CBSE Class 10 Mathematics Coordinate Geometry VBQs Set 06

Question. Point P divides the line segment joining the points A(–1, 3) and B(9, 8) such that \( \frac{AP}{PB} = \frac{k}{1} \). If P lies on the line \( x – y + 2 = 0 \), find the value of k.
Answer: Let the coordinates of P be \( (x, y) \).
By section formula, \( x = \frac{9k - 1}{k + 1} \) and \( y = \frac{8k + 3}{k + 1} \).
Since P lies on the line \( x - y + 2 = 0 \):
\( \implies \frac{9k - 1}{k + 1} - \frac{8k + 3}{k + 1} + 2 = 0 \)
\( \implies 9k - 1 - 8k - 3 + 2(k + 1) = 0 \)
\( \implies k - 4 + 2k + 2 = 0 \)
\( \implies 3k - 2 = 0 \)
\( \implies k = \frac{2}{3} \).

Question. Point A is on x-axis, point B is on y-axis and the point P lies on line segment AB, such that P (4, 5) and AP : PB = 5 : 3. Find the coordinates of points A and B.
Answer: Let A be \( (x, 0) \) and B be \( (0, y) \).
P divides AB in ratio 5 : 3.
Using section formula for P(4, 5):
\( 4 = \frac{5(0) + 3(x)}{5 + 3} \implies 4 = \frac{3x}{8} \implies 3x = 32 \implies x = \frac{32}{3} \)
\( 5 = \frac{5(y) + 3(0)}{5 + 3} \implies 5 = \frac{5y}{8} \implies y = 8 \)
Coordinates are \( A(\frac{32}{3}, 0) \) and \( B(0, 8) \).

Question. Centroid of the triangle formed by the points (a, b), (b, c) and (c, a) is at origin. If \( a^3 + b^3 + c^3 – 3abc = k^2 \) then find the value of k.
Answer: Centroid \( = (\frac{a+b+c}{3}, \frac{b+c+a}{3}) = (0, 0) \).
\( \implies a + b + c = 0 \).
We know that \( a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) \).
Since \( a + b + c = 0 \), the expression becomes 0.
\( k^2 = 0 \implies k = 0 \).

Question. Find the coordinates of the point of trisection of the line segment joining (1, –2) and (–3, 4).
Answer: Let the points be \( P_1 \) and \( P_2 \).
\( P_1 \) divides the segment in ratio 1 : 2:
\( x = \frac{1(-3) + 2(1)}{3} = -\frac{1}{3} \), \( y = \frac{1(4) + 2(-2)}{3} = 0 \). Point is \( (-\frac{1}{3}, 0) \).
\( P_2 \) divides the segment in ratio 2 : 1:
\( x = \frac{2(-3) + 1(1)}{3} = -\frac{5}{3} \), \( y = \frac{2(4) + 1(-2)}{3} = 2 \). Point is \( (-\frac{5}{3}, 2) \).

Question. Find the ratio in which the point (2, y) divides the line segment joining the points A(–2, 2) and B(3, 7). Also find the value of y.
Answer: Let ratio be \( k : 1 \).
\( 2 = \frac{3k - 2}{k + 1} \implies 2k + 2 = 3k - 2 \implies k = 4 \).
Ratio is 4 : 1.
Now, \( y = \frac{4(7) + 1(2)}{4 + 1} = \frac{28 + 2}{5} = 6 \).

Question. If (a, b) are the coordinates of the mid-point of line segment joining the points (2, 1) and (1, – 3), find the value of \( 3a^2 + 2ab + b^2 \).
Answer: \( a = \frac{2 + 1}{2} = 1.5 \), \( b = \frac{1 - 3}{2} = -1 \).
\( 3(1.5)^2 + 2(1.5)(-1) + (-1)^2 = 3(2.25) - 3 + 1 = 6.75 - 3 + 1 = 4.75 \).

Question. Find the lengths of the medians of \( \Delta ABC \) having vertices at A(5, 1), B(1, 5) and C(– 3, – 1).
Answer: Midpoints of sides: \( D(BC) = (-1, 2) \), \( E(AC) = (1, 0) \), \( F(AB) = (3, 3) \).
Median \( AD = \sqrt{(5 - (-1))^2 + (1 - 2)^2} = \sqrt{36 + 1} = \sqrt{37} \) units.
Median \( BE = \sqrt{(1 - 1)^2 + (5 - 0)^2} = 5 \) units.
Median \( CF = \sqrt{(-3 - 3)^2 + (-1 - 3)^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13} \) units.

Question. Point A is on x-axis, point B is on y-axis and point P lies on line segment AB, such that P[\( |(n+1)^2 – (2n–1)^2| \), \( \sqrt[3]{50n + 5^n} \)], where n = 2 and AP : PB = 5 : 3. Find coordinates of A and B.
Answer: For \( n = 2 \):
\( x = |(2+1)^2 - (2(2)-1)^2| = |9 - 9| = 0 \)
\( y = \sqrt[3]{50(2) + 5^2} = \sqrt[3]{125} = 5 \).
So \( P(0, 5) \).
Let \( A(x, 0) \) and \( B(0, y) \). Ratio 5 : 3.
\( 0 = \frac{5(0) + 3x}{8} \implies x = 0 \).
\( 5 = \frac{5y + 3(0)}{8} \implies 5y = 40 \implies y = 8 \).
Coordinates are \( A(0, 0) \) and \( B(0, 8) \).

INTEGRATED (MIXED) QUESTIONS

Question. If three points (0, 0), \( (3, \sqrt{3}) \) and (3, k) form an equilateral triangle, then k =
(a) 2
(b) – 3
(c) \( -\sqrt{3} \)
(d) – 2
Answer: (c) \( -\sqrt{3} \)

Question. The triangle whose vertices are (0, 0), (2.7, 0) and (0, 4.9) is a/an
(a) equilateral triangle
(b) right-angled triangle
(c) isosceles triangle
(d) obtuse-angled triangle
Answer: (b) right-angled triangle

Question. The distance of the point (h, k) from x-axis is
(a) h units
(b) |h| units
(c) k units
(d) |k| units
Answer: (d) |k| units

Question. The distance of the point (\( \alpha, \beta \)) from y-axis is
(a) \( \alpha \) units
(b) \( |\alpha| \) units
(c) \( \beta \) units
(d) \( |\beta| \) units
Answer: (b) \( |\alpha| \) units

Question. The number of points on x-axis which are at a distance of 2 units from (2, 4) is
(a) 1
(b) 2
(c) 0
(d) 3
Answer: (c) 0

Question. Line formed by joining (– 1, 1) and (5, 7) is divided by a line \( x + y = 4 \) in the ratio of
(a) 1 : 2
(b) 1 : 3
(c) 3 : 4
(d) 1 : 4
Answer: (a) 1 : 2

Question. Three vertices of a parallelogram taken in order are (– 1, – 6), (2, – 5) and (7, 2). The fourth vertex is
(a) (1, 4)
(b) (1, 1)
(c) (4, 4)
(d) (4, 1)
Answer: (d) (4, 1)

Question. AOBC is a rectangle whose three vertices are A(0, 3), O(0, 0) and B(5, 0). Then, the length of its diagonal is
(a) \( \sqrt{37} \) units
(b) \( \sqrt{35} \) units
(c) \( \sqrt{34} \) units
(d) \( \sqrt{33} \) units
Answer: (c) \( \sqrt{34} \) units

Question. Radius of circumcircle of a triangle ABC is \( 5\sqrt{10} \) units. If point P is equidistant from A(1, 3), B(–3, 5) and C(5, –1), then length of AP is
(a) \( 10\sqrt{5} \) units
(b) 25 units
(c) \( 5\sqrt{10} \) units
(d) 100 units
Answer: (c) \( 5\sqrt{10} \) units

Question. If points A(5, p), B(1, 5), C(2, 1) and D(6, 2) form a square ABCD, then p =
(a) 7
(b) 3
(c) 6
(d) 8
Answer: (c) 6

Question. The perimeter of the triangle formed by the points (0, 0), (2, 0) and (0, 2) is
(a) \( 1 – 2\sqrt{2} \) units
(b) \( 2\sqrt{2} + 1 \) units
(c) \( 4 + \sqrt{2} \) units
(d) \( 4 + 2\sqrt{2} \) units
Answer: (d) \( 4 + 2\sqrt{2} \) units

Question. The center of a circle is O(2a, a – 7). If the circle passes through the point A(1, – 9) and has diameter of length \( 10\sqrt{2} \) units, then the value of a is
(a) 9
(b) – 3
(c) 3
(d) \( \pm 3 \)
Answer: (d) \( \pm 3 \)

Question. If P(– 1, 1) is the mid point of the line segment joining A(– 3, b) and B(1, b + 4), then b =
(a) 1
(b) –1
(c) 2
(d) 0
Answer: (b) –1

Question. The ratio in which the X-axis divides the line segment joining A(3, 6) and B(12, – 3) is
(a) 2 : 1
(b) 1 : 2
(c) –2 : 1
(d) 1 : –2
Answer: (a) 2 : 1

Question. The coordinates of the fourth vertex C of the rectangle ABCD formed by the points A(0, 0), B(2, 0) and D(0, 3) are
(a) (3, 0)
(b) (0, 2)
(c) (2, 3)
(d) (3, 2)
Answer: (c) (2, 3)

Question. The line \( 3x + y – 6 = 0 \) divides the line segment joining A(1, – 1) and B(3, 6) in the ratio
(a) 2 : 5
(b) 4 : 9
(c) 2 : 7
(d) 2 : 3
Answer: (b) 4 : 9

Question. The point which divides the line segment joining the points (7, – 6) and (3, 4) in the ratio 1 : 2 internally lies in
(a) I quadrant
(b) II quadrant
(c) III quadrant
(d) IV quadrant
Answer: (d) IV quadrant

Question. Find the distance between the points \( A(2a, 6a) \) and \( B(2a + \sqrt{3}a, 5a) \)
(a) a
(b) 2a
(c) 3a
(d) –a
Answer: (b) 2a

Question. If the point C(–1, 2) divides internally the line segment joining A(2, 5) and B in the ratio 3 : 4, then the coordinates of B are
(a) (2, – 5)
(b) (2, 5)
(c) (5, 2)
(d) (–5, – 2)
Answer: (d) (–5, – 2)

Question. y-axis divides the join of P(–4, 2) and Q (8, 3) in the ratio
(a) 3 : 1
(b) 1 : 3
(c) 2 : 1
(d) 1 : 2
Answer: (d) 1 : 2

Question. If the mid-point of the line joining (3, 4) and (k, 7) is (x, y) and \( 2x + 2y + 1 = 0 \), then the value of k is
(a) 10
(b) –15
(c) 15
(d) –10
Answer: (b) –15

Question. The three consecutive vertices of a parallelogram are \( (a + b, a – b) \); \( (2a + b, 2a – b) \); \( (a – b, a + b) \). The fourth vertex is
(a) (a, b)
(b) (b, b)
(c) (–b, b)
(d) (–a, –b)
Answer: (c) (–b, b)

Question. The values of k for which the distance between the points A(k, – 5) and B(2, 7) is 13 units are
(a) 7, 3
(b) –7, 3
(c) 7, – 3
(d) – 7, –3
Answer: (c) 7, – 3

Question. If the distance of the point (4, a) from X-axis is double its distance from Y-axis, then the value of a is
(a) 5
(b) 8
(c) 16
(d) 4
Answer: (b) 8

Question. The positive value of x, for which the distance between the points (– 2, 5) and (x, 19) be \( \sqrt{205} \) units is
(a) 0
(b) 1
(c) 5
(d) 6
Answer: (b) 1

Question. Determine the ratio in which the point P(a, – 2) divides the joining of A (– 4, 3) and B (2, – 4). Also find the value of a.
Answer: Let ratio be \( k : 1 \).
\( -2 = \frac{-4k + 3}{k + 1} \implies -2k - 2 = -4k + 3 \implies 2k = 5 \implies k = 2.5 \). Ratio is 5 : 2.
\( a = \frac{2.5(2) + 1(-4)}{3.5} = \frac{5 - 4}{3.5} = \frac{1}{3.5} = \frac{2}{7} \).

Question. If ‘a’ is the length of one of the sides of an equilateral triangle ABC, base BC lies on x-axis and vertex B is at the origin, find the coordinates of the vertices of the triangle ABC.
Answer: B is at origin \( (0,0) \). C is on x-axis at distance a, so \( C(a, 0) \).
A is the third vertex of equilateral triangle: Coords are \( (\frac{a}{2}, \frac{\sqrt{3}a}{2}) \) or \( (\frac{a}{2}, -\frac{\sqrt{3}a}{2}) \).

Question. If A(5, 2), B(2, –2) and C(–2, t) are the vertices of a right angled triangle with \( \angle B = 90^\circ \), then find the value of t.
Answer: Slope \( AB \times \) Slope \( BC = -1 \).
\( \frac{-2-2}{2-5} \times \frac{t - (-2)}{-2 - 2} = -1 \implies \frac{-4}{-3} \times \frac{t + 2}{-4} = -1 \implies \frac{t+2}{3} = -1 \implies t = -5 \).

Question. The x-coordinate of a point P is twice its y-coordinate. If P is equidistant from Q(2, –5) and R(–3, 6), find the coordinates of P. 
Answer: Let \( P(2y, y) \).
\( (2y - 2)^2 + (y + 5)^2 = (2y + 3)^2 + (y - 6)^2 \)
\( 4y^2 + 4 - 8y + y^2 + 25 + 10y = 4y^2 + 9 + 12y + y^2 + 36 - 12y \)
\( 2y + 29 = 45 \implies 2y = 16 \implies y = 8 \).
Point P is (16, 8).

Question. Find the coordinates of the centre of a circle passing through the points (0, 0), (– 2, 1) and (– 3, 2). Also, find its radius.
Answer: (Solution requires solving simultaneous quadratic equations for equidistant points from origin and others).
Center: (13, 11), Radius: \( \sqrt{13^2 + 11^2} = \sqrt{290} \).

Question. Points P, Q, R and S divide the line segment joining the points A(1, 2) and B(6, 7) in 5 equal parts. Find the coordinates of the points P, Q and R. 
Answer: Total difference in x is 5, in y is 5.
P is \( (1+1, 2+1) = (2, 3) \).
Q is \( (1+2, 2+2) = (3, 4) \).
R is \( (1+3, 2+3) = (4, 5) \).

Question. A(0, 3), B(–1, –2) and C(4, 2) are vertices of a \( \Delta ABC \). D is a point on the side BC such that \( \frac{BD}{DC} = \frac{1}{2} \). P is a point on AD such that \( AP = \frac{2\sqrt{5}}{3} \) units. Find coordinates of P.
Answer: D is \( (\frac{1(4) + 2(-1)}{3}, \frac{1(2) + 2(-2)}{3}) = (\frac{2}{3}, -\frac{2}{3}) \).
Length \( AD = \sqrt{(\frac{2}{3})^2 + (-\frac{2}{3}-3)^2} = \sqrt{\frac{4}{9} + \frac{121}{9}} = \frac{\sqrt{125}}{3} = \frac{5\sqrt{5}}{3} \).
Ratio AP : PD = \( \frac{2\sqrt{5}}{3} : (\frac{5\sqrt{5}}{3} - \frac{2\sqrt{5}}{3}) = 2 : 3 \).
P divides AD in ratio 2 : 3.
\( x = \frac{2(2/3) + 3(0)}{5} = \frac{4}{15} \), \( y = \frac{2(-2/3) + 3(3)}{5} = \frac{-4/3 + 9}{5} = \frac{23}{15} \).
Point P is \( (\frac{4}{15}, \frac{23}{15}) \).

Question. The mid-point P of the line segment joining the points A(– 10, 4) and B(– 2, 0) lies on the line segment joining the points C(– 9, – 4) and D(– 4, y). Find the ratio in which P divides CD. Also find the value of y. 
Answer: Midpoint P \( = (\frac{-10-2}{2}, \frac{4+0}{2}) = (-6, 2) \).
P divides CD in ratio \( k : 1 \).
\( -6 = \frac{k(-4) + 1(-9)}{k + 1} \implies -6k - 6 = -4k - 9 \implies 2k = 3 \implies k = 1.5 \). Ratio is 3 : 2.
\( 2 = \frac{1.5(y) + 1(-4)}{2.5} \implies 5 = 1.5y - 4 \implies 1.5y = 9 \implies y = 6 \).

Question. The base BC of an equilateral triangle ABC lies on y-axis. The coordinates of point C are (0, –3). The origin is the mid-point of the base. Find the coordinates of the points A and B. Also find the coordinates of another point D such that BACD is a rhombus. 
Answer: C is (0, -3), origin is midpoint, so B is (0, 3).
Length \( BC = 6 \). Height of triangle \( = \frac{\sqrt{3}}{2} \times 6 = 3\sqrt{3} \).
A is \( (3\sqrt{3}, 0) \) or \( (-3\sqrt{3}, 0) \).
Rhombus point D is the reflection: \( A(3\sqrt{3}, 0) \) and \( D(-3\sqrt{3}, 0) \).

Question. Preeti and Arun are both driving to their respective offices from the same home. Preeti drives towards the east at an average speed of 30 km per hour for 12 minutes and then towards the south at an average speed of 60 km per hour for 3 minutes. Arun drives towards the west at an average speed of 30 km per hour for 4 minutes and then towards the north at an average speed of 45 km per hour for 4 minutes. What is the straight-line distance between Preeti’s office and Arun’s office? Show your steps and represent the given scenario on the coordinate plane. 
Answer: Preeti's movement:
East: \( 30 \text{ km/h} \times (12/60) \text{ h} = 6 \text{ km} \).
South: \( 60 \text{ km/h} \times (3/60) \text{ h} = 3 \text{ km} \).
Preeti's Office Coords: \( (6, -3) \).
Arun's movement:
West: \( 30 \text{ km/h} \times (4/60) \text{ h} = 2 \text{ km} \).
North: \( 45 \text{ km/h} \times (4/60) \text{ h} = 3 \text{ km} \).
Arun's Office Coords: \( (-2, 3) \).
Distance \( = \sqrt{(-2 - 6)^2 + (3 - (-3))^2} = \sqrt{(-8)^2 + 6^2} = \sqrt{64 + 36} = 10 \text{ km} \).

Question. Three players are standing on the circle at points A(–5, 0), B(1, 0) and C(3, 4). A ball is placed at a point that is equidistant from all 3 players.
(i) What are the coordinates of the ball?
(ii) The fourth player is standing at the point D(–5, 4). Is he/she standing on the circle?
Show your steps and give valid reasons. [CFPQ, CBSE]
Answer: (i) Ball is at circumcenter. Let it be \( (x, y) \).
\( (x + 5)^2 + y^2 = (x - 1)^2 + y^2 \implies 10x + 25 = -2x + 1 \implies 12x = -24 \implies x = -2 \).
\( (-2 - 1)^2 + y^2 = (-2 - 3)^2 + (y - 4)^2 \implies 9 + y^2 = 25 + y^2 - 8y + 16 \)
\( 9 = 41 - 8y \implies 8y = 32 \implies y = 4 \).
Ball coordinates are \( (-2, 4) \).
Radius \( R = \sqrt{(-2 - 1)^2 + 4^2} = 5 \).
(ii) Distance from \( (-2, 4) \) to \( D(-5, 4) \):
\( \text{Dist} = \sqrt{(-5 - (-2))^2 + (4 - 4)^2} = \sqrt{(-3)^2} = 3 \).
Since distance 3 \( \neq \) radius 5, the fourth player is not standing on the circle.

ASSERTION AND REASON QUESTIONS

Question. Assertion (A): The distance of the point (2, 11) from the x-axis is 11 units.
Reason (R): The distance of a point (x, y) from x-axis is its ordinate, i.e., y-units.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer: (a) Both A and R are true and R is the correct explanation of A.

Question. Assertion (A): The distance between the points \( A(\tan \theta, 0) \) and \( B(1, \sqrt{2}\tan \theta) \) is 2 units.
Reason (R): The distance between two points \( A(x_1, y_1) \) and \( B(x_2, y_2) \) is given by \( AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer: (d) A is false but R is true.