CBSE Class 10 Mathematics Coordinate Geometry VBQs Set D

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 \(B(-7, 4)\). \(P\) and \(Q\) are points of trisection, then \(P\) divides \(AB\) in ratio \(1:2\) and \(Q\) 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 \(\frac{4 - 4}{3}\)
\(\Rightarrow P(x, y) = (-1, 0)\)
For \(Q(x', y') = \frac{2 \times (-7) + 1 \times 2}{3}\) and \(y' = \frac{2 \times 4 + 1 \times (-2)}{3}\)
\(Q(x', y') = \frac{-14 + 2}{3}\) and \(\frac{8 - 2}{3}\)
\(\Rightarrow Q(x', y') = (-4, 2)\) 

Question. Find the coordinates of the points of trisection of the line segment joining the points \((3, -1)\) and \((6, 8)\). 
Answer: Let the given points be \(A\) and \(B\) and \(P\) and \(Q\) be the points of trisection of \(AB\), as shown in the figure.
\(A(3, -1)\) — \(P\) — \(Q\) — \(B(6, 8)\)
Here, \(P\) divides \(AB\) in the ratio \(1 : 2\) and \(Q\) divides \(AB\) in the ratio \(2 : 1\).
By section formula \(\left[ \frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \frac{m_1y_2 + m_2y_1}{m_1 + m_2} \right]\)
So, \(P\left( \frac{6 + 6}{3}, \frac{8 - 2}{3} \right)\) and \(Q\left( \frac{12 + 3}{3}, \frac{16 - 1}{3} \right)\)
i.e., \(P(4, 2)\) and \(Q(5, 5)\)

Question. Find the points on the x-axis which are at a distance of \(2\sqrt{5}\) from the point \((7, -4)\). How many such points are there? 
Answer: We know that any point on x-axis is of the form \((x, 0)\).
Let \(P(x, 0)\) be the point on x-axis having \(2\sqrt{5}\) distance from the point \(Q(7, -4)\).
Distance between \(P(x, 0)\) and \(Q(7, -4)\) using distance formula,
\(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)
\(PQ = \sqrt{(7 - x)^2 + (-4 - 0)^2} = \sqrt{(7 - x)^2 + 16}\)
According to given condition \(PQ = 2\sqrt{5}\)
\(\Rightarrow (PQ)^2 = (2\sqrt{5})^2\)
\(\Rightarrow (7 - x)^2 + 4^2 = (2\sqrt{5})^2\)
\(\Rightarrow 49 + x^2 - 14x + 16 = 20\)
\(\Rightarrow x^2 - 14x + 45 = 0\)
\(\Rightarrow x^2 - 9x - 5x + 45 = 0\) [using factorisation method]
\(\Rightarrow x(x - 9) - 5(x - 9) = 0\)
\(\Rightarrow (x - 9)(x - 5) = 0\)
\(\Rightarrow x = 9, 5\).
Hence, there are two points that lie on x-axis, which are \((5, 0)\) and \((9, 0)\), having a distance of \(2\sqrt{5}\) from the point \((7, -4)\).

Question. What type of a quadrilateral do the points \((2, -2)\), \(B(7, 3)\), \(C(11, -1)\) and \(D(6, -6)\), taken in that order, form? 
Answer: To find the type of quadrilateral, we will find the length of all four sides and the length of diagonals.
We know that distance between points \((x_1, y_1)\) and \((x_2, y_2)\)
\(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)
Here, \(A(2, -2), B(7, 3), C(11, -1)\) and \(D(6, -6)\)
\(AB = \sqrt{(7 - 2)^2 + (3 + 2)^2} = \sqrt{5^2 + 5^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2}\)
\(BC = \sqrt{(11 - 7)^2 + (-1 - 3)^2} = \sqrt{4^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}\)
\(CD = \sqrt{(6 - 11)^2 + (-6 + 1)^2} = \sqrt{(-5)^2 + (-5)^2} = \sqrt{25 + 25} = 5\sqrt{2}\)
\(AD = \sqrt{(6 - 2)^2 + (-6 + 2)^2} = \sqrt{4^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}\)
Diagonal, \(AC = \sqrt{(11 - 2)^2 + (-1 + 2)^2} = \sqrt{9^2 + 1^2} = \sqrt{81 + 1} = \sqrt{82}\)
Diagonal \(BD = \sqrt{(6 - 7)^2 + (-6 - 3)^2} = \sqrt{(-1)^2 + (-9)^2} = \sqrt{1 + 81} = \sqrt{82}\)
Here, we see that length of opposite sides are equal i.e. \(AB = DC = 5\sqrt{2}\) and \(AD = BC = 4\sqrt{2}\)
Also, length of diagonals \(AC = BD = \sqrt{82}\)
This shows that the given quadrilateral is a rectangle.

Question. Find a point which is equidistant from the points \(A(-5, 4)\) and \(B(-1, 6)\)? How many such points are there? 
Answer: Let \(P(r, s)\) be the point which is equidistant from points \(A(-5, 4)\) and \(B(-1, 6)\).
We know that distance between the points \((x_1, y_1)\) and \((x_2, y_2)\),
\(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)
\(\therefore PA = PB \Rightarrow (PA)^2 = (PB)^2\)
\(\Rightarrow (-5 - r)^2 + (4 - s)^2 = (-1 - r)^2 + (6 - s)^2\)
\(\Rightarrow 25 + r^2 + 10r + 16 + s^2 - 8s = 1 + r^2 + 2r + 36 + s^2 - 12s\)
\(\Rightarrow 25 + 10r + 16 - 8s = 1 + 2r + 36 - 12s\)
\(\Rightarrow 8r + 4s + 4 = 0\)
\(\Rightarrow 2r + s + 1 = 0\) ...(i)
Midpoint of \(AB = \left( \frac{-5 - 1}{2}, \frac{4 + 6}{2} \right) = (-3, 5)\)
At point \((-3, 5)\) from eqn (i), we get
\(\Rightarrow 2(-3) + 5 = -6 + 5 = -1\)
\(\Rightarrow 2r + s + 1 = 0\)
Hence, midpoint of \(AB\) satisfies eqn (i). This implies that there are infinite number of points which satisfy eqn (i) are equidistant from points \(A\) and \(B\).
Replacing \(r, s\) with \(x\) and \(y\) in the above eqn. we get \(2x + y + 1 = 0\)

Question. In what ratio does the point \(P(-4, y)\) divide the line segment joining the points \(A(-6, 10)\) and \(B(3, -8)\)? Find the value of \(y\).
Answer: Let the point \(P\) divides the line segment \(AB\) in the ratio of \(k : 1\).
By the section formula, the coordinates of \(P\) are: \(P(x, y) = \left( \frac{mx_2 + nx_1}{m + n}, \frac{my_2 + ny_1}{m + n} \right)\)
Here \(m = k, n = 1, x_1 = -6, y_1 = 10, x_2 = 3, y_2 = -8\) and \(x = -4, y = y\)
\(\therefore P(-4, y) = \left( \frac{3k - 6}{k + 1}, \frac{-8k + 10}{k + 1} \right)\)
On comparing 'x' coordinates \(\Rightarrow -4 = \frac{3k - 6}{k + 1}\)
\(\Rightarrow -4k - 4 = 3k - 6 \Rightarrow -7k = -2 \Rightarrow k = \frac{2}{7}\)
\(\therefore\) The ratio is \(2 : 7\).
And \(y = \frac{-8k + 10}{k + 1}\)
Put the value of '\(k\)':
\(\Rightarrow y = \frac{-8 \times \frac{2}{7} + 10}{\frac{2}{7} + 1} = \frac{-16 + 70}{9} = \frac{54}{9} = 6\)
Hence, the value of '\(y\)' is \(6\).

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 \(k : 1\) at point \(P(x, y)\).
By using the section formula, coordinates of \(P(x, y)\) are:
\(P(x, y) = \left( \frac{6k - 2}{k + 1}, \frac{3k - 5}{k + 1} \right)\)
But, \(P\) lies on \(x - 3y = 0 \Rightarrow x = 3y\)
\(\Rightarrow \frac{6k - 2}{k + 1} = 3 \left( \frac{3k - 5}{k + 1} \right)\)
\(\Rightarrow 6k - 2 = 9k - 15 \Rightarrow 3k = 13 \Rightarrow k = \frac{13}{3}\)
Coordinates of \(P\) are: \(\left( \frac{6 \times \frac{13}{3} - 2}{\frac{13}{3} + 1}, \frac{3 \times \frac{13}{3} - 5}{\frac{13}{3} + 1} \right) = \left( \frac{24}{16} \times 3, \frac{8}{16} \times 3 \right) = \left( \frac{9}{2}, \frac{3}{2} \right)\)
Hence, the coordinates of point of intersection '\(P\)' are \(\frac{9}{2}\) and \(\frac{3}{2}\).

Question. Find the coordinates of the point \(Q\) on the x-axis which lies on the perpendicular bisector of the line segment joining the points \(A(-5, -2)\) and \(B(4, -2)\). Name the type of triangle formed by the points \(Q\), \(A\) and \(B\). 
Answer: Let \(Q(x, 0)\) be the point on the x-axis which lies on the perpendicular bisector of \(AB\).
\(\therefore QA = QB \Rightarrow (QA)^2 = (QB)^2\)
\(\Rightarrow (-5 - x)^2 + (-2 - 0)^2 = (4 - x)^2 + (-2 - 0)^2\)
\(\Rightarrow 25 + x^2 + 10x + 4 = 16 + x^2 - 8x + 4\)
\(\Rightarrow 10x + 8x = 16 - 25 \Rightarrow 18x = -9 \Rightarrow x = -\frac{1}{2}\)
Hence, the point \(Q\) is \(\left( -\frac{1}{2}, 0 \right)\).
Now \(QA^2 = \left( -5 + \frac{1}{2} \right)^2 + (-2 - 0)^2 = \left( -\frac{9}{2} \right)^2 + 4 = \frac{81}{4} + 4 = \frac{97}{4}\)
\(\Rightarrow QA = \frac{\sqrt{97}}{2}\) units
Now \(QB^2 = \left( 4 + \frac{1}{2} \right)^2 + (-2 - 0)^2 = \left( \frac{9}{2} \right)^2 + 4 = \frac{81}{4} + 4 = \frac{97}{4}\)
\(\Rightarrow QB = \frac{\sqrt{97}}{2}\) units
\(AB = \sqrt{(4 + 5)^2 + (-2 + 2)^2} = \sqrt{9^2} = 9\) units
\(\Rightarrow AB = 9\) units and \(QA = QB = \frac{\sqrt{97}}{2}\) units
Hence, \(\Delta QAB\) is an isosceles \(\Delta\).

Question. Find the point on y-axis which is equidistant from the points \((5, -2)\) and \((-3, 2)\). 
Answer: Let the required point on y-axis be \(P(0, b)\).
Given, points are \(A(5, -2)\) and \(B(-3, 2)\).
Since the points \(A\) and \(B\) are equidistant from point \(P\), the distance \(AP = \text{distance } BP\).
Applying the distance formula, we get:
\(\sqrt{(x_1 - x)^2 + (y_1 - y)^2} = \sqrt{(x_2 - x)^2 + (y_2 - y)^2}\)
Here, \(x = 0, y = b, x_1 = 5, y_1 = -2, x_2 = -3, y_2 = 2\)
\(\Rightarrow \sqrt{(5 - 0)^2 + (-2 - b)^2} = \sqrt{(-3 - 0)^2 + (2 - b)^2}\)
On squaring both sides, we get
\(\Rightarrow 25 + (-2 - b)^2 = 9 + (2 - b)^2\)
\(\Rightarrow 25 + 4 + b^2 + 4b = 9 + 4 + b^2 - 4b\)
\(\Rightarrow 8b = -16 \Rightarrow b = -2\)
Hence, the required point is \((0, -2)\).

Question. The line segment joining the points \(A(2, 1)\) and \(B(5, - 8)\) is trisected at the points \(P\) and \(Q\) such that \(P\) is nearer to \(A\). If \(P\) also lies on the line given by \(2x - y + k = 0\), find the value of \(k\). 
Answer: Here, line \(AB\) is trisected at point \(P\) and \(Q\). Let the co-ordinates of \(P\) be \((x, y)\).
Then, \(AP : PB = 1 : 2\).
By the section formula, the co-ordinates of \(P\) are:
\(P(x, y) = \left( \frac{mx_2 + nx_1}{m + n}, \frac{my_2 + ny_1}{m + n} \right)\)
Here, \(m = 1, n = 2, x_1 = 2, y_1 = 1, x_2 = 5, y_2 = -8\)
Then \(P(x, y) = \left( \frac{1 \times 5 + 2 \times 2}{1 + 2}, \frac{1 \times (-8) + 2 \times 1}{1 + 2} \right) = \left( \frac{9}{3}, \frac{-6}{3} \right) = (3, -2)\)
So, the co-ordinates of \(P\) are \(3\) and \(-2\).
Since, point \(P\) lies on the line \(2x - y + k = 0\), its co-ordinates will satisfy the equation of the line.
\(2(3) - (-2) + k = 0 \Rightarrow 6 + 2 + k = 0 \Rightarrow k = -8\)
Hence, the value of \(k\) is \(-8\).

Question. If the point \(A(2, -4)\) is equidistant from \(P(3, 8)\) and \(Q(-10, y)\), find the values of \(y\). Also find distance \(PQ\). 
Answer: It is given that \(A(2, -4)\) is equidistant from \(P(3, 8)\) and \(Q(-10, y)\).
We know that distance between the points \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)
\(\Rightarrow \text{Distance between } P(3, 8) \text{ and } A(2, -4) = \text{Distance between } A(2, -4) \text{ and } Q(-10, y)\)
\(\sqrt{(2 - 3)^2 + (-4 - 8)^2} = \sqrt{(-10 - 2)^2 + (y + 4)^2}\)
\(\sqrt{1^2 + 12^2} = \sqrt{12^2 + (y + 4)^2}\)
Squaring both sides, we get \(1^2 + 12^2 = 12^2 + (y + 4)^2 \Rightarrow 1 = (y + 4)^2\)
\(\Rightarrow y^2 + 8y + 15 = 0 \Rightarrow (y + 5)(y + 3) = 0 \Rightarrow y = -5, -3\)
Distance between \(P(3, 8)\) and \(Q(-10, y)\) when \(y = -3\):
\(PQ = \sqrt{(-10 - 3)^2 + (-3 - 8)^2} = \sqrt{(-13)^2 + (-11)^2} = \sqrt{169 + 121} = \sqrt{290}\)
Distance between \(P(3, 8)\) and \(Q(-10, y)\) when \(y = -5\):
\(PQ = \sqrt{(-10 - 3)^2 + (-5 - 8)^2} = \sqrt{(-13)^2 + (-13)^2} = \sqrt{169 + 169} = \sqrt{338} = 13\sqrt{2}\).
Hence, the values of \(y\) are \(-3\) and \(-5\) and the corresponding values of \(PQ\) are \(\sqrt{290}\) and \(13\sqrt{2}\).

Question. If \(A(-2, 1)\), \(B(a, 0)\), \(C(4, b)\) and \(D(1, 2)\) are the vertices of a parallelogram \(ABCD\), find the values of \(a\) and \(b\). Also, find the lengths of its sides.
Answer: Given, \(ABCD\) is a parallelogram, in which diagonals \(AC\) and \(BD\) bisect each other at \(O\).
\(O\) is the mid-point of \(AC\) and \(BD\).
By the mid point formula: Coordinates of \(O = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)\)
For line \(BD, O(x, y) = \left( \frac{a + 1}{2}, \frac{0 + 2}{2} \right) = \left( \frac{a + 1}{2}, 1 \right)\) ...(i)
For line \(AC, O(x, y) = \left( \frac{-2 + 4}{2}, \frac{1 + b}{2} \right) = \left( 1, \frac{1 + b}{2} \right)\) ...(ii)
From (i) and (ii), we get: \(\frac{a + 1}{2} = 1 \Rightarrow a = 1\) and \(\frac{1 + b}{2} = 1 \Rightarrow b = 1\)
Length of side \(AB = \sqrt{(1 + 2)^2 + (0 - 1)^2} = \sqrt{9 + 1} = \sqrt{10}\) units.
\(\therefore AB = CD = \sqrt{10}\) units. Side \(BC = \sqrt{(4-1)^2 + (1-0)^2} = \sqrt{10}\) units.
Hence, all the four sides of parallelogram is \(\sqrt{10}\) units.

Question. Find the coordinates of the points of trisection of the line segment joining the points \((3, - 2)\) and \((-3, -4)\). 
Answer: The given line segment is \(A(3, - 2)\) and \(B(-3, -4)\). Let \(C(x, y)\) and \(C'(x', y')\) are the points of trisection of \(AB\).
Then, \(AC : CB = 1 : 2\). By section formula, \(C(x, y) = \left( \frac{1 \times (-3) + 2 \times 3}{1 + 2}, \frac{1 \times (-4) + 2 \times (-2)}{1 + 2} \right) = \left( \frac{3}{3}, \frac{-8}{3} \right) = (1, -8/3)\).
Now, coordinates of \(C'\) are (where ratio is \(2 : 1\)):
\(C'(x', y') = \left( \frac{2 \times (-3) + 1 \times 3}{1 + 2}, \frac{2 \times (-4) + 1 \times (-2)}{1 + 2} \right) = \left( \frac{-3}{3}, \frac{-10}{3} \right) = (-1, -10/3)\).
Hence, the coordinates of the trisection are \((1, - 8/3)\) and \((-1, - 10/3)\).

Question. If \(P(9a - 2, -b)\) divides line segment joining \(A(3a + 1, -3)\) and \(B(8a, 5)\) in the ratio \(3 : 1\), find the values of \(a\) and \(b\). 
Answer: It is given that \(P(9a - 2, -b)\) divides \(AB\) in the ratio \(3 : 1\).
By section formula, \(P(x, y) = \left( \frac{3(8a) + 1(3a + 1)}{3 + 1}, \frac{3(5) + 1(-3)}{3 + 1} \right) = \left( \frac{24a + 3a + 1}{4}, \frac{15 - 3}{4} \right) = \left( \frac{27a + 1}{4}, \frac{12}{4} \right)\)
\(\Rightarrow 9a - 2 = \frac{27a + 1}{4} \Rightarrow 36a - 8 = 27a + 1 \Rightarrow 9a = 9 \Rightarrow a = 1\)
And \(-b = \frac{12}{4} = 3 \Rightarrow b = -3\).
Hence, the required values of \(a\) and \(b\) are \(1\) and \(-3\).

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

Question. Find the ratio in which line \(2x + 3y - 5 = 0\) divides the line segment joining the points \((8, -9)\) and \((2, 1)\). Also find the coordinates of the point of division. 
Answer: Let the line \(2x + 3y - 5 = 0\) divide the segment joining \(A(8, -9)\) and \(B(2, 1)\) in the ratio \(m : 1\) at point \(P\).
Coordinates of \(P = \left( \frac{2m + 8}{m + 1}, \frac{m - 9}{m + 1} \right)\).
Since \(P\) lies on \(2x + 3y - 5 = 0\):
\(2\left( \frac{2m + 8}{m + 1} \right) + 3\left( \frac{m - 9}{m + 1} \right) - 5 = 0 \Rightarrow 4m + 16 + 3m - 27 - 5m - 5 = 0\)
\(\Rightarrow 2m - 16 = 0 \Rightarrow m = 8\). Ratio is \(8 : 1\).
Coordinates of \(P = \left( \frac{2(8) + 8}{9}, \frac{8 - 9}{9} \right) = \left( \frac{24}{9}, \frac{-1}{9} \right) = \left( \frac{8}{3}, -\frac{1}{9} \right)\).

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, the coordinates of points on x-axis be \(P(x, 0)\).
Point \(A(2, - 5)\) and point \(B(- 2, 9)\) are equidistant from point \(P\).
\(PA = PB \Rightarrow (2 - x)^2 + (- 5)^2 = (- 2 - x)^2 + 9^2\)
\(4 + x^2 - 4x + 25 = 4 + x^2 + 4x + 81 \Rightarrow - 8x = 56 \Rightarrow x = - 7\).
Hence, the coordinates of point on x-axis is \((- 7, 0)\).

Question. Write the coordinates of a point \(P\) on x-axis which is equidistant from the points \(A(-2, 0)\) and \(B(6, 0)\). 
Answer: Let the co-ordinates of \(P\) be \((x, 0)\).
\(PA = PB \Rightarrow (x + 2)^2 + 0^2 = (6 - x)^2 + 0^2 \Rightarrow 4 + x^2 + 4x = 36 + x^2 - 12x \Rightarrow 16x = 32 \Rightarrow x = 2\).
Hence, the co-ordinate of \(P\) are \((2, 0)\).

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: \(AP = BP \Rightarrow AP^2 = BP^2\)
\([x - (a+b)]^2 + [y - (b-a)]^2 = [x - (a-b)]^2 + [y - (a+b)]^2\)
\(x^2 + (a+b)^2 - 2x(a+b) + y^2 + (b-a)^2 - 2y(b-a) = x^2 + (a-b)^2 - 2x(a-b) + y^2 + (a+b)^2 - 2y(a+b)\)
\((b-a)^2 - 2x(a+b) - 2y(b-a) = (a-b)^2 - 2x(a-b) - 2y(a+b)\)
\(-2xa - 2xb - 2yb + 2ya = -2xa + 2xb - 2ya - 2yb\)
\(-4xb = -4ya \Rightarrow bx = ay\). Hence proved.

Question. If the coordinates of points A and B are (–2, –2) and (2, –4) respectively, find the coordinates of P such that \(AP = \frac{3}{7}AB\), where P lies on the line segment AB. 
Answer: \(AP = \frac{3}{7}AB \Rightarrow AP : PB = 3 : 4\)
\(\therefore x = \frac{6 - 8}{7} = -\frac{2}{7}\)
\(y = \frac{-12 - 8}{7} = -\frac{20}{7}\)
\(P\left(-\frac{2}{7}, -\frac{20}{7}\right)\)

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: Given, AB line segment with coordinates of A as (– 4, 0) and coordinates of B as (0, 6).
and \(AR = \frac{3}{4}AB\)
\(\Rightarrow 4AR = 3(AR + RB)\)
\(\Rightarrow AR = 3RB\)
\(\Rightarrow \frac{AR}{RB} = \frac{3}{1}\)
Let the coordinates of R be \((x, y)\).
By the section formula,
\(R(x, y) = \left( \frac{mx_2 + nx_1}{m + n}, \frac{my_2 + ny_1}{m + n} \right)\)
Here, \(m = 3, n = 1\)
\(x_1 = -4, y_1 = 0\)
\(x_2 = 0, y_2 = 6\)
\(\therefore R(x, y) = \left( \frac{3 \times 0 + 1 \times (-4)}{3 + 1}, \frac{3 \times 6 + 1 \times 0}{3 + 1} \right)\)
\(\therefore R(x, y) = \left( -1, \frac{9}{2} \right)\)
Hence, the coordinates of R are: \(\left( -1, \frac{9}{2} \right)\).

Question. Find the ratio in which the line segment joining the points (1, – 3) and (4, 5) is divided by x-axis? Also find the coordinates of this point on x-axis. 
Answer: Let, the required point be \((a, 0)\) and the required ratio be \(k : 1\).
By the section formula, the co-ordinates of P are given by:
\(P(x, y) = \left[ \frac{mx_2 + nx_1}{m + n}, \frac{my_2 + ny_1}{m + n} \right]\)
Here, \(m = k, n = 1\)
\(x_1 = 1, x_2 = 4\)
\(y_1 = -3, y_2 = 5\)
\(x = a, y = 0\)
\(\therefore a = \frac{4k + 1}{k + 1}\), and \(0 = \frac{5k - 3}{k + 1}\)
\(\Rightarrow 5k - 3 = 0\)
\(\Rightarrow k = \frac{3}{5}\)
Now, \(a = \frac{4\left(\frac{3}{5}\right) + 1}{\frac{3}{5} + 1}\)
\(\Rightarrow a = \frac{\frac{12}{5} + 1}{\frac{8}{5}}\)
\(\Rightarrow a = \frac{17}{8}\)
Hence, the ratio in which the line segment is divided is \(3 : 5\) and the coordinates of the point are \(\left( \frac{17}{8}, 0 \right)\)

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 \(k : 1\).
Here, the coordinates of A are (2, 3) B are (6, – 3) and division point are (4, m).
By the section formula
\(P(x, y) = \left( \frac{mx_2 + nx_1}{m + n}, \frac{my_2 + ny_1}{m + n} \right)\)
Here \(x_1 = 2, y_1 = 3\)
\(x_2 = 6, y_2 = -3\)
\(x = 4, y = m\)
\(m = k, n = 1\)
\(P(4, m) = \left( \frac{k \times 6 + 1 \times 2}{k + 1}, \frac{k \times (-3) + 1 \times (3)}{k + 1} \right)\)
\(\therefore 4 = \frac{6k + 2}{k + 1}\)
\(\Rightarrow 4k + 4 = 6k + 2\)
\(\Rightarrow 2k = 2\)
\(\Rightarrow k = 1\)
\(\therefore\) Ratio in which P divides AB is \(1 : 1\).
Now \(m = \frac{-3k + 3}{k + 1}\)
\(\Rightarrow m = \frac{-3(1) + 3}{1 + 1} = 0\)
Hence, the value of m is 0.

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

Question. If the line segment joining the points A(2, 1) and B(5, –8) is trisected at the point P and Q, find the coordinates of P.
Answer: Given, line segment A(2, 1) and B(5, –8) is trisected at point P and Q.
Then \(AP : PB = 1 : 2\) and \(AQ : QB = 2 : 1\)
Let the coordinates of P be \((x, y)\) and Q be \((x', y')\)
Now, on applying the section formula,
\(P(x, y) = \left( \frac{mx_2 + nx_1}{m + n}, \frac{my_2 + ny_1}{m + n} \right)\)
Here, \(x_1 = 2, y_1 = 1\)
\(x_2 = 5, y_2 = -8\)
\(m = 1, n = 2\)
\(\therefore P(x, y) = \left[ \frac{1 \times 5 + 2 \times 2}{1 + 2}, \frac{1 \times (-8) + 2 \times 1}{1 + 2} \right]\)
\( = \left[ \frac{9}{3}, \frac{-6}{3} \right] = [3, -2]\)
Now, coordinates \(Q(x', y')\)
\(Q(x', y') = \left[ \frac{2 \times 5 + 1 \times 2}{1 + 2}, \frac{2 \times (-8) + 1 \times 1}{1 + 2} \right]\)
\( = \left[ \frac{12}{3}, \frac{-15}{3} \right] = (4, -5)\)
Hence, the coordinates of P are (3, –2) and Q are (4, –5)

LONG ANSWER (LA) Type Questions

Question. Show that \(\Delta ABC\), where A(–2, 0), B(2, 0), C(0, 2) and \(\Delta PQR\) where P(–4, 0), Q(4, 0), R(0, 4) are similar triangles. 
Answer: Given: \(\Delta ABC\) with coordinates of vertices are A(–2, 0), B(2, 0), C(0, 2) and P(–4, 0) Q(4, 0) and R(0, 4).
To prove: \(\Delta ABC \sim \Delta PQR\)
Proof: In \(\Delta ABC\), find the length of sides AB, BC and CA by distance formula.
\(\therefore AB = \sqrt{(2 + 2)^2 + (0 - 0)^2} = 4\)
\(BC = \sqrt{(0 - 2)^2 + (2 - 0)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}\)
\(CA = \sqrt{(0 + 2)^2 + (2 - 0)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2}\)
Now, in \(\Delta PQR\), the length of side PQ, QR and PR will be calculate by distance formula.
\(PQ = \sqrt{(4 + 4)^2 + (0 - 0)^2} = \sqrt{8^2} = 8\)
\(QR = \sqrt{(0 - 4)^2 + (4 - 0)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}\)
\(PR = \sqrt{(0 + 4)^2 + (4 - 0)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2}\)
Now, for \(\Delta ABC\) and \(\Delta PQR\) to be similar, their corresponding sides should be proportional.
i.e., \(\frac{AB}{PQ} = \frac{4}{8} = \frac{1}{2}\)
\(\frac{BC}{QR} = \frac{2\sqrt{2}}{4\sqrt{2}} = \frac{1}{2}\)
\(\frac{CA}{PR} = \frac{2\sqrt{2}}{4\sqrt{2}} = \frac{1}{2}\)
So \(\frac{AB}{PQ} = \frac{BC}{QR} = \frac{CA}{PR}\)
\(\therefore \Delta ABC \sim \Delta PQR\)
Hence proved.