CBSE Class 10 Coordinate geometry Sure Shot Questions Set H

SECTION FORMULAE

EXAMPLES

Type I : On finding the section point when the section ratio is given

Question. Find the coordinates of the point which divides the line segment joining the points (6, 3) and (– 4, 5) in the ratio 3 : 2 internally.
Answer: Let P (x, y) be the required point. Then, \( x = \frac{3 \times (-4) + 2 \times 6}{3 + 2} = 0 \) and \( y = \frac{3 \times 5 + 2 \times 3}{3 + 2} = \frac{21}{5} \) So, the coordinates of P are (0, 21/5).

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. 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 \( P (\frac{1 \times (-3) + 2 \times 1}{1 + 2}, \frac{1 \times 4 + 2 \times (-2)}{1 + 2}) = P (-\frac{1}{3}, 0) \) \( Q (\frac{2 \times (-3) + 1 \times 1}{2 + 1}, \frac{2 \times 4 + 1 \times (-2)}{2 + 1}) = Q (-\frac{5}{3}, 2) \) Hence, the two points of trisection are (–1/3, 0) and (–5/3, 2).

Question. If the point C (–1, 2) divides internally the line segment joining A (2, 5) and B in ratio 3 : 4, find the coordinates of B.
Answer: Let the coordinates of B be \( (\alpha, \beta) \). It is given that AC : BC = 3 : 4. So, the coordinates of C are \( (\frac{3\alpha + 4 \times 2}{3 + 4}, \frac{3\beta + 4 \times 5}{3 + 4}) = (\frac{3\alpha + 8}{7}, \frac{3\beta + 20}{7}) \). But, the coordinates of C are (–1, 2) \( \therefore \frac{3\alpha + 8}{7} = -1 \text{ and } \frac{3\beta + 20}{7} = 2 \implies \alpha = -5 \text{ and } \beta = -2 \). Thus, the coordinates of B are (–5, –2).

Question. Determine the ratio in which the line 3x + y – 9 = 0 divides the segment joining the points (1, 3) and (2, 7).
Answer: Suppose the line 3x + y – 9 = 0 divides the line segment joining A (1, 3) and B(2, 7) in the ratio k : 1 at point C. Then, the coordinates of C are \( (\frac{2k + 1}{k + 1}, \frac{7k + 3}{k + 1}) \). But, C lies on 3x + y – 9 = 0. Therefore, \( 3 (\frac{2k + 1}{k + 1}) + \frac{7k + 3}{k + 1} - 9 = 0 \implies 6k + 3 + 7k + 3 - 9k - 9 = 0 \implies k = \frac{3}{4} \). So, the required ratio is 3 : 4 internally.

Type III : On determination of the type of a given quadrilateral

Question. Prove that the points (–2, –1), (1, 0), (4, 3) and (1, 2) are the vertices of a parallelogram. Is it a rectangle ?
Answer: Let the given points be A, B, C and D respectively. Coordinates of the mid-point of AC are \( (\frac{-2 + 4}{2}, \frac{-1 + 3}{2}) = (1, 1) \). Coordinates of the mid-point of BD are \( (\frac{1 + 1}{2}, \frac{0 + 2}{2}) = (1, 1) \). Thus, AC and BD have the same mid-point. Hence, ABCD is a parallelogram. Now, \( AC = \sqrt{(4 - (-2))^2 + (3 - (-1))^2} = \sqrt{36 + 16} = \sqrt{52} \) and \( BD = \sqrt{(1 - 1)^2 + (0 - 2)^2} = 2 \). Clearly, \( AC \neq BD \). So, ABCD is not a rectangle.

Question. Prove that (4, – 1), (6, 0), (7, 2) and (5, 1) are the vertices of a rhombus. Is it a square ?
Answer: Coordinates of the mid-point of AC are \( (\frac{4 + 7}{2}, \frac{-1 + 2}{2}) = (\frac{11}{2}, \frac{1}{2}) \). Coordinates of the mid-point of BD are \( (\frac{6 + 5}{2}, \frac{0 + 1}{2}) = (\frac{11}{2}, \frac{1}{2}) \). ABCD is a parallelogram. Now, \( AB = \sqrt{(6 - 4)^2 + (0 + 1)^2} = \sqrt{5} \), \( BC = \sqrt{(7 - 6)^2 + (2 - 0)^2} = \sqrt{5} \). \( \therefore AB = BC \). So, ABCD is a rhombus. We have, \( AC = \sqrt{(7 - 4)^2 + (2 + 1)^2} = 3\sqrt{2} \) and \( BD = \sqrt{(6 - 5)^2 + (0 - 1)^2} = \sqrt{2} \). Clearly, \( AC \neq BD \). So, ABCD is not a square.

Type IV : On finding the unknown vertex from given points

Question. The three vertices of a parallelogram taken in order are (–1, 0), (3, 1) and (2, 2) respectively. Find the coordinates of the fourth vertex.
Answer: Let A(–1, 0), B(3, 1), C(2, 2) and D(x, y) be the vertices. Mid-point of AC = Mid-point of BD. \( \implies (\frac{-1 + 2}{2}, \frac{0 + 2}{2}) = (\frac{3 + x}{2}, \frac{1 + y}{2}) \implies (\frac{1}{2}, 1) = (\frac{3 + x}{2}, \frac{y + 1}{2}) \) \( \implies \frac{3 + x}{2} = \frac{1}{2} \text{ and } \frac{y + 1}{2} = 1 \implies x = -2 \text{ and } y = 1 \). Hence, the fourth vertex is (–2, 1).

Question. If the points A (6, 1), B (8, 2), C(9, 4) and D (p, 3) are vertices of a parallelogram, taken in order, find the value of p.
Answer: Coordinates of the mid point of diagonal AC = coordinates of the mid-point of diagonal BD. \( \therefore (\frac{6 + 9}{2}, \frac{1 + 4}{2}) = (\frac{8 + p}{2}, \frac{2 + 3}{2}) \implies (\frac{15}{2}, \frac{5}{2}) = (\frac{8 + p}{2}, \frac{5}{2}) \implies \frac{15}{2} = \frac{8 + p}{2} \implies p = 7 \).

Question. If A(–2, –1), B(a, 0), C(4, b) and D(1, 2) are the vertices of a parallelogram, find the values of a and b.
Answer: mid-point of AC = mid-point of BD. \( (\frac{-2 + 4}{2}, \frac{-1 + b}{2}) = (\frac{a + 1}{2}, \frac{0 + 2}{2}) \implies (1, \frac{b - 1}{2}) = (\frac{a + 1}{2}, 1) \implies \frac{a + 1}{2} = 1 \text{ and } \frac{b - 1}{2} = 1 \implies a = 1 \text{ and } b = 3 \).

APPLICATION OF SECTION FORMULA

Theorem : The coordinates of the centroid of the triangle whose vertices are \( (x_1, y_1), (x_2, y_2) \text{ and } (x_3, y_3) \) are \( (\frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}) \)

EXAMPLES

Question. Find the coordinates of the centroid of a triangle whose vertices are (–1, 0), (5, –2) and (8, 2).
Answer: Coordinates of the centroid = \( (\frac{-1 + 5 + 8}{3}, \frac{0 - 2 + 2}{3}) = (4, 0) \).

Question. If the coordinates of the mid points of the sides of a triangle are (1, 1), (2, – 3) and (3, 4) Find its centroid.
Answer: Sum of mid-points \( x_1 + x_2 + x_3 = 1+2+3 = 6 \) and \( y_1 + y_2 + y_3 = 1-3+4 = 2 \). Centroid \( = (\frac{6}{3}, \frac{2}{3}) = (2, 2/3) \).

Question. Two vertices of a triangle are (3, –5) and (–7, 4). If its centroid is (2, –1). Find the third vertex.
Answer: \( \frac{x + 3 - 7}{3} = 2 \text{ and } \frac{y - 5 + 4}{3} = -1 \implies x = 10 \text{ and } y = -2 \). Third vertex is (10, –2).

AREA OF A TRIANGLE

Theorem : The area of a triangle, the coordinates of whose vertices are \( (x_1, y_1), (x_2, y_2) \text{ and } (x_3, y_3) \) is \[ \frac{1}{2} | x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) | \]

EXAMPLES

Type I : On finding the area of a triangle when coordinates of its vertices are given.

Question. Find the area of a triangle whose vertices are A(3, 2), B (11, 8) and C(8, 12).
Answer: Area of \( \triangle ABC = \frac{1}{2} | 3(8 - 12) + 11(12 - 2) + 8(2 - 8) | = \frac{1}{2} | -12 + 110 - 48 | = 25 \text{ sq. units} \)

Question. Prove that the area of triangle whose vertices are (t, t – 2), (t + 2, t + 2) and (t + 3, t) is independent of t.
Answer: Area of \( \triangle ABC = \frac{1}{2} | t(t + 2 - t) + (t + 2)(t - (t - 2)) + (t + 3)(t - 2 - (t + 2)) | = \frac{1}{2} | 2t + 2t + 4 - 4t - 12 | = 4 \text{ sq. units} \), which is independent of t.

Type II : On finding the area of a quadrilateral when coordinates of its vertices are given

Question. Find the area of the quadrilateral ABCD whose vertices are respectively A(1, 1), B(7, –3), C(12, 2) and D(7, 21).
Answer: Area = Area of \( \triangle ABC + \text{ Area of } \triangle ACD \). Area of \( \triangle ABC = 25 \text{ sq. units} \). Area of \( \triangle ACD = 107 \text{ sq. units} \). Total area = 132 sq. units.

Type III : On collinearity of three points

FORMULA: Three points \( (x_1, y_1), (x_2, y_2) \text{ and } (x_3, y_3) \) are collinear iff \( x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) = 0 \).

Question. Prove that the points (2, – 2), (–3, 8) and (–1, 4) are collinear.
Answer: \( \Delta = \frac{1}{2} | 2(8 - 4) + (-3)(4 + 2) + (-1)(-2 - 8) | = \frac{1}{2} | 8 - 18 + 10 | = 0 \). Hence points are collinear.

Question. Prove that the points (a, b + c), (b, c + a) and (c, a + b) are collinear.
Answer: \( \Delta = \frac{1}{2} | a(c + a - a - b) + b(a + b - b - c) + c(b + c - c - a) | = \frac{1}{2} | a(c - b) + b(a - c) + c(b - a) | = 0 \). Hence collinear.

Type IV : On Finding the desired result or unknown when three points are collinear

Question. For what value of k are the points (k, 2 – 2k) (–k + 1, 2k) and (–4 – k, 6 – 2k) are collinear ?
Answer: Area = 0 \( \implies | 2k^2 + (-k + 1)(6 - 2k) + (-4 - k)(2 - 2k) - \dots | = 0 \implies 2k^2 + k - 1 = 0 \implies (2k - 1)(k + 1) = 0 \implies k = 1/2 \) or \( k = -1 \).

Question. For what value of x will the points (x, –1), (2, 1) and (4, 5) lie on a line ?
Answer: Area = 0 \( \implies | x(1 - 5) + 2(5 + 1) + 4(-1 - 1) | = 0 \implies -4x + 4 = 0 \implies x = 1 \).

Type V : Mixed problems based upon the concept of area of a triangle

Question. If the coordinates of two points A and B are (3, 4) and (5, – 2) respectively. Find the coordinates of any point P, if PA = PB and Area of \( \triangle PAB = 10 \).
Answer: \( PA = PB \implies x - 3y - 1 = 0 \dots (1) \). Area = 10 \( \implies 3x + y - 23 = 0 \) or \( 3x + y - 3 = 0 \). Solving with (1) gives P(7, 2) or (1, 0).

Question. The coordinates of A, B, C are (6, 3), (–3, 5) and (4, – 2) respectively and P is any point (x, y). Show that the ratio of the areas of triangle PBC and ABC is \( \frac{x + y - 2}{7} \).
Answer: Area of \( \triangle PBC = \frac{7}{2} | x + y - 2 | \). Area of \( \triangle ABC = \frac{49}{2} \). Ratio \( = \frac{| x + y - 2 |}{7} \).

EXERCISE 

Question. Find the distance between the points \( (\cos\theta, \sin\theta) \) and \( (\sin\theta, – \cos\theta) \).
Answer: \( \sqrt{2} \)

Question. Find the distance between the points \( (a \cos 35^\circ, 0) \) and \( (0, a \cos 65^\circ) \).
Answer: a

Question. Find the distance between the points \( (a \cos \theta + b \sin \theta, 0) \) and \( (0, a \sin \theta – b \cos \theta) \)
Answer: \( \sqrt{a^2 + b^2} \)

Question. If the distance between the points \( (4, p) \) and \( (1, 0) \) is 5, then find p.
Answer: \( \pm 4 \)

Question. A line segment is of length 10 units. If the coordinates of its one end are \( (2, –3) \) and the abscissa of the other end is 10, then find its ordinate.
Answer: \( (3, –9) \)

Question. Find the perimeter of the triangle formed by the points \( (0, 0), (1, 0) \) and \( (0, 1) \).
Answer: \( 2 + \sqrt{2} \)

Question. If A \( (2, 2) \), B \( (–4, –4) \) and C \( (5, –8) \) are the vertices of a triangle, then find the length of the median through vertex C.
Answer: \( \sqrt{85} \)

Question. If three points \( (0, 0) (3, \sqrt{3}) \) and \( (3, \lambda) \) form an equilateral triangle, then find \( \lambda \).
Answer: \( -\sqrt{3} \) (Calculated by setting side lengths equal)

Question. If the points \( (k, 2k), (3k, 3k) \) and \( (3, 1) \) are collinear, then find k.
Answer: –1/3

Question. Find the coordinates of the point of X-axis which are equidistant from the points \( (–3, 4) \) and \( (2, 5) \).
Answer: \( (4/5, 0) \)

Question. If \( (–2, –1), (a, 0), (4, b) \) and \( (1, 2) \) vertices of a parallelogram then find value of a and b.
Answer: a = –2, b = 6

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

Question. Find the area of the triangle formed by \( (a, b + c), (b, c + a) \) and \( (c, a + b) \).
Answer: 0

Question. If \( (x, 2), (–3, –4) \) and \( (7, –5) \) are collinear, then find x.
Answer: –63

Question. If points \( (t, 2t), (–2, 6) \) and \( (3, 1) \) are collinear then find t.
Answer: 4/3

Question. If the area of the triangle formed by the points \( (x, 2x), (–2, 6) \) and \( (3, 1) \) is 5 square units, then find x.
Answer: 2/3

Question. If points \( (a, 0), (0, b) \) and \( (1, 1) \) are collinear, then find \( 1/a + 1/b \).
Answer: 1

Question. If the centroid of a triangle is \( (1, 4) \) and two of its vertices are \( (4, –3) \) and \( (–9, 7) \), then find the area of the triangle.
Answer: 183/2

Question. Find the ratio in which line segment joining points \( (–3, –4) \) and \( (1, –2) \) is divided by y-axis.
Answer: 3 : 1

Question. Find the ratio in which \( (4, 5) \) divides the join of \( (2, 3) \) and \( (7, 8) \).
Answer: 2 : 3

Question. The ratio in which the x-axis divides the segment joining \( (3, 6) \) and \( (12, –3) \).
Answer: 2 : 1

Question. If the centroid of the triangle formed by the points \( (a, b), (b, c) \) and \( (c, a) \) is at the origin, then find \( a^3 + b^3 + c^3 = \)
Answer: 3abc

Question. If the centroid of the triangle formed by \( (7, x), (y, –6) \) and \( (9, 10) \) is at \( (6, 3) \) then find \( (x, y) \)
Answer: \( (5, 2) \)

Question. The line joining the points A\( (4, –5) \) and B\( (4, 5) \) is divided by the point P such that AP : AB = 2 : 5, find the coordinates of P.
Answer: \( (4, –1) \)

Question. The line segment joining A\( (–3, 1) \) and B\( (5, –4) \) is a diameter of a circle whose centre is C. Find the coordinates of the point C.
Answer: \( (1, -3/2) \)

Question. The mid-point of the line joining \( (a, 2) \) and \( (3, 6) \) is \( (2, b) \). Find the values of a and b.
Answer: a = 1, b = 4

Question. The mid-point of the line segment joining \( (2a, 4) \) and \( (–2, 3b) \) is \( (1, 2a + 1) \). Find the values of a and b.
Answer: a = 2, b = 2

Question. The centre of a circle is \( (2, –3) \) and one end of a diameter is \( (1, 4) \), find the other end.
Answer: \( (3, –10) \)

Question. The point P\( (–4, 1) \) divides the line segment joining the points A\( (2, –2) \) and B in the ratio 3 : 5. Find the point B.
Answer: \( (–14, 6) \)

Question. If A\( (1, 1) \) and B\( (–2, 3) \) are two points and C is a point on AB produced such that AC = 3AB, find the co-ordinates of C.
Answer: \( (–8, 7) \)

Question. In what ratio does the point \( (–4, 6) \) divide the line segment joining the points A\( (–6, 10) \) and B\( (3, –8) \) ?
Answer: 2 : 7

Question. The line segment joining A \( (-1, -5/3) \) and B \( (a, 5) \) is divided in the ratio 1 : 3 at P, the point where the line segment AB intersects y-axis. Find (i) the value of a (ii) the co-ordinates of P.
Answer: (i) 3, (ii) \( (0, 5/2) \)

Question. Find the ratio in which the y-axis divides the line segment joining the points \( (5, – 6,) \) and \( (– 1, – 4) \). Also find the co-ordinates of the point of intersection.
Answer: 5 : 1, \( (0, –13/3) \)

Question. Calculate the ratio in which the line joining A\( (6, 5) \) and B\( (4, –3) \) is divided by the line \( y = 2 \). Also find the co-ordinates of the point of division.
Answer: 3 : 5, \( (21/4, 2) \)

Question. Determine the ratio in which the line \( 2x + y – 4 = 0 \) divide the line segment joining the points A\( (2, –2) \) and B\( (3, 7) \). Also find the coordinates of the point of division.
Answer: 2 : 9, \( (24/11, -4/11) \)

Question. A\( (10, 5) \), B\( (6, –3) \) and C\( (2, 1) \) are the vertices of a triangle ABC. L is the mid-point of AB and M is the mid-point of AC. Write down the co-ordinates of L and M. Show that LM = 1/2 BC.
Answer: \( L(8, 1), M(6, 3) \). LM = \( \sqrt{(8-6)^2 + (1-3)^2} = \sqrt{8} = 2\sqrt{2} \); BC = \( \sqrt{(6-2)^2 + (-3-1)^2} = \sqrt{32} = 4\sqrt{2} \). LM = 1/2 BC Proved.

Question. Find the third vertex of a triangle if its two vertices are \( (–1, 4) \) and \( (5, 2) \) and mid-point of one side is \( (0, 3) \).
Answer: \( (–5, 4) \) or \( (1, 2) \)