \)

Question. If \( P\left(\frac{a}{3}, 4\right) \) is the mid-point of the line segment joining the points \( Q(- 6, 5) \) and \( R(- 2, 3) \), then find the value of \( a \).
Answer: \( a = -12 \)

Question. A line intersects the y-axis and x-axis at the points \( P \) and \( Q \) respectively. If \( (2, -5) \) is the mid-point of \( PQ \), then find the coordinates of \( P \) and \( Q \) 
Answer: \( (0, -10) \) and \( (4, 0) \)

Question. Where does the perpendicular bisector of the line segment joining the points \( A(1, 5) \) and \( B(4, 6) \) cut the y-axis? 
Answer: \( (0, 13) \)

Question. In what ratio is the line segment joining the points \( P(3, - 6) \) and \( Q(5, 3) \) divided by x-axis?
Answer: \( 2 : 1 \)

Question. The point which divides the line segment joining the points \( (7, -6) \) and \( (3, 4) \) in ratio \( 1 : 2 \) internally lies in which quadrant? [Competency Based Question] 
Answer: IVth

Question. If the points \( (k, 2k), (3k, 3k) \) and \( (3, 1) \) are collinear, then find the values of \( k \).
Answer: \( k = -\frac{1}{3} \) or 0

Question. If points \( (a, 0), (0, b) \) and \( (1, 1) \) are collinear, then find the value of \( \frac{1}{a} + \frac{1}{b} \). 
Answer: 1

Question. If the centroid of a triangle formed by the points \( (a, b), (b, c) \) and \( (c, a) \) is at the origin, then find the value of \( a^3 + b^3 + c^3 \).
Answer: \( 3abc \)

Short Answer Questions-I: 

Question. Find the linear relation between \( x \) and \( y \) such that \( P(x, y) \) is equidistant from the points \( A(1, 4) \) and \( B(-1, 2) \). 
Answer: \( x + y - 3 = 0 \)

Question. Prove that the points \( (3, 0), (6, 4) \) and \( (-1, 3) \) are the vertices of a right angled isosceles triangle. 
Answer: Verification: \( AB = \sqrt{(6-3)^2 + (4-0)^2} = 5 \) units \( BC = \sqrt{(-1-6)^2 + (3-4)^2} = \sqrt{50} = 5\sqrt{2} \) units \( AC = \sqrt{(-1-3)^2 + (3-0)^2} = 5 \) units Since \( AB = AC \) and \( AB^2 + AC^2 = BC^2 \), it is a right angled isosceles triangle.

Question. Name the type of triangle \( ABC \) formed by the points \( A(\sqrt{2}, \sqrt{2}), B(-\sqrt{2}, -\sqrt{2}) \) and \( C(-\sqrt{6}, \sqrt{6}) \).
Answer: Equilateral

Question. Find a point which is equidistant from the points \( P(-5, 4) \) and \( Q(-1, 6) \). How many such points are there?
Answer: \( (-3, 5) \), infinitely many

Question. If \( A(-2, 2), B(5, 2) \) and \( C(k, 8) \) are the vertices of a right-angled triangle \( ABC \) with \( \angle B = 90^\circ \), then find the value of \( k \). 
Answer: \( k = 5 \)

Question. Points \( A(3, 1), B(5, 1), C(a, b) \) and \( D(4, 3) \) are vertices of a parallelogram \( ABCD \). Find the values of \( a \) and \( b \). 
Answer: \( a = 6, b = 3 \)

Question. If the mid-point of a segment joining \( A\left(\frac{x}{2}, \frac{y+1}{2}\right) \) and \( B(x+1, y-3) \) is \( C(5, -2) \), find \( x, y \).
Answer: \( x = 6, y = -1 \)

Question. The point \( R \) divides the line segment \( AB \), where \( A(-4, 0) \) and \( B(0, 6) \) such that \( AR = \frac{3}{4} AB \). Find the coordinates of \( R \). 
Answer: \( R\left(-1, \frac{9}{2}\right) \)

Question. Find the relation between \( x \) and \( y \) such that the points \( (x, y), (1, 2) \) and \( (7, 0) \) are collinear.
Answer: \( x + 3y - 7 = 0 \)

Question. The vertices of a triangle are \( (a, b - c), (b, c - a) \) and \( (c, a - b) \). Prove that its centroid lies on x-axis.
Answer: Let centroid be \( (X, Y) \).
\( Y = \frac{(b-c) + (c-a) + (a-b)}{3} = \frac{0}{3} = 0 \). Since the y-coordinate is 0, the centroid lies on the x-axis.