CBSE Class 10 Maths HOTs Co-Ordinate Geometry Set 03

Question. Find the distance between the following pairs of points: (2, 3), (4, 1)
Answer: Let \( A(2, 3) \) and \( B(4, 1) \) are given points.
The distance between two points is
\( AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)
\( \implies \) \( AB = \sqrt{(4 - 2)^2 + (1 - 3)^2} \)
\( \implies \) \( AB = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \) units.

Question. Find the distance between the following pairs of points: (– 5, 7), (– 1, 3)
Answer: Let \( A(-5, 7) \) and \( B(-1, 3) \) are given points.
\( AB = \sqrt{(-1 + 5)^2 + (3 - 7)^2} \)
\( \implies \) \( AB = \sqrt{16 + 16} \)
\( \implies \) \( AB = \sqrt{32} = \sqrt{16 \times 2} \)
\( \implies \) \( AB = 4\sqrt{2} \) units.

Question. Find the distance between the following pairs of points: (a, b), (– a, – b)
Answer: Let \( A(a, b) \) and \( B(-a, -b) \) are given points.
\( AB = \sqrt{(-a - a)^2 + (-b - b)^2} \)
\( \implies \) \( AB = \sqrt{(-2a)^2 + (-2b)^2} \)
\( \implies \) \( AB = \sqrt{4a^2 + 4b^2} = 2\sqrt{a^2 + b^2} \) units.

Question. Determine if the points (1, 5), (2, 3) and (–2, –11) are collinear.
Answer: Let points be \( A(1, 5), B(2, 3) \) and \( C(-2, -11) \)
\( AB = \sqrt{(2 - 1)^2 + (3 - 5)^2} = \sqrt{1 + 4} = \sqrt{5} \) units.
\( BC = \sqrt{(-2 - 2)^2 + (-11 - 3)^2} = \sqrt{16 + 196} = \sqrt{212} = 2\sqrt{53} \) units.
\( AC = \sqrt{(-2 - 1)^2 + (-11 - 5)^2} = \sqrt{9 + 256} = \sqrt{265} = \sqrt{5 \times 53} \) units.
\( AB + BC \neq AC \)
Hence, the given points are not collinear.

Question. Check whether (5, –2), (6, 4) and (7, –2) are the vertices of an isosceles triangle.
Answer: Let points be \( A(5, -2), B(6, 4) \) and \( C(7, -2) \)
\( AB = \sqrt{(6 - 5)^2 + (4 + 2)^2} = \sqrt{1 + 36} = \sqrt{37} \) units
\( BC = \sqrt{(7 - 6)^2 + (-2 - 4)^2} = \sqrt{1 + 36} = \sqrt{37} \) units
\( AC = \sqrt{(7 - 5)^2 + (-2 + 2)^2} = \sqrt{4 + 0} = 2 \) units
Here, \( AB = BC \)
\( \therefore \) \( \Delta ABC \) is an isosceles triangle.

Question. Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer: (–1, –2), (1, 0), (–1, 2), (–3, 0)
Answer: Let \( A(-1, -2), B(1, 0), C(-1, 2) \) and \( D(-3, 0) \).
The distance between two points is \( \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \).
\( AB = \sqrt{(1 + 1)^2 + (0 + 2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \) units
\( BC = \sqrt{(-1 - 1)^2 + (2 - 0)^2} = \sqrt{4 + 4} = 2\sqrt{2} \) units
\( CD = \sqrt{(-3 + 1)^2 + (0 - 2)^2} = \sqrt{4 + 4} = 2\sqrt{2} \) units
\( AD = \sqrt{(-3 + 1)^2 + (0 + 2)^2} = \sqrt{4 + 4} = 2\sqrt{2} \) units
\( AC = \sqrt{(-1 + 1)^2 + (2 + 2)^2} = \sqrt{0 + 16} = 4 \) units
\( BD = \sqrt{(-3 - 1)^2 + (0 - 0)^2} = 16 = 4 \) units
Here \( AC = BD \), \( AB = BC = CD = AD \)
Hence, the quadrilateral ABCD is a square.

Question. Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer: (–3, 5), (3, 1), (0, 3), (–1, –4)
Answer: Let points be \( A(-3, 5), B(3, 1), C(0, 3) \) and \( D(-1, -4) \).
\( AB = \sqrt{(3 + 3)^2 + (1 - 5)^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13} \) units
\( BC = \sqrt{(0 - 3)^2 + (3 - 1)^2} = \sqrt{9 + 4} = \sqrt{13} \) units
\( CD = \sqrt{(-1 - 0)^2 + (-4 - 3)^2} = \sqrt{1 + 49} = \sqrt{50} = 5\sqrt{2} \) units
\( AD = \sqrt{(-3 + 1)^2 + (5 + 4)^2} = \sqrt{4 + 81} = \sqrt{85} \) units
The given points do not form any quadrilateral.

Question. Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer: (4, 5), (7, 6), (4, 3), (1, 2)
Answer: Let points be \( A(4, 5), B(7, 6), C(4, 3) \) and \( D(1, 2) \).
\( AB = \sqrt{(7 - 4)^2 + (6 - 5)^2} = \sqrt{9 + 1} = \sqrt{10} \) units
\( BC = \sqrt{(4 - 7)^2 + (3 - 6)^2} = \sqrt{9 + 9} = 3\sqrt{2} \) units
\( CD = \sqrt{(1 - 4)^2 + (2 - 3)^2} = \sqrt{9 + 1} = \sqrt{10} \) units
\( AD = \sqrt{(1 - 4)^2 + (2 - 5)^2} = \sqrt{9 + 9} = 3\sqrt{2} \) units
\( AC = \sqrt{(4 - 4)^2 + (3 - 5)^2} = \sqrt{4} = 2 \) units
\( BD = \sqrt{(1 - 7)^2 + (2 - 6)^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13} \) units
Here, \( AB = CD \), \( BC = AD \) and \( AC \neq BD \)
\( \therefore \) The quadrilateral ABCD is a parallelogram.

Question. Find the point on the x-axis which is equidistant from (2, –5), and (–2, 9).
Answer: Let points be \( A(2, -5) \) and \( B(-2, 9) \).
Let \( P(x, 0) \) be the point on x-axis.
\( \therefore \) \( PA = PB \)
\( \implies \) \( \sqrt{(x - 2)^2 + (0 + 5)^2} = \sqrt{(x + 2)^2 + (0 - 9)^2} \)
\( \implies \) \( (x - 2)^2 + 25 = (x + 2)^2 + 81 \)
\( \implies \) \( x^2 + 4 - 4x + 25 = x^2 + 4 + 4x + 81 \)
\( \implies \) \( -4x - 4x = 81 - 25 \)
\( \implies \) \( -8x = 56 \)
\( \implies \) \( x = -7 \)
\( \therefore \) The required point is \( (-7, 0) \).

Question. Find the values of y for which the distance between the points P(2, –3) and Q(10, y) is 10 units.
Answer: Points \( P(2, -3), Q(10, y) \) and \( PQ = 10 \) units.
The distance between two points is
\( \sqrt{(10 - 2)^2 + (y + 3)^2} = 10 \)
\( \implies \) \( 64 + y^2 + 9 + 6y = 100 \)
\( \implies \) \( y^2 + 6y + 73 - 100 = 0 \)
\( \implies \) \( y^2 + 6y - 27 = 0 \)
\( \implies \) \( y^2 + 9y - 3y - 27 = 0 \)
\( \implies \) \( y(y + 9) - 3(y + 9) = 0 \)
\( \implies \) \( (y - 3)(y + 9) = 0 \)
\( \implies \) \( y - 3 = 0 \) or \( y + 9 = 0 \)
\( \implies \) \( y = 3 \) or \( -9 \).

Question. If Q(0, 1) is equidistant from P(5, –3), and R(x, 6), find the values of x. Also find the distances QR and PR.
Answer: Points are \( P(5, -3) \) and \( R(x, 6) \).
Point \( Q(0, 1) \) is equidistant from points P and R.
\( \therefore \) \( QP = QR \)
\( \implies \) \( \sqrt{(5 - 0)^2 + (-3 - 1)^2} = \sqrt{(x - 0)^2 + (6 - 1)^2} \)
\( \implies \) \( 25 + 16 = x^2 + 25 \)
\( \implies \) \( 25 + 16 = x^2 + 25 \)
\( \implies \) \( x^2 = 16 \)
\( \implies \) \( x = \pm 4 \).
\( QR = \sqrt{(x - 0)^2 + (6 - 1)^2} = \sqrt{x^2 + 25} = \sqrt{16 + 25} = \sqrt{41} \)
\( \therefore \) \( QR = \sqrt{41} \) units.
When \( x = 4 \),
\( PR = \sqrt{(4 - 5)^2 + (6 + 3)^2} = \sqrt{(-1)^2 + 81} = \sqrt{1 + 81} = \sqrt{82} \) units.
When \( x = -4 \), \( PR = 9\sqrt{2} \) units.

Question. Find a relation between x and y such that the point (x, y) is equidistant from the points (3, 6) and (–3, 4).
Answer: Points \( A(3, 6), B(-3, 4) \) and point \( P(x, y) \) is equidistant from points A and B.
\( \therefore \) \( AP = BP \)
\( \implies \) \( \sqrt{(x - 3)^2 + (y - 6)^2} = \sqrt{(x + 3)^2 + (y - 4)^2} \)
\( \implies \) \( (x - 3)^2 + (y - 6)^2 = (x + 3)^2 + (y - 4)^2 \)
\( \implies \) \( x^2 + 9 - 6x + y^2 + 36 - 12y = x^2 + 9 + 6x + y^2 + 16 - 8y \)
\( \implies \) \( -6x - 6x - 12y + 8y + 45 - 25 = 0 \)
\( \implies \) \( -12x - 4y + 20 = 0 \)
Divided by \( -4 \), we get \( 3x + y - 5 = 0 \).

Question. Find the coordinates of the point which divides the join of (–1, 7) and (4, –3) in the ratio 2 : 3.
Answer: Let \( P(x, y) \) be the point.
\( A(-1, 7), B(4, -3) \), Ratio \( 2 : 3 \).
\( x = \frac{m_1x_2 + m_2x_1}{m_1 + m_2} = \frac{2(4) + 3(-1)}{2 + 3} = \frac{8 - 3}{5} = 1 \)
\( x = 1 \).
\( y = \frac{m_1y_2 + m_2y_1}{m_1 + m_2} = \frac{2(-3) + 3(7)}{2 + 3} = \frac{-6 + 21}{5} = \frac{15}{5} = 3 \).
\( y = 3 \).
Then, the coordinates of point are \( (1, 3) \).

Question. Find the coordinates of the points of trisection of the line segment joining (4, –1) and (–2, –3).
Answer: Let points P and Q trisect the line joining the points.
\( A(4, -1), B(-2, -3) \).
\( \therefore \) \( AP = PQ = QB \).
P divides AB in the ratio 1: 2 and Q divides AB in the ratio 2:1.
\( P (\text{coordinate } x) = \frac{1(-2) + 2(4)}{1 + 2} = \frac{6}{3} = 2 \);
\( P (\text{coordinate } y) = \frac{1(-3) + 2(-1)}{1 + 2} = -\frac{5}{3} \).
The coordinates of P are \( (2, -\frac{5}{3}) \).
\( Q (\text{x-coordinate}) = \frac{2(-2) + 1(4)}{2 + 1} = \frac{-4 + 4}{3} = 0 \);
\( Q (\text{y-coordinate}) = \frac{2(-3) + 1(-1)}{2 + 1} = -\frac{7}{3} \).
The coordinates of Q are \( (0, -\frac{7}{3}) \).

Question. To conduct Sports Day Activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1 m each. 100 flower pots have been placed at a distance of 1 m from each other along AD, as shown in given figure. Niharika runs \( \frac{1}{4} \) th the distance AD on the 2nd line and posts a green flag. Preet runs \( \frac{1}{5} \) th distance AD on the eighth line and posts a red flag. What is the distance between both the flags?
Answer: The distance AD is 100 m.
Niharika's distance \( = \frac{1}{4} \times 100 = 25 \) m. Her position is \( (2, 25) \).
Preet's distance \( = \frac{1}{5} \times 100 = 20 \) m. Her position is \( (8, 20) \).
Distance between both flags \( = \sqrt{(8 - 2)^2 + (20 - 25)^2} \)
\( = \sqrt{6^2 + (-5)^2} = \sqrt{36 + 25} = \sqrt{61} \) m.