CBSE Class 10 Mathematics Coordinate Geometry VBQs Set 05

Short Answer Type Questions 

Question. Show that the points A(1, 2), B(5, 4), C(3, 8) and D (–1, 6) are the vertices of a square.
Answer: A(1, 2), B (5, 4), C(3, 8) and D (–1, 6)
\( AB = \sqrt{4^2 + 2^2} = \sqrt{16 + 4} = \sqrt{20} \);
\( BC = \sqrt{(-2)^2 + (4)^2} = \sqrt{4 + 16} = \sqrt{20} \)
\( CD = \sqrt{(-4)^2 + (-2)^2} = \sqrt{16 + 4} = \sqrt{20} \);
\( DA = \sqrt{(-2)^2 + (4)^2} = \sqrt{4 + 16} = \sqrt{20} \)
Here AB = BC = CA = DA
\( AC = \sqrt{2^2 + 6^2} = \sqrt{40} \)
and \( BD = \sqrt{(-6)^2 + (2)^2} = \sqrt{36 + 4} = \sqrt{40} \)
All sides of quadrilateral are equal and diagonals are equal.
\( \implies \) ABCD is square.

Question. Show that points A(7, 5), B(2, 3) and C(6, – 7) are the vertices of a right triangle. Also find its area.
Answer: AB = \( \sqrt{(2 - 7)^2 + (3 - 5)^2} = \sqrt{25 + 4} = \sqrt{29} \)
BC = \( \sqrt{(6 - 2)^2 + (-7 - 3)^2} = \sqrt{16 + 100} = \sqrt{116} \)
CA = \( \sqrt{(7 - 6)^2 + (5 - (-7))^2} = \sqrt{1 + 144} = \sqrt{145} \)
Since, \( AB^2 + BC^2 = 29 + 116 = 145 = CA^2 \)
\( \implies \) \( \Delta ABC \) is right angled at B.
Area = \( \frac{1}{2} AB \times BC \)
\( = \frac{1}{2} \sqrt{29} \sqrt{116} \)
\( = \frac{1}{2} \sqrt{29} \times 2\sqrt{29} \)
\( = 29 \) sq. units

Question. Two points A(1, 0) and B(–1, 0) with a variable point P(x, y) satisfy the relation AP – BP = 1. Show that \( 12x^2 - 4y^2 = 3 \).
Answer: Given: AP – BP = 1
\( \implies \sqrt{(x - 1)^2 + (y - 0)^2} - \sqrt{(x + 1)^2 + (y - 0)^2} = 1 \)
\( \implies \sqrt{(x - 1)^2 + y^2} = 1 + \sqrt{(x + 1)^2 + y^2} \)
Squaring both sides
\( (x - 1)^2 + y^2 = 1 + (x + 1)^2 + y^2 + 2\sqrt{(x + 1)^2 + y^2} \)
\( \implies x^2 + 1 - 2x = 1 + x^2 + 1 + 2x + 2\sqrt{(x + 1)^2 + y^2} \)
\( \implies - 1 - 4x = 2\sqrt{(x + 1)^2 + y^2} \)
Squaring both sides
\( (-1 - 4x)^2 = 4((x + 1)^2 + y^2) \)
\( \implies 1 + 16x^2 + 8x = 4(x^2 + 1 + 2x + y^2) \)
\( \implies 12x^2 - 4y^2 = 3 \) Hence proved.

PRACTICE QUESTIONS

Question. The points (2, 5), (4, – 1) and (6, – 7) are vertices of an/a
(a) isosceles triangle
(b) equilateral triangle
(c) right-angled triangle
(d) None of the options
Answer: (d) None of the options

Question. If the segment joining the points (a, b) and (c, d) subtends a right angle at the origin, then
(a) ac – bd = 0
(b) ac + bd = 0
(c) ab + cd = 0
(d) ab – cd = 0
Answer: (b) ac + bd = 0

Question. If A = \( (a^2, 2a) \) and B = \( (\frac{1}{a^2}, -\frac{2}{a}) \) and S = (1, 0), then find \( \frac{1}{SA} + \frac{1}{SB} \).
Answer: 1

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

Question. Point A is on the y-axis at a distance of 4 units from the origin. If coordinates of point B are (–3, 0) then find the length of AB. 
Answer: 5 units

Question. Find the number of points on x-axis which are at a distance of 2 units from (2, 4).
Answer: 0

Question. Find the value of x such that PQ = QR where the coordinates of P, Q and R are (6, –1), (1, 3) and (x, 8) respectively.
Answer: x = -3 or 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: (5, 0) and (9, 0). Two such points.

Question. The centre of a circle is (2a, a – 7). Find the values of a if the circle passes through the point (11, –9) and has diameter \( 10\sqrt{2} \) units.
Answer: a = 5, 3

Question. If A(5, 3), B(11, –5) and C(12, y) are vertices of a right triangle right angled at C, then find the value of y.
Answer: y = -4 or 2

Question. Show that A(–3, 2), B(–5, –5), C(2, – 3) and D(4, 4) are the vertices of a rhombus.
Answer: Verify that all sides are equal but diagonals are not equal.

Question. Show that the points A(3, 5), B(6, 0), C(1, – 3) and D(–2, 2) are the vertices of a square ABCD.
Answer: Verify that all sides and diagonals are equal.

Question. An equilateral triangle has two vertices at the points (3, 4) and (– 2, 3). Find the coordinates of the third vertex.
Answer: \( (\frac{1 \pm \sqrt{3}}{2}, \frac{7 \pm 5\sqrt{3}}{2}) \)

PRACTICE QUESTIONS

Question. The point P which divides the line segment joining the points A(2, –5) and B(5, 2) in the ratio 2 : 3 lies in the quadrant
(a) I
(b) II
(c) III
(d) IV
Answer: (d) IV

Question. The vertices of a triangle are (0, 0), (3, 0) and (0, 4). The centroid of the triangle is
(a) \( (\frac{1}{2}, 2) \)
(b) \( (1, \frac{4}{3}) \)
(c) (1, 1)
(d) (0, 3)
Answer: (b) \( (1, \frac{4}{3}) \)

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

Question. The line segment joining the points P(–3, 2) and Q(5, 7) is divided by the y-axis in the ratio
(a) 3 : 1
(b) 3 : 7
(c) 3 : 2
(d) 3 : 5
Answer: (d) 3 : 5

Question. The radius of a circle with centre at the origin is 1/2 units. Find all the points on the circle which are of the form(–y, y). Show your steps.
Answer: \( \sqrt{(-y)^2 + y^2} = \frac{1}{2} \implies 2y^2 = \frac{1}{4} \implies y = \pm \frac{1}{\sqrt{8}} \).
Points are \( (\mp \frac{1}{\sqrt{8}}, \pm \frac{1}{\sqrt{8}}) \).

Question. If P(2, p) is the mid-point of the line segment joining the points A(6, –5) and B(–2, 11), find the value of p.
Answer: p = 3

Question. Find the value of k if P(4, –2) is the mid-point of the line segment joining the points A(5k, 3) and B(–k, –7).
Answer: k = 2

Question. Find the ratio in which the y-axis divides the line segment joining the points (5, – 6) and (–1, –4). 
Answer: 5 : 1

Question. Find the fourth vertex of a rectangle whose three vertices taken in order are (4, 1), (7, 4) and (13, –2).
Answer: (10, -5)

Question. In what ratio does the line x – y – 2 = 0 divide the line segment joining (3, –1) and (8, 9)?
Answer: 2 : 3

Question. Find the value of m for which the points with coordinates (3, 5), (m, 6) and \( (\frac{1}{2}, \frac{15}{2}) \) are collinear.
Answer: m = 2

Question. If the points (10, 5), (8, 4) and (6, 6) are the mid points of the sides of a triangle, find its vertices.
Answer: (8, 7), (12, 3), (4, 5)

Multiple Choice Questions 

Question. If P(1, 2), Q(4, 6), R(5, 7) and S(a, b) are the vertices of a parallelogram PQRS, then
(a) a = 2, b = 4
(b) a = 3, b = 4
(c) a = 2, b = 3
(d) a = 3, b = 5
Answer: (c) a = 2, b = 3

Question. A straight line is drawn joining the points (3, 4) and (5, 6). If the line is extended, the ordinate of the point on the line, whose abscissa is –1 is
(a) 1
(b) 0
(c) –1
(d) 2
Answer: (b) 0

Question. A(5, 1); B(1, 5) and C(–3, – 1) are the vertices of \( \Delta ABC \). The length of median AD is
(a) \( \sqrt{35} \) units
(b) 6 units
(c) \( \sqrt{41} \) units
(d) \( \sqrt{37} \) units
Answer: (d) \( \sqrt{37} \) units

Question. If A(1, 2), B(4, 3) and C(6, 6) are the three vertices of a parallelogram ABCD, find the coordinates of the fourth vertex D.
(a) \( (\frac{1}{2}, 4) \)
(b) \( (\frac{7}{2}, 5) \)
(c) (3, 4)
(d) (3, 5)
Answer: (d) (3, 5)

Question. \( \Delta ABC \) is a triangle such that AB : BC = 1 : 2. Point A lies on the y-axis and the coordinates of B and C are known. Which of the following formula can DEFINITELY be used to find the coordinates of A?
(i) Section formula
(ii) Distance formula
(a) only (i)
(b) only (ii)
(c) both (i) and (ii)
(d) neither (i) nor (ii)
Answer: (b) only (ii)

Question. The coordinates of one end point of a diameter of a circle are (4, –1) and the coordinates of the centre are (1, –3). Then the coordinates of the other end of the diameter are
(a) (–2, 5)
(b) (2, 5)
(c) (–5, –2)
(d) (–2, –5)
Answer: (d) (–2, –5)

Question. Point P divides the line segment joining the points A(2, –5) and B(5, 2) in the ratio 2 : 3. Name the quadrant in which P lies.
(a) Ist quadrant
(b) IInd quadrant
(c) IIIrd quadrant
(d) IVth quadrant
Answer: (d) IVth quadrant

Question. P(5, –3) and Q(3, y) are the points of trisection of the line segment joining A(7, – 2) and B(1, –5). Then, the value of y is
(a) –2
(b) –3
(c) –4
(d) 4
Answer: (c) –4

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 be k:1.
By section formula,
\( P(4, m) = (\frac{6k + 2}{k + 1}, \frac{-3k + 3}{k + 1}) \)
\( \implies \frac{6k + 2}{k + 1} = 4 \)
\( \implies 6k + 2 = 4k + 4 \)
\( \implies 2k = 2 \)
\( \implies k = 1 \).
The ratio is 1 : 1.
Now, \( m = \frac{-3k + 3}{k + 1} \)
\( \implies m = \frac{-3(1) + 3}{1 + 1} = 0 \).
\( \implies \) Value of m is 0.

Question. The point R divides the line segment AB where A(–4, 0), B(0, 6) are such that AR = \( \frac{3}{4} \)AB. Find the coordinates of R.
Answer: AR = \( \frac{3}{4} \)AB.
RB = AB - AR = \( \frac{1}{4} \)AB.
Ratio AR : RB = 3 : 1.
x = \( \frac{3 \times 0 + 1 \times (-4)}{3 + 1} = -1 \)
y = \( \frac{3 \times 6 + 1 \times 0}{3 + 1} = \frac{18}{4} = \frac{9}{2} \)
Thus, coordinates of R are \( (-1, \frac{9}{2}) \).

Question. If C is a point lying on the line segment AB joining A(1, 1) and B(2, – 3) such that 3AC = CB, then find the coordinates of C.
Answer: AC/CB = 1/3
x = \( \frac{2 + 3}{4} = \frac{5}{4} \) and y = \( \frac{-3 + 3}{1 + 3} = 0 \)
\( \therefore (x, y) = (\frac{5}{4}, 0) \)

Question. The coordinates of the mid-point of the line joining the points (3p, 4) and (– 2, 2q) are (5, p). Find the values of p and q.
Answer: \( \frac{3p - 2}{2} = 5 \implies 3p - 2 = 10 \implies 3p = 12 \implies p = 4 \)
\( \frac{4 + 2q}{2} = p \implies \frac{4 + 2q}{2} = 4 \implies 4 + 2q = 8 \implies 2q = 4 \implies q = 2 \)
\( \therefore p = 4 \) and q = 2

Question. Find the ratio in which the line segment joining (2, – 3) and (5, 6) is divided by x-axis.
Answer: Let ratio be k : 1.
Ordinate \( y = \frac{6k - 3}{k + 1} = 0 \implies 6k = 3 \implies k = \frac{1}{2} \).
\( \therefore \) Ratio is 1 : 2 internally.

Question. In given figure BD bisects \( \angle B \). Find the length of BD.
Answer: Using angle bisector property: AD/CD = AB/BC
\( AB = \sqrt{1^2 + 7^2} = \sqrt{50} \)
\( BC = \sqrt{(2-1)^2 + (0-7)^2} = \sqrt{50} \)
\( \implies \) AD/CD = 1 \( \implies \) D bisects AC.
Coordinates of D are \( (\frac{0 + 2}{2}, \frac{0 + 0}{2}) = (1, 0) \)
BD = \( \sqrt{(1 - 1)^2 + (0 - 7)^2} = 7 \)
\( \therefore \) BD = 7 units.

Question. The line segment AB joining the points A(3, –4) and B(1, 2) is trisected at the points P(p, –2) and Q(5/3, q). Find the values of p and q.
Answer: For P, ratio is 1:2.
p = \( \frac{1 \times 1 + 2 \times 3}{1 + 2} = \frac{7}{3} \)
For Q, ratio is 2:1.
q = \( \frac{2 \times 2 + 1 \times (-4)}{1 + 2} = 0 \).