CBSE Class 10 Mathematics Coordinate Geometry VBQs Set 07

Question. The points \( A(0, \frac{8}{3}) \), \( B(1, 3) \) and \( C(82, 30) \)
(a) lie on a circle
(b) are vertices of a right-angled triangle
(c) are vertices of an obtuse-angled triangle
(d) None of the options
Answer: (d) None of the options

Question. If A and B are the points (– 3, 4) and (2, 1) respectively, then the coordinates of the point C on AB produced such that AC = 2BC are
(a) (2, 4)
(b) (3, 7)
(c) (7, – 2)
(d) None of the options
Answer: (c) (7, – 2)

Question. If the distance of the point (4, a) from x-axis is half its distance from y-axis, then a =
(a) 4 units
(b) 8 units
(c) 2 units
(d) 6 units
Answer: (c) 2 units

Question. If the distance between the points (8, p) and (4, 3) is 5 units, then value of p is
(a) 6
(b) 0
(c) both (a) and (b)
(d) None of the options
Answer: (c) both (a) and (b)

Question. Three vertices of a parallelogram ABCD are A(1, 4), B(–2, 3) and C(5, 8). The ordinate of the fourth vertex D is
(a) 8
(b) 9
(c) 7
(d) 6
Answer: (b) 9

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 AP =
(a) 5 units
(b) \( 5\sqrt{5} \) units
(c) 25 units
(d) \( 5\sqrt{10} \) units
Answer: (d) \( 5\sqrt{10} \) units

Question. If the origin is the mid-point of the line segment joined by the points (2, 3) and (x, y), then the value of (x, y) is
(a) (2, 3)
(b) (– 2, 3)
(c) (– 2, – 3)
(d) (2, – 3)
Answer: (c) (– 2, – 3)

Question. If four vertices of a parallelogram taken in order are (– 3, – 1), (a, b), (3, 3) and (4, 3), then a : b =
(a) 1 : 4
(b) 4 : 1
(c) 1 : 2
(d) 2 : 1
Answer: (b) 4 : 1

Question. If the points (–2, 1), (a, b) and (4, –1) are collinear and a – b = 1, then find the values of a and b.
Answer: Since the points are collinear, the slope between (–2, 1) and (a, b) must equal the slope between (a, b) and (4, –1).
\( \frac{b - 1}{a + 2} = \frac{-1 - b}{4 - a} \)
\( \implies (b - 1)(4 - a) = (a + 2)(-1 - b) \)
\( \implies 4b - ab - 4 + a = -a - ab - 2 - 2b \)
\( \implies 6b + 2a = 2 \implies a + 3b = 1 \).
Given \( a - b = 1 \).
Subtracting the two equations: \( 4b = 0 \implies b = 0 \).
Substituting \( b = 0 \) in \( a - b = 1 \) gives \( a = 1 \).
The values are \( a = 1, b = 0 \).

Question. Three vertices of a parallelogram taken in order are (1, 2), (2, 4) and (3, 7). Find its fourth vertex.
Answer: Let the fourth vertex be \( D(x, y) \).
In a parallelogram, midpoints of diagonals AC and BD coincide.
Midpoint of AC \( = (\frac{1+3}{2}, \frac{2+7}{2}) = (2, 4.5) \).
Midpoint of BD \( = (\frac{2+x}{2}, \frac{4+y}{2}) \).
Equating them: \( \frac{2+x}{2} = 2 \implies x = 2 \) and \( \frac{4+y}{2} = 4.5 \implies y = 5 \).
The fourth vertex is (2, 5).

Question. Find the coordinates of a point R which divides the line segment joining the points P(– 2, 3) and Q(4, 7) internally in the ratio 4 : 7.
Answer: Using section formula: \( R = (\frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}) \)
\( x = \frac{4(4) + 7(-2)}{4+7} = \frac{16 - 14}{11} = \frac{2}{11} \)
\( y = \frac{4(7) + 7(3)}{4+7} = \frac{28 + 21}{11} = \frac{49}{11} \)
The coordinates of R are \( (\frac{2}{11}, \frac{49}{11}) \).

Question. Determine k, so that the following points are collinear: (k, 2 – 2k), (– k + 1, 2k) and (– 4 – k, 6 – 2k).
Answer: For collinear points, the area of the triangle formed is zero.
\( x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) = 0 \)
\( k(2k - (6 - 2k)) + (-k + 1)((6 - 2k) - (2 - 2k)) + (-4 - k)((2 - 2k) - 2k) = 0 \)
\( k(4k - 6) + (-k + 1)(4) + (-4 - k)(2 - 4k) = 0 \)
\( 4k^2 - 6k - 4k + 4 + (-8 + 16k - 2k + 4k^2) = 0 \)
\( 8k^2 + 4k - 4 = 0 \implies 2k^2 + k - 1 = 0 \)
\( \implies (2k - 1)(k + 1) = 0 \)
\( \implies k = 1/2 \) or \( k = -1 \).

Question. Show that the points A(5, 6), B(1, 5), C(2, 1) and D(6, 2) are the vertices of a square.
Answer: Check distances:
\( AB = \sqrt{(1-5)^2 + (5-6)^2} = \sqrt{16+1} = \sqrt{17} \)
\( BC = \sqrt{(2-1)^2 + (1-5)^2} = \sqrt{1+16} = \sqrt{17} \)
\( CD = \sqrt{(6-2)^2 + (2-1)^2} = \sqrt{16+1} = \sqrt{17} \)
\( DA = \sqrt{(5-6)^2 + (6-2)^2} = \sqrt{1+16} = \sqrt{17} \)
Check diagonals:
\( AC = \sqrt{(2-5)^2 + (1-6)^2} = \sqrt{9+25} = \sqrt{34} \)
\( BD = \sqrt{(6-1)^2 + (2-5)^2} = \sqrt{25+9} = \sqrt{34} \)
Since all sides are equal and diagonals are equal, it is a square.

Question. Prove that the points (0, 0), (5, 5) and (–5, 5) are vertices of a right isosceles triangle.
Answer: Let points be O, A, B.
\( OA = \sqrt{5^2+5^2} = \sqrt{50} \)
\( OB = \sqrt{(-5)^2+5^2} = \sqrt{50} \)
\( AB = \sqrt{(5-(-5))^2 + (5-5)^2} = \sqrt{10^2} = 10 = \sqrt{100} \)
Since \( OA = OB \), it is isosceles.
Since \( OA^2 + OB^2 = 50 + 50 = 100 = AB^2 \), it is a right-angled triangle. Hence proved.

Question. Find the coordinates of the points which divide the line segment joining the points (–4, 0) and (0, 6) in three equal parts.
Answer: Ratio 1:2 and 2:1.
Point 1 (1:2): \( x = \frac{1(0) + 2(-4)}{3} = -8/3 \), \( y = \frac{1(6) + 2(0)}{3} = 2 \). Coordinate: \( (-8/3, 2) \).
Point 2 (2:1): \( x = \frac{2(0) + 1(-4)}{3} = -4/3 \), \( y = \frac{2(6) + 1(0)}{3} = 4 \). Coordinate: \( (-4/3, 4) \).

Question. Prove that (– 2, 3), (8, 3) and (6, 7) are the vertices of a right-angled triangle. 
Answer: Let points be A, B, C.
\( AB = \sqrt{(8+2)^2 + (3-3)^2} = 10 \implies AB^2 = 100 \)
\( BC = \sqrt{(6-8)^2 + (7-3)^2} = \sqrt{4+16} = \sqrt{20} \implies BC^2 = 20 \)
\( AC = \sqrt{(6+2)^2 + (7-3)^2} = \sqrt{64+16} = \sqrt{80} \implies AC^2 = 80 \)
Since \( BC^2 + AC^2 = 20 + 80 = 100 = AB^2 \), it is a right-angled triangle.

Question. The points A(2, 0), B(9, 1), C(11, 6) and D(4, 4) are the vertices of a quadrilateral ABCD. Determine whether ABCD is a rhombus or not.
Answer: Check distances:
\( AB = \sqrt{7^2+1^2} = \sqrt{50} \)
\( BC = \sqrt{2^2+5^2} = \sqrt{29} \)
Since adjacent sides are not equal, it is not a rhombus.

Question. If G be the centroid of a triangle ABC and P be any other point in the plane, prove that \( PA^2 + PB^2 + PC^2 = GA^2 + GB^2 + GC^2 + 3GP^2 \).
Answer: Let origin be G. Then \( \vec{GA} + \vec{GB} + \vec{GC} = 0 \).
\( PA^2 = |\vec{GA} - \vec{GP}|^2 = GA^2 + GP^2 - 2\vec{GA} \cdot \vec{GP} \)
Similarly for \( PB^2 \) and \( PC^2 \).
Summing gives \( GA^2 + GB^2 + GC^2 + 3GP^2 - 2\vec{GP} \cdot (\vec{GA} + \vec{GB} + \vec{GC}) \).
Since the sum in brackets is 0, the relation holds. Hence proved.

Question. If G be the centroid of a triangle ABC, prove that \( AB^2 + BC^2 + CA^2 = 3(GA^2 + GB^2 + GC^2) \). 
Answer: Let \( G \) be the origin \( (0,0) \). Then \( \sum x_i = 0 \) and \( \sum y_i = 0 \).
\( AB^2 = (x_1-x_2)^2 + (y_1-y_2)^2 = x_1^2 + x_2^2 + y_1^2 + y_2^2 - 2(x_1x_2 + y_1y_2) \).
Summing for all sides gives \( 2 \sum x_i^2 + 2 \sum y_i^2 - 2 \sum_{ij} \vec{GA} \cdot \vec{GB} \).
Since \( (\sum \vec{GA})^2 = 0 = \sum GA^2 + 2 \sum \vec{GA} \cdot \vec{GB} \), the cross terms are \( -\sum GA^2 \).
Substituting gives \( 2 \sum GA^2 - (- \sum GA^2) = 3 \sum GA^2 \). Hence proved.

Question. Find the coordinates of the centroid of a triangle whose vertices are \( (x_1, y_1), (x_2, y_2) \) and \( (x_3, y_3) \). 
Answer: The centroid \( G \) is given by the average of the coordinates of the vertices:
\( G = (\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}) \).

Question. If the points (a, b), \( (a_1, b_1) \) and \( (a – a_1, b – b_1) \) are collinear show that \( ab_1 = a_1b \). Also, show that the line joining the given points passes through the origin. 
Answer: For collinearity, area = 0:
\( a(b_1 - (b - b_1)) + a_1((b - b_1) - b) + (a - a_1)(b - b_1) = 0 \)
\( a(2b_1 - b) + a_1(-b_1) + ab - ab_1 - a_1b + a_1b_1 = 0 \)
\( 2ab_1 - ab - a_1b_1 + ab - ab_1 - a_1b + a_1b_1 = 0 \)
\( ab_1 - a_1b = 0 \implies ab_1 = a_1b \).
Since the ratio \( a/b = a_1/b_1 \), the coordinates are proportional, and the line satisfies \( y = (b/a)x \), which passes through (0,0). Hence proved.

Question. If the points (a, 0), (0, b) and (1, 1) are collinear, show that \( \frac{1}{a} + \frac{1}{b} = 1 \). 
Answer: For collinearity, the slope between (a, 0) and (1, 1) equals the slope between (0, b) and (1, 1).
\( \frac{1 - 0}{1 - a} = \frac{1 - b}{1 - 0} \)
\( \implies 1 = (1 - b)(1 - a) \)
\( \implies 1 = 1 - a - b + ab \)
\( \implies a + b = ab \).
Dividing by \( ab \) gives: \( \frac{1}{b} + \frac{1}{a} = 1 \). Hence proved.

Question. Find the angle subtended at the origin by the line segment whose end points are (0, 10) and (10, 0). 
Answer: The point (0, 10) lies on the y-axis and (10, 0) lies on the x-axis. Since the coordinate axes are perpendicular to each other, the angle subtended at the origin is \( 90^\circ \).

Question. Using section formula, show that the points (1, –1), (2, 1) and (4, 5) are collinear.
Answer: Let point \( B(2, 1) \) divide the segment joining \( A(1, -1) \) and \( C(4, 5) \) in the ratio \( k : 1 \).
Using the x-coordinate:
\( 2 = \frac{4k + 1}{k + 1} \)
\( \implies \) \( 2k + 2 = 4k + 1 \)
\( \implies \) \( 2k = 1 \)
\( \implies \) \( k = \frac{1}{2} \)
Using the y-coordinate with \( k = \frac{1}{2} \):
\( y = \frac{5(\frac{1}{2}) + (-1)}{\frac{1}{2} + 1} = \frac{\frac{3}{2}}{\frac{3}{2}} = 1 \)
Since the coordinates match, the points are collinear.

Question. Prove that the straight line joining the points (–1, 3) and (4, –2) passed through the point (a, b) if a + b = 2.
Answer: The equation of the line passing through \( (-1, 3) \) and \( (4, -2) \) is:
\( y - 3 = \frac{-2 - 3}{4 - (-1)}(x - (-1)) \)
\( y - 3 = \frac{-5}{5}(x + 1) \)
\( y - 3 = -x - 1 \)
\( \implies \) \( x + y = 2 \)
If the point \( (a, b) \) lies on this line, it must satisfy the equation:
\( \implies \) \( a + b = 2 \). Hence proved.

Question. Let the opposite angular points of a square be (3, 4) and (1, –1). Find the coordinates of the remaining angular points.
Answer: Let the given points be \( A(3, 4) \) and \( C(1, -1) \). The midpoint of the diagonal \( AC \) is \( M(2, 1.5) \).
Let the other vertices be \( B(x, y) \). Since \( AB = BC \) and \( \angle ABC = 90^\circ \):
Using rotation or geometric properties, the remaining coordinates are found to be \( (4.5, 0.5) \) and \( (-0.5, 2.5) \).

Question. Find the distance between the points \( (\sqrt{5} - 2, \sqrt{3} + 2) \) and \( (\sqrt{5} + 1, \sqrt{3} - 1) \).
Answer: Using distance formula:
\( D = \sqrt{[(\sqrt{5} + 1) - (\sqrt{5} - 2)]^2 + [(\sqrt{3} - 1) - (\sqrt{3} + 2)]^2} \)
\( D = \sqrt{(3)^2 + (-3)^2} \)
\( D = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2} \) units.

Question. If A(–2, 5) and B(3, 2) are the two points on a straight line. If AB is extended to ‘C’ such that AC = 2BC, then find the coordinates of C.
Answer: \( AC = 2BC \) implies that \( B \) is the midpoint of \( AC \).
Let \( C \) be \( (x, y) \).
\( 3 = \frac{-2 + x}{2} \)
\( \implies \) \( 6 = -2 + x \)
\( \implies \) \( x = 8 \)
\( 2 = \frac{5 + y}{2} \)
\( \implies \) \( 4 = 5 + y \)
\( \implies \) \( y = -1 \)
The coordinates of C are (8, –1).

Question. If (3, 3), (h, 0) and (0, k) are collinear, then find the value of \( \frac{1}{h} + \frac{1}{k} \)?
Answer: Since the points are collinear, the slope between \( (h, 0) \) and \( (3, 3) \) equals the slope between \( (0, k) \) and \( (3, 3) \).
\( \frac{3 - 0}{3 - h} = \frac{3 - k}{3 - 0} \)
\( \frac{3}{3 - h} = \frac{3 - k}{3} \)
\( \implies \) \( 9 = (3 - k)(3 - h) \)
\( \implies \) \( 9 = 9 - 3h - 3k + hk \)
\( \implies \) \( 3h + 3k = hk \)
Dividing by \( 3hk \):
\( \implies \) \( \frac{1}{k} + \frac{1}{h} = \frac{1}{3} \).

Question. If (3, – 4) and (–6, 5) are the extremities of the diagonal of a parallelogram and (–2, 1) is its third vertex, then find the fourth vertex?
Answer: Let the diagonal extremities be \( A(3, -4) \) and \( C(-6, 5) \). The midpoint of diagonal \( AC \) is \( M(-1.5, 0.5) \).
Let the other vertices be \( B(-2, 1) \) and \( D(x, y) \). Diagonals of a parallelogram bisect each other, so \( M \) is also the midpoint of \( BD \).
\( -1.5 = \frac{-2 + x}{2} \)
\( \implies \) \( -3 = -2 + x \)
\( \implies \) \( x = -1 \)
\( 0.5 = \frac{1 + y}{2} \)
\( \implies \) \( 1 = 1 + y \)
\( \implies \) \( y = 0 \)
The fourth vertex is (–1, 0).

Question. If a point A lies on x-axis and has abscissa (p + q) and another point B lies on y-axis and has ordinate (p – q), find distance between A and B.
Answer: The coordinates are \( A(p+q, 0) \) and \( B(0, p-q) \).
Using distance formula:
\( AB = \sqrt{(0 - (p+q))^2 + ((p-q) - 0)^2} \)
\( AB = \sqrt{(p+q)^2 + (p-q)^2} \)
\( AB = \sqrt{p^2 + q^2 + 2pq + p^2 + q^2 - 2pq} \)
\( AB = \sqrt{2p^2 + 2q^2} = \sqrt{2(p^2 + q^2)} \) units.

Question. A triangle ABC, right angled at A, has points A and B as (2, 3) and (0, –1) respectively. If BC = 5 units, then find the coordinates of point.
Answer: Let \( C \) be \( (x, y) \). Since \( \angle A = 90^\circ \), \( AB^2 + AC^2 = BC^2 \).
\( AB^2 = (2-0)^2 + (3-(-1))^2 = 4 + 16 = 20 \).
\( AC^2 = (x-2)^2 + (y-3)^2 \).
\( 20 + (x-2)^2 + (y-3)^2 = 25 \)
\( \implies \) \( (x-2)^2 + (y-3)^2 = 5 \).
Also, the slope of \( AB \times \) slope of \( AC = -1 \).
\( \frac{-1-3}{0-2} \times \frac{y-3}{x-2} = -1 \)
\( \implies \) \( 2 \times \frac{y-3}{x-2} = -1 \)
\( \implies \) \( 2y - 6 = -x + 2 \)
\( \implies \) \( x = 8 - 2y \).
Substitute \( x \) in the distance equation to find \( (x, y) \).

Question. Find the coordinates of the point which divides internally the line joining the points (p, q) and (q, p) in the ratio (p – q): (p + q)
Answer: Let the ratio be \( m : n \) where \( m = p-q \) and \( n = p+q \).
\( x = \frac{(p-q)q + (p+q)p}{(p-q) + (p+q)} = \frac{pq - q^2 + p^2 + pq}{2p} = \frac{p^2 + 2pq - q^2}{2p} \)
\( y = \frac{(p-q)p + (p+q)q}{(p-q) + (p+q)} = \frac{p^2 - pq + pq + q^2}{2p} = \frac{p^2 + q^2}{2p} \)
The point is \( (\frac{p^2 + 2pq - q^2}{2p}, \frac{p^2 + q^2}{2p}) \).

Question. Two consecutive vertices of a rectangle of area 10 sq. units are (1, 3) and (–2, –1). Find the coordinates of other two vertices.
Answer: Let the given points be \( A(1, 3) \) and \( B(-2, -1) \).
Length \( AB = \sqrt{(-2-1)^2 + (-1-3)^2} = \sqrt{9 + 16} = 5 \) units.
Area \( = AB \times BC = 10 \). Since \( AB = 5 \), \( BC = 2 \).
Other vertices lie on lines perpendicular to \( AB \). By solving, we find coordinates like \( (-18/5, 11/5) \) and \( (-3/5, 31/5) \) or other pairs based on orientation.

Question. Assertion (A): The point P(–4, 6) divides the join of A(–6, 10) and B(3, –8) in the ratio 2 : 7.
Reason (R): If the point C(x, y) divides the join \( A(x_1, y_1) \) and \( B(x_2, y_2) \) in the ratio m : n, then \( x = \frac{mx_2 + nx_1}{m + n} \) and \( y = \frac{my_2 + ny_1}{m + n} \).
(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): Point P \( (1, \frac{5}{2}) \) is equidistant from the points A(–5, 3) and B(7, 2).
Reason (R): If a point P is equidistant from the points A and B, then AP = BP.
(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): If the centre of a circle is at the origin and its radius = 5 units, then a point on the circle is (0, 5).
Reason (R): The centre of the circle is the mid point of the line joining the end points of its diameter.
(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: (b) Both A and R are true but R is not the correct explanation of A.