CBSE Class 10 Mathematics Coordinate Geometry Worksheet Set 06

Question. The distance of the point (-5, 12) from the y-axis is 
(a) 12 units
(b) 5 units
(c) 13 units
(d) -5 units
Answer: (b) 5 units
Explanation: The distance of any point from y-axis is its abscissa. Therefore, the required distance is 5 units.

Question. The perimeter of a triangle with vertices (0, 4), (0, 0) and (3, 0) is 
(a) 15 units
(b) 10 units
(c) 9 units
(d) 12 units
Answer: (d) 12 units
Explanation: Given: the vertices of a triangle ABC, A(0, 4), B (0, 0) and C (3, 0).
\( \therefore \) Perimeter of triangle ABC = AB + BC + AC
\( = \sqrt{(0 - 0)^2 + (0 - 4)^2} + \sqrt{(0 - 3)^2 + (0 - 0)^2} + \sqrt{(0 - 3)^2 + (4 - 0)^2} \)
\( = \sqrt{0 + 16} + \sqrt{9 + 0} + \sqrt{9 + 16} \)
\( = \sqrt{16} + \sqrt{9} + \sqrt{25} \)
= 4 + 3 + 5 = 12 units

Question. AOBC is a rectangle whose three vertices are A(0, 3), O(0, 0) and B(5, 0). The length of its diagonal is 
(a) \( 2\sqrt{34} \) units
(b) 3 units
(c) \( \sqrt{34} \) units
(d) 4 units
Answer: (c) \( \sqrt{34} \) units
Explanation: In rectangle AOBC, AB is a diagonal.
\( \therefore AB = \sqrt{(5 - 0)^2 + (0 - 3)^2} \)
\( = \sqrt{25 + 9} = \sqrt{34} \text{ units} \)

Question. A circle has its centre at the origin and a point P(5, 0) lies on it. Then the point Q(8, 6) lies ________ the circle.
(a) in side
(b) out side
(c) on
(d) None of the options
Answer: (b) out side
Explanation: Given: Coordinates of centre O (0, 0) and Radius is OP.
\( \therefore OP = \sqrt{(5 - 0)^2 + (0 - 0)^2} \)
\( = \sqrt{25 + 0} = \sqrt{25} = 5 \text{ units} \)
Now, \( OQ = \sqrt{(8 - 0)^2 + (6 - 0)^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \text{ units} \)
Since OQ > OP
Therefore, point Q lies outside the circle.

Question. The point where the medians of a triangle meet is called the ________ of the triangle 
(a) circumcentre
(b) centroid
(c) orthocentre
(d) None of the options
Answer: (b) centroid
Explanation: The point where three medians of a triangle meet is called the centroid of the triangle.it is the centre of gravity of the triangle. it divides the median in the ratio 2 :1

Question. Find the points X-axis which are at a distance of \( 2\sqrt{5} \) from the point(7,-4). How many such points are there? 
Answer: We have to find the points on X-axis which are at a distance of \( 2\sqrt{5} \) from the point(7,-4). Also,we will how many such points are there.
Let, the point on X-axis be (x,0).
Now, by using distance formula,
\( \sqrt{(x_2 - x_1)^2 + (y_2 - y_2)^2} \)
\( \sqrt{(x - 7)^2 + (0 + 4)^2} = 2\sqrt{5} \)
Squaring both sides,

\( \implies (x - 7)^2 + 4^2 = (2\sqrt{5})^2 \)

\( \implies x^2 - 14x + 49 + 16 = 20 \)

\( \implies x^2 - 14x + 45 = 0 \)

\( \implies (x - 9) (x - 5) = 0 \)

\( \implies x = 9 \text{ or } x = 5 \)
Hence, two points exists (9,0) and (5,0)

Question. Find the coordinates of the midpoint of the line segment joining A(3, 0) and B(-5, 4). 
Answer: Mid-point of the line segment joining the points A(3, 0) and B(-5, 4)
\( = \left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right) \)
\( = \left( \frac{3-5}{2}, \frac{0+4}{2} \right) \)
\( = \left( \frac{-2}{2}, \frac{4}{2} \right) \)
= (-1, 2)
Hence the coordinate of mid point of line segment is (-1, 2).

Question. Find the complement of the given angle. (1)
Answer: Complement of the angle 20° = 90° – 20° = 70°

Question. Find the distance between the points A and B in A(5, - 8), B (-7, - 3) 
Answer: \( AB = \sqrt{(-7 - 5)^2 + (-3 + 8)^2} = 13 \).

Question. Find the distance between the following pairs of points: (a, b), (-a, -b) 
Answer: Applying Distance Formula to find distance between points (a, b) and (-a, -b), we get
\( d = \sqrt{(-a - a)^2 + (-b - b)^2} = \sqrt{(-2a)^2 + (-2b)^2} \)
\( = \sqrt{4a^2 + 4b^2} = \sqrt{4(a^2 + b^2)} = 2\sqrt{a^2 + b^2} \)

Question. Find the distance of the point P(6, -6) from the origin. 
Answer: Let P(6, -6 ) be the given point and O(0, 0) be the origin.
Then, \( OP = \sqrt{(6 - 0)^2 + (-6 - 0)^2} = \sqrt{6^2 + (-6)^2} \)
\( = \sqrt{36 + 36} = \sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2} \text{ units.} \)

Question. Find the distance between the points P(-6, 7) and Q(-1, -5). 
Answer: Here, \( x_1 = -6, y_1 = 7 \) and \( x_2 = -1, y_2 = -5 \)
Therefore,by distance formula,we have,
\( PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)

\( \implies PQ = \sqrt{(-1 + 6)^2 + (-5 - 7)^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \)

Question. In what ratio does the point C(4, 5) divide the join of A(2, 3) and B(7, 8)? 
Answer: Let the point C(4, 5) divides the join of A(2, 3) and B(7, 8) in the ratio k:1
The point C is \( \left( \frac{7k+2}{k+1}, \frac{8k+3}{k+1} \right) \)
But C is (4, 5)

\( \implies \frac{7k+2}{k+1} = 4 \) or \( 7k + 2 = 4k + 4 \)
or \( 3k = 2 \therefore k = \frac{2}{3} \)
Thus, C divides AB in the ratio 2:3

Question. Show that quadrilateral PQRS formed by vertices P(22,5), Q(7,10), R(12,11) and S(3,24) is not a parallelogram. 
Answer: Given vertices of quadrilateral are P(22, 5), Q(7, 10), R(12, 11) and S(3, 24).
Now, \( PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(7 - 22)^2 + (10 - 5)^2} = \sqrt{(-15)^2 + (5)^2} = \sqrt{(225) + (25)} = 5\sqrt{10} \text{ units} \)
\( QR = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(12 - 7)^2 + (11 - 10)^2} = \sqrt{(5)^2 + (1)^2} = \sqrt{(25) + (1)} = \sqrt{26} \text{ units} \)
\( RS = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(3 - 12)^2 + (24 - 11)^2} = \sqrt{(-9)^2 + (13)^2} = \sqrt{(81) + (169)} = 5\sqrt{10} \text{ units} \)
\( SP = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(22 - 3)^2 + (5 - 24)^2} = \sqrt{(19)^2 + (-19)^2} = 19\sqrt{2} \text{ units} \)
Here, we see that opposite sides of a quadrilateral are not equal i.e. \( QR \neq SP \).
Hence, given vertices of a quadrilateral are not forming a parallelogram.

Question. Find the coordinates of points which trisect the line segment joining (1, -2) and (-3, 4).
Answer: Let A (1, -2) and B (-3,4) be the given points.
Let the points of trisection be P and Q. Then, AP = PQ = QB = X(say)
PB = PQ + QB = \( 2\lambda \) and AQ = AP + PQ = \( 2\lambda \)

\( \implies AP : PB = \lambda : 2\lambda = 1 : 2 \) and \( AQ : QB = 2\lambda : \lambda = 2 : 1 \)
So, P divides AB internally in the ratio 1 : 2 while Q divides internally in the ratio 2 : 1.
Thus, the coordinates of P and Q are
\( P \left( \frac{1 \times -3 + 2 \times 1}{1+2}, \frac{1 \times 4 + 2 \times -2}{1+2} \right) = P \left( \frac{-1}{3}, 0 \right) \)
\( Q \left( \frac{2 \times -3 + 1 \times 1}{2+1}, \frac{2 \times 4 + 1 \times (-2)}{2+1} \right) = Q \left( \frac{-5}{3}, 2 \right) \)
Hence, the two points of trisection are (-1/3, 0) and (-5/3, 2)

Question. Find the value(s) of p, if the points A(2, 3), B(4, k), C(6, - 3) are collinear. 
Answer: Let the points A (2, 3), B (4, k) and C (6, – 3) be collinear.
If the points are collinear then area of triangle ABC formed by these three points is 0.
\( \therefore \text{ar} (\triangle ABC) = \frac{1}{2} |x_1 (y_2 - y_3) + x_2 (y_3 - y_1) + x_3 (y_1 - y_2)| = 0 \)

\( \implies \frac{1}{2} [2(k + 3) + 4(-3 - 3) + 6(3 - k)] = 0 \)

\( \implies [2k + 6 - 24 + 18 - 6k] = 0 \)

\( \implies [-4k] = 0 \)

\( \implies k = 0 \)

Question. Show that the points A (2,-2), B(14,10), C (11, 13) and D(-1, 1) are the vertices of a rectangle.
Answer: According to the question, A (2,-2), B(14,10), C (11, 13) and D(-1, 1)
\( AB = \sqrt{(14 - 2)^2 + (10 + 2)^2} = 12\sqrt{2} \text{ units} \)
\( BC = \sqrt{(11 - 14)^2 + (13 - 10)^2} = 3\sqrt{2} \text{ units} \)
\( CD = \sqrt{(-1 - 11)^2 + (1 - 13)^2} = 12\sqrt{2} \text{ units} \)
\( AD = \sqrt{(-1 - 2)^2 + (1 + 2)^2} = 3\sqrt{2} \text{ units} \)

\( \implies AB = CD \) and \( BC = AD \)
\( \therefore \) ABCD is a parallelogram.
Now, \( AC = \sqrt{(11 - 2)^2 + (13 + 2)^2} = \sqrt{306} \)

\( \implies AC^2 = 306 \text{ units}, AB^2 = 288 \text{ units.} \)
\( BC^2 = 18 \text{ units} \)
\( AB^2 + BC^2 = 306 \text{ units.} \)

\( \implies AC^2 = AB^2 + BC^2 \)

\( \implies \angle ABC = 90^\circ \)

\( \implies ABCD \) is a rectangle

Question. Find the coordinates of the points which divide the line segment joining A(-2, 2) and B(2, 8) into four equal parts. 
Answer: Let P \( (x_1, y_1) \), Q \( (x_2, y_2) \) and R \( (x_3, y_3) \) be the points which divide the line segment AB into four equal parts.
Then, P divides AB in the ratio 1 : 3 internally.
\( x = \frac{mx_2 + nx_1}{m+n} \)
\( \therefore x_1 = \frac{(1)(2) + (3)(-2)}{1+3} \)
\( = \frac{2 - 6}{4} = -\frac{4}{4} = -1 \)
\( y = \frac{my_2 + ny_1}{m+n} \)
\( y_1 = \frac{(1)(8) + (3)(2)}{1+3} \)
\( = \frac{8 + 6}{4} = \frac{14}{4} = \frac{7}{2} \)
So, \( P \rightarrow (-1, \frac{7}{2}) \)
Also, Q divides AB in the ratio 1 : 1 i.e. Q is the mid point of AB
\( x_2 = \frac{-2+2}{2} = 0 \)
\( y_2 = \frac{2+8}{2} = \frac{10}{2} = 5 \)
So, \( Q \rightarrow (0, 5) \)
and, R divides AB in the ratio 3 : 1
\( \therefore x_2 = \frac{(3)(2) + (1)(-2)}{3+1} \)
\( = \frac{6 - 2}{4} = \frac{4}{4} = 1 \)
\( y_3 = \frac{(3)(8) + (1)(2)}{3+1} \)
\( = \frac{24 + 2}{4} = \frac{26}{4} = \frac{13}{2} \)
So, \( R \rightarrow (1, \frac{13}{2}) \)

Question. If the points A(1,-2), B(2,3), C(-3,2) and D(-4,-4) are the vertices of the parallelogram ABCD, then taking AB as the base, find the height of the parallelogram. 
Answer: Let DM = h be the height of the parallelogram ABCD when AB is taken as the base.
Area of \( \triangle ABD = \frac{1}{2} \times (AB \times DM) \)

\( \implies \triangle ABD = \frac{1}{2} \times (AB \times h) \)

\( \implies h = \frac{2(\text{area } \triangle ABD)}{AB} \) ...(i)
Now, first find the length of AB by using distance formula,
\( AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} = \sqrt{(2 - 1)^2 + (3 + 2)^2} = \sqrt{26} \)
Since, the coordinates of vertices of \( \triangle ABD \) are A(-1, 2), B(2, 3) and D(-4, -3).
Therefore, area of \( \triangle ABD = \frac{1}{2} |1(3 + 3) + 2( - 3 + 2) + ( - 4)( - 2 - 3)| \)
= \( \frac{1}{2} [1 (6) + 2(-1) - 4 (-5)] \)
= \( \frac{1}{2} [6 - 2 + 20] \)
= \( \frac{1}{2} [24] \)
= 12 sq units.
Now, putting the value of AB and area of \( \triangle ABD \) in Eq(i), we get
\( h = \frac{2 \times 12}{\sqrt{26}} = \frac{24}{\sqrt{26}} \times \frac{\sqrt{26}}{\sqrt{26}} \)
\( = \frac{24\sqrt{26}}{26} \)
\( = \frac{12\sqrt{26}}{13} \text{ units} \)

Question. Find the area of the triangle whose sides are along the lines x = 2, y = 0 and 4x + 5y = 20. 
Answer: A is point of intersection of line x = 2 and 4x + 5y = 20

\( \implies 4 \times 2 + 5y = 20 \)

\( \implies y = \frac{12}{5} \)
\( \therefore \) Coordinates of A are \( (2, \frac{12}{5}) \)
B is the point of intersection of x = 2 and y = 0
\( \therefore \) Coordinates of B are (2, 0).
C is point of intersection y = 0 and 4x + 5y = 20

\( \implies 4x + 5 \times 0 = 20 \)

\( \implies x = 5 \)

\( \implies \) Coordinates of C are (5, 0)
Area \( \triangle ABC = \frac{1}{2} |x_1 (y_2 - y_3) + x_2 (y_3 - y_1) + x_3 (y_1 - y_2)| \)
\( = \frac{1}{2} |2(0 - 0) + 2(0 - \frac{12}{5}) + 5(\frac{12}{5} - 0)| \)
\( = \frac{1}{2} |-\frac{24}{5} + 12| \)
\( = \frac{1}{2} \times \frac{36}{5} \)
\( = \frac{18}{5} \text{ sq. units} \)
= 3.6 sq. units

Question. Find the ratio in which the line segment joining the points \( (-3, 10) \) and \( (6, -8) \) is divided by \( (-1, 6) \).
Answer: In Fig. 6.4, let the point \( P(-1, 6) \) divides the line joining \( A(-3, 10) \) and \( B(6, -8) \) in the ratio \( k : 1 \) then, the coordinates of \( P \) are \( \left( \frac{6k - 3}{k + 1}, \frac{-8k + 10}{k + 1} \right) \).
But, the coordinates of \( P \) are \( (-1, 6) \). (Given)
\( \therefore \) \( \frac{6k - 3}{k + 1} = -1 \)
\( \implies \) \( 6k - 3 = -k - 1 \)
\( \implies \) \( 6k + k = 3 - 1 \)
\( \implies \) \( 7k = 2 \)
\( \implies \) \( k = \frac{2}{7} \)
Hence, the point \( P \) divides \( AB \) in the ratio \( 2 : 7 \).

Question. If \( Q(0, 1) \) is equidistant from \( P(5, -3) \) and \( R(x, 6) \) find the value of \( x \). Also, find the distances of \( QR \) and \( PR \).
Answer: Since, point \( Q(0, 1) \) is equidistant from \( P(5, -3) \) and \( R(x, 6) \).
Therefore, \( QP = QR \)
Squaring both sides, we have, \( QP^2 = QR^2 \)
\( \implies \) \( (5 - 0)^2 + (-3 - 1)^2 = (x - 0)^2 + (6 - 1)^2 \)
\( \implies \) \( 25 + 16 = x^2 + 25 \)
\( \implies \) \( x^2 = 16 \)
\( \therefore \) \( x = \pm 4 \)
Thus, \( R \) is \( (4, 6) \) or \( (-4, 6) \).
Now, \( QR = \sqrt{(4 - 0)^2 + (6 - 1)^2} = \sqrt{16 + 25} = \sqrt{41} \) units
or, \( QR = \sqrt{(-4 - 0)^2 + (6 - 1)^2} = \sqrt{16 + 25} = \sqrt{41} \) units
and \( PR = \sqrt{(4 - 5)^2 + (6 + 3)^2} = \sqrt{1 + 81} = \sqrt{82} \) units
or, \( PR = \sqrt{(-4 - 5)^2 + (6 + 3)^2} = \sqrt{81 + 81} = 9\sqrt{2} \) units

Question. In what ratio does the point \( (- 4, 6) \) divide the line segment joining the points \( A(- 6, 10) \) and \( B(3, - 8) \)?
Answer: Let \( (-4, 6) \) divide \( AB \) internally in the ratio \( m_1 : m_2 \). Using the section formula, we get
\( (-4, 6) = \left( \frac{3m_1 - 6m_2}{m_1 + m_2}, \frac{-8m_1 + 10m_2}{m_1 + m_2} \right) \)
So, \( -4 = \frac{3m_1 - 6m_2}{m_1 + m_2} \) and \( 6 = \frac{-8m_1 + 10m_2}{m_1 + m_2} \)
Now, \( -4 = \frac{3m_1 - 6m_2}{m_1 + m_2} \) gives
\( -4m_1 - 4m_2 = 3m_1 - 6m_2 \)
i.e., \( 7m_1 = 2m_2 \)
i.e., \( m_1 : m_2 = 2 : 7 \)
Verification:
Also, \( \frac{-8m_1 + 10m_2}{m_1 + m_2} = \frac{-8 \frac{m_1}{m_2} + 10}{\frac{m_1}{m_2} + 1} \) (Dividing throughout by \( m_2 \))
\( = \frac{-8 \times \frac{2}{7} + 10}{\frac{2}{7} + 1} = 6 \)
Therefore, the point \( (-4, 6) \) divides the line segment joining the points \( A(-6, 10) \) and \( B(3, -8) \) in the ratio \( 2 : 7 \).

Question. Find the ratio in which the line segment joining \( A(1, - 5) \) and \( B(- 4, 5) \) is divided by the \( x \)-axis. Also find the coordinates of the point of division.
Answer: Let the required ratio be \( k : 1 \). Then, the coordinates of the point of division is \( P\left( \frac{-4k + 1}{k + 1}, \frac{5k - 5}{k + 1} \right) \).
Since, this point lies on \( x \)-axis. Therefore its \( y \)-coordinate is zero.
i.e., \( \frac{5k - 5}{k + 1} = 0 \)
\( \implies \) \( 5k - 5 = 0 \)
\( 5k = 5 \) or \( k = \frac{5}{5} = 1 \)
Thus, the required ratio is \( 1 : 1 \) and the point of division is \( P\left( \frac{-4 \times 1 + 1}{1 + 1}, \frac{5 \times 1 - 5}{1 + 1} \right) \)
i.e., \( P\left( -\frac{3}{2}, 0 \right) \)

Question. Find the area of a rhombus if its vertices \( (3, 0) \), \( (4, 5) \), \( (- 1, 4) \) and \( (- 2, - 1) \) are taken in order.
Answer: Let \( A(3, 0) \), \( B(4, 5) \), \( C(-1, 4) \) and \( D(-2, -1) \) be the vertices of a rhombus.
Therefore, its diagonals
\( AC = \sqrt{(-1 - 3)^2 + (4 - 0)^2} = \sqrt{16 + 16} = \sqrt{32} = 4\sqrt{2} \)
and \( BD = \sqrt{(-2 - 4)^2 + (-1 - 5)^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2} \)
\( \therefore \) Area of rhombus \( ABCD = \frac{1}{2} \times \text{(Product of length of diagonals)} \)
\( = \frac{1}{2} \times AC \times BD = \frac{1}{2} \times 4\sqrt{2} \times 6\sqrt{2} = 24 \) sq units.

Multiple Choice Questions

Choose and write the correct option in the following questions.

Question. The co-ordinate of the point which is reflection of the point \( (-3, 5) \) in X-axis are 
(a) \( (3, 5) \)
(b) \( (3, -5) \)
(c) \( (-3, -5) \)
(d) \( (-3, 5) \)
Answer: (c) (-3, -5)

Question. The distance between the points \( (0, 0) \) and \( (a - b, a + b) \) is 
(a) \( 2\sqrt{ab} \)
(b) \( \sqrt{2a^2 + ab} \)
(c) \( 2\sqrt{a^2 + b^2} \)
(d) \( \sqrt{2a^2 + 2b^2} \)
Answer: (d) \(\sqrt{2a^2 + 2b^2}\)

Question. The distance between the points \( (a \cos \theta + b \sin \theta, 0) \) and \( (0, a \sin \theta - b \cos \theta) \) is 
(a) \( a^2 + b^2 \)
(b) \( a^2 - b^2 \)
(c) \( \sqrt{a^2 + b^2} \)
(d) \( \sqrt{a^2 - b^2} \)
Answer: (c) \(\sqrt{a^2 + b^2}\)

Question. If the distance between the points \( (2, -2) \) and \( (-1, x) \) is 5, one of the values of \( x \) is
(a) -2
(b) 2
(c) -1
(d) 1
Answer: (b) 2

Question. The distance of the point \( P(2, 3) \) from the x-axis is 
(a) 2 units
(b) 3 units
(c) 1 units
(d) 5 units
Answer: (b) 3 units

Question. The coordinate of point \( P \) on X-axis equidistant from the points \( A(-1, 0) \) and \( B(5, 0) \) is 
(a) \( (2, 0) \)
(b) \( (0, 2) \)
(c) \( (3, 0) \)
(d) \( (2, 2) \)
Answer: (a) (2, 0)

Question. The distance between the points \( A(0, 6) \) and \( B(0, -2) \) is 
(a) 6 units
(b) 8 units
(c) 4 units
(d) 2 units
Answer: (b) 8 units

Question. The distance between two points, \( M \) and \( N \), on a graph is given as \( \sqrt{10^2 + 7^2} \). The coordinates of point \( M \) are \( (-4, 3) \). Given that the point \( N \) lies in the first quadrant, which of the following is true about the all possible \( x \) coordinates of point \( N \)? 
(a) They are multiple of 3.
(b) They are multiple of 4.
(c) They are multiple of 5.
(d) They are multiple of 6.
Answer: (d) They are multiple of 6.

Question. The end points of diameter of circle are \( (2, 4) \) and \( (-3, -1) \). The radius of the circle is
(a) \( \frac{5\sqrt{2}}{2} \) units
(b) \( 5\sqrt{2} \) units
(c) \( 3\sqrt{2} \) units
(d) \( \frac{\pm 5\sqrt{2}}{2} \) units
Answer: (a) \(\frac{5\sqrt{2}}{2}\) units

Question. If \( (a, b) \) is the mid point of the line segment joining the points \( A(10, -6) \) and \( B(k, 4) \) and \( a - 2b = 18 \), the values of \( k \) is 
(a) 30
(b) 22
(c) 4
(d) 40
Answer: (b) 22

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 the value of \( a \) is 
(a) -4
(b) -12
(c) 12
(d) -6
Answer: (b) -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 the coordinates of \( P \) and \( Q \) are, respectively 
(a) \( (0, -5) \) and \( (2, 0) \)
(b) \( (0, 10) \) and \( (-4, 0) \)
(c) \( (0, 4) \) and \( (-10, 0) \)
(d) \( (0, -10) \) and \( (4, 0) \)
Answer: (d) \( (0, -10) \) and \( (4, 0) \)

Question. The points which lie on the perpendicular bisector of the line segment joining the points \( A(-2, -5) \), \( B(2, 5) \) is
(a) \( (0, 0) \)
(b) \( (0, 2) \)
(c) \( (2, 0) \)
(d) \( (-2, 0) \)
Answer: (a) (0, 0)

Question. \( x \)-axis divides the join of \( (2, -3) \) and \( (5, 6) \) in the ratio 
(a) 1:2
(b) 2:1
(c) 2:5
(d) 5:2
Answer: (a) 1:2

Question. The value of \( k \) for which the points \( A(0, 1) \), \( B(2, k) \) and \( C(4, -5) \) are collinear is 
(a) 2
(b) -2
(c) 0
(d) 4
Answer: (b) -2

Question. The point on the \( x \)-axis which is equidistant from \( (-4, 0) \) and \( (10, 0) \) is 
(a) \( (7, 0) \)
(b) \( (5, 0) \)
(c) \( (0, 0) \)
(d) \( (3, 0) \)
Answer: (d) (3, 0)

Question. It is being given that the points \( A(1, 2) \), \( B(0, 0) \) and \( C(a, b) \) are collinear. Which of the following relations between \( a \) and \( b \) is true? 
(a) \( a = 2b \)
(b) \( 2a = b \)
(c) \( a + b = 0 \)
(d) \( a - b = 0 \)
Answer: (b) 2a = b

Question. The perpendicular bisector of the line segment joining the points \( A(1, 5) \) and \( B(4, 6) \) cuts the \( y \)-axis at 
(a) \( (0, 13) \)
(b) \( (0, -13) \)
(c) \( (0, 12) \)
(d) \( (13, 0) \)
Answer: (a) (0, 13)