CBSE Class 10 Coordinate geometry Sure Shot Questions Set C

CBSE Class 10 Coordinate geometry Sure Shot Questions Set C. There are many more useful educational material which the students can download in pdf format and use them for studies. Study material like concept maps, important and sure shot question banks, quick to learn flash cards, flow charts, mind maps, teacher notes, important formulas, past examinations question bank, important concepts taught by teachers. Students can download these useful educational material free and use them to get better marks in examinations.  Also refer to other worksheets for the same chapter and other subjects too. Use them for better understanding of the subjects.

1. Find the area of a triangle formed by the points A(5, 2), B(4, 7) and C(7, – 4).

2. Find the area of a triangle formed by the points A(1, –1), B(– 4, 6) and C(– 3, – 5).

3. Find the area of a triangle formed by the points A(2, 3), B(– 1, 0) and C(2, –4).

4. Find the area of a triangle formed by the points A(10, –6), B(2, 5) and C(– 1, 3).

5. Determine if the points (1, 5), (2, 3) and (– 2, – 11) are collinear.

6. Show that the points ( -3 /2, 3), (6, –2), (–3, 4) are collinear by using area of triangle.

7. By using area of triangle show that the points (a, b + c), (b, c + a) and (c, a + b) are collinear.

8. Find the value of k if the points A(8, 1), B(k, –4) and C(2, –5) are collinear.

9. Find the value of k if the points A(7, –2), B(5, 1) and C(3, k) are collinear.

10. If A(3, 2), B(–1, 0) and C(1, –12) are the vertices of a triangle and D is midpoint of BC, find the coordinates of the point D. Also find the areas of ΔABD and ΔACD. Hence verify that the median AD divides the triangle ABC into two triangles of equal areas.

11. Find the value of k if the points A(2, 3), B(4, k) and C(6, –3) are collinear.

12. If A(–5, 7), B(– 4, –5), C(–1, –6) and D(4, 5) are the vertices of a quadrilateral, find the area of the quadrilateral ABCD.

13. If A(2, 1), B(6, 0), C(5, –2) and D(–3, –1) are the vertices of a quadrilateral, find the area of the quadrilateral ABCD.

14. If A(–4, 5), B(0, 7), C(5, –5) and D(–4, –2) are the vertices of a quadrilateral, find the area of the quadrilateral ABCD.

15. If A(0, 0), B(6, 0), C(4, 3) and D(0, 3) are the vertices of a quadrilateral, find the area of the quadrilateral ABCD.

16. If A(–4, –2), B(–3, –5), C(3, –2) and D(2, 3) are the vertices of a quadrilateral, find the area of the quadrilateral ABCD.

17. Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, –1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle.

18. Find a relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear. 

19. If A(2, 1), B(–2, 3) and C(4, –3) are the vertices of a △ABC and D, E are the midpoints of the sides AB, AC respectively, find the coordinates of D and E. Prove that the area of △ABC is four times the area of △ADE.

20. If A(4, 4), B(3, –16) and C(3, –2) are the vertices of a △ABC and D, E, F are the midpoints of the sides BC, CA and AB respectively. Prove that the area of △ABC is four times the area of △DEF.

21. Find the point P on the x – axis which is equidistant from the points A(5, 4) and B(–2, 3). Also find the area of △PAB.

22. If P(x, y) is any point on the line joining the points A(a, 0) and B(0, b), then show that x/a + y/b =1 

23. If the vertices of a triangle are (1, k), (4, –3) and (–9, 7) and its area is 15 sq. units, find the value(s) of k.

24. Find the value of m for which the points with coordinates (3, 5), (m, 6) and 1/2 ,15/2 ) are collinear.

25. Find the value of k for which the points with coordinates (2, 5), (4, 6) and ,(k , 11/2) are collinear.

26. Find the point P on x-axis which is equidistant from A(–2, 5) and B(2, –3). Also find the area of △PAB.

27. Find the point P on x-axis which is equidistant from A(7, 6) and B(–3, 4). Also find the area of △PAB.

28. Find the point P on the x-axis which is equidistant from A(2, –5) and B(–2, 9). Also find the area of △PAB.

29. Find a point P on the y-axis which is equidistant from the points A(6, 5) and B(– 4, 3). Also find the area of △PAB.

30. Find a point P on the y-axis which is equidistant from the points A(5, 2) and B(– 4, 3). Also find the area of △PAB.

31. Find a point P on the y-axis which is equidistant from the points A(5, – 2) and B(– 3, 2). Also find the area of △PAB.

32. Find the value of k for which the area formed by the triangle with vertices A(k, 2k), (–2, 6) and C(3, 1) is 5 square units.

33. Find the value of k for which the area formed by the triangle with vertices A(1, 2), (–2, 3) and C(–3, k) is 11 square units.

34. Find the value of k for which the area formed by the triangle with vertices A(4, 4), (3, k) and C(3, –2) is 7 square units.

35. For what value of p are the points A(–3, 9), B(2, p) and C(4, –5) are collinear.

36. Prove that the area of triangle whose vertices are (t, t – 2), (t + 2, t + 2) and (t + 3, t) is independent of t.

37. For what value of k are the points (k, 2 – 2k), (–k + 1, 2k) and (–4 – k, 6 – 2k) are collinear.

38. Find the condition that the point (x, y) may lie on the line joining (3, 4) and (–5, –6).

39. 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 △PAB = 10 sq. units.

40. 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 triangles PBC and ABC is

41. If (x, y) be on the line joining the two points (1, –3) and (–4, 2), prove that x + y + 2 = 0.

42. Prove that the points (a, b), (x, y) and (a – x, b – y) are collinear if ay = bx.

43. Four points A(6, 3), B(–3, 5), C(4, –2) and D(x, 3x) are given in such a way that DBC /ABC = 1/2 find the value of x.

44. If three points (a, b), (c, d) and (e, f) are collinear, prove that d-f/ce +f-b/ea + b-d /ac =0

45. The area of triangle is 5 sq. units. Two of its vertices are (2, 1) and (3, –2). The third vertex lie on y = x + 3. Find the third vertex.

Facts that Matter

• Distance Formula
  • I. The distance between two points \(P(x_1, y_1)\) and \(Q(x_2, y_2)\) is given by \(PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)
  • II. The distance of \(P(x, y)\) from the origin is \(OP = \sqrt{x^2 + y^2}\)
• Section Formula
  • I. If \(P(x, y)\) divides the line segment \(AB\), joining the points \(A(x_1, y_1)\) and \(B(x_2, y_2)\) in the ratio \(m : n\), then \(x = \frac{mx_2 + nx_1}{m + n}\), \(y = \frac{my_2 + ny_1}{m + n}\)
  • II. The coordinates \((x, y)\) of the mid-point of \(PQ\) are given by: \(x = \frac{x_1 + x_2}{2}\), \(y = \frac{y_1 + y_2}{2}\)
  • III. If \(A(x_1, y_1)\), \(B(x_2, y_2)\) and \(C(x_3, y_3)\) be the vertices of \(\Delta ABC\), then the coordinates \(G(x, y)\) of the centroid of \(\Delta ABC\) are: \(x = \frac{x_1 + x_2 + x_3}{3}\) and \(y = \frac{y_1 + y_2 + y_3}{3}\)

Question. Find the distance between the following pairs of points: (i) (2, 3), (4, 1) (ii) (− 5, 7), (− 1, 3) (iii) (a, b), (− a, − b)
Answer: (i) Here \(x_1 = 2, y_1 = 3, x_2 = 4\) and \(y_2 = 1\)
\(\therefore\) The required distance \( = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)
\( = \sqrt{(4 - 2)^2 + (1 - 3)^2} \)
\( = \sqrt{2^2 + (-2)^2} \)
\( = \sqrt{4 + 4} = \sqrt{8} \)
\( = \sqrt{2 \times 4} = 2\sqrt{2} \)
(ii) Here, \(x_1 = -5, y_1 = 7, x_2 = -1, y_2 = 3\)
\(\therefore\) The required distance \( = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)
\( = \sqrt{[-1 - (-5)]^2 + (3 - 7)^2} \)
\( = \sqrt{(-1 + 5)^2 + (-4)^2} \)
\( = \sqrt{4^2 + 16} \)
\( = \sqrt{16 + 16} \)
\( = \sqrt{32} = \sqrt{2 \times 16} = 4\sqrt{2} \)
(iii) Here, \(x_1 = a, y_1 = b, x_2 = -a, y_2 = -b\)
\(\therefore\) The required distance \( = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)
\( = \sqrt{(-a - a)^2 + (-b - b)^2} \)
\( = \sqrt{(-2a)^2 + (-2b)^2} \)
\( = \sqrt{4a^2 + 4b^2} \)
\( = \sqrt{4(a^2 + b^2)} = 2\sqrt{a^2 + b^2} \)

Question. Determine if the points (1, 5), (2, 3) and (− 2, − 11) are collinear.
Answer: Let the points be \(A(1, 5), B(2, 3)\) and \(C(-2, -11)\)
\(A, B\) and \(C\) are collinear, if
\(AB + BC = AC\)
\(AC + CB = AB\)
\(BA + AC = BC\)
\(\because AB = \sqrt{(2 - 1)^2 + (3 - 5)^2} \)
\( = \sqrt{1^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5} \)
\(BC = \sqrt{(-2 - 2)^2 + (-11 - 3)^2} \)
\( = \sqrt{(-4)^2 + (-14)^2} = \sqrt{16 + 196} = \sqrt{212} \)
\(AC = \sqrt{(-2 - 1)^2 + (-11 - 5)^2} \)
\( = \sqrt{(-3)^2 + (-16)^2} = \sqrt{9 + 256} = \sqrt{265} \)
But \(AB + BC \neq AC\)
\(AC + CB \neq AB\)
\(BA + AC \neq BC\)
\(\therefore A, B\) and \(C\) are not collinear.

Question. Check whether (5, − 2), (6, 4) and (7, − 2) are the vertices of an isosceles triangle.
Answer: Let the points be \(A(5, -2), B(6, 4)\) and \(C(7, -2)\).
\(\therefore AB = \sqrt{(6 - 5)^2 + [4 - (-2)]^2} \)
\( = \sqrt{1^2 + (6)^2} \)
\( = \sqrt{1 + 36} = \sqrt{37} \)
\(BC = \sqrt{(7 - 6)^2 + (-2 - 4)^2} \)
\( = \sqrt{1^2 + (-6)^2} = \sqrt{1 + 36} = \sqrt{37} \)
\(AC = \sqrt{(7 - 5)^2 + [-2 - (-2)]^2} \)
\( = \sqrt{2^2 + (0)^2} = \sqrt{4 + 0} = 2 \)
We have \(AB = BC \neq AC\)
\(\therefore \Delta ABC\) is an isosceles triangle.

Question. Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:
(i) (− 1, − 2), (1, 0), (− 1, 2), (− 3, 0)
(ii) (− 3, 5), (3, 1), (0, 3), (− 1, − 4)
(iii) (4, 5), (7, 6), (4, 3), (1, 2)
Answer: (i) Let the points be: \(A(-1, -2), B(1, 0), C(-1, 2)\) and \(D(-3, 0)\).
\(\therefore AB = \sqrt{[1 - (-1)]^2 + [0 - (-2)]^2} \)
\( = \sqrt{2^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} \)
\(BC = \sqrt{(-1 - 1)^2 + (2 - 0)^2} = \sqrt{(-2)^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} \)
\(CD = \sqrt{[-3 - (-1)]^2 + (0 - 2)^2} = \sqrt{(-2)^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} \)
\(DA = \sqrt{[-1 - (-3)]^2 + [-2 - 0]^2} = \sqrt{2^2 + (-2)^2} = \sqrt{4 + 4} = \sqrt{8} \)
\(AC = \sqrt{[-1 - (-1)]^2 + [2 - (-2)]^2} = \sqrt{0^2 + 4^2} = 4 \)
\(BD = \sqrt{(-3 - 1)^2 + (0 - 0)^2} = \sqrt{(-4)^2 + 0^2} = 4 \)
\(\Rightarrow AB = BC = CD = AD\)
i.e., All the sides are equal. And \(AC = BD\) (the diagonals are equal).
\(\therefore ABCD\) is a square.
(ii) Let the points be \(A(-3, 5), B(3, 1), C(0, 3)\) and \(D(-1, -4)\).
\(\therefore AB = \sqrt{[3 - (-3)]^2 + (1 - 5)^2} \)
\( = \sqrt{6^2 + (-4)^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13} \)
\(BC = \sqrt{(0 - 3)^2 + (3 - 1)^2} = \sqrt{(-3)^2 + 2^2} = \sqrt{9 + 4} = \sqrt{13} \)
\(AC = \sqrt{[0 - (-3)]^2 + (3 - 5)^2} = \sqrt{3^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13} \)
We see that: \(\sqrt{13} + \sqrt{13} = 2\sqrt{13}\)
i.e., \(AC + BC = AB\)
\(\Rightarrow A, B, C\) and \(D\) are collinear. Thus, \(ABCD\) is not a quadrilateral.
(iii) Let the points be \(A(4, 5), B(7, 6), C(4, 3)\) and \(D(1, 2)\).
\(\therefore AB = \sqrt{(7 - 4)^2 + (6 - 5)^2} = \sqrt{3^2 + 1^2} = \sqrt{10} \)
\(BC = \sqrt{(4 - 7)^2 + (3 - 6)^2} = \sqrt{(-3)^2 + (-3)^2} = \sqrt{18} \)
\(CD = \sqrt{(1 - 4)^2 + (2 - 3)^2} = \sqrt{(-3)^2 + (-1)^2} = \sqrt{10} \)
\(DA = \sqrt{(4 - 1)^2 + (5 - 2)^2} = \sqrt{3^2 + 3^2} = \sqrt{18} \)
\(AC = \sqrt{(4 - 4)^2 + (3 - 5)^2} = \sqrt{0^2 + (-2)^2} = 2 \)
\(BD = \sqrt{(1 - 7)^2 + (2 - 6)^2} = \sqrt{(-6)^2 + (-4)^2} = \sqrt{36 + 16} = \sqrt{52} \)
Since, \(AB = CD\) and \(BC = DA\) [opposite sides are equal] and \(AC \neq BD \Rightarrow\) Diagonals are unequal.
\(\therefore ABCD\) is a parallelogram.

Question. Find the point on the x-axis which is equidistant from (2, − 5) and (− 2, 9).
Answer: We know that any point on x-axis has its ordinate \( = 0\).
Let the required point be \(P(x, 0)\).
Let the given points be \(A(2, -5)\) and \(B(-2, 9)\).
\(\therefore PA = \sqrt{(x - 2)^2 + [0 - (-5)]^2} = \sqrt{x^2 - 4x + 4 + 25} = \sqrt{x^2 - 4x + 29} \)
\(PB = \sqrt{[x - (-2)]^2 + (0 - 9)^2} = \sqrt{(x + 2)^2 + (-9)^2} = \sqrt{x^2 + 4x + 4 + 81} = \sqrt{x^2 + 4x + 85} \)
Since, \(A\) and \(B\) are equidistant from \(P\),
\(\therefore PA = PB \)
\(\Rightarrow \sqrt{x^2 - 4x + 29} = \sqrt{x^2 + 4x + 85} \)
\(\Rightarrow x^2 - 4x + 29 = x^2 + 4x + 85 \)
\(\Rightarrow -4x - 4x = 85 - 29 \)
\(\Rightarrow -8x = 56 \)
\(\Rightarrow x = \frac{56}{-8} = -7 \)
\(\therefore\) The required point is \((-7, 0)\).

Question. Find the values of y for which the distance between the points P (2, − 3) and Q (10, y) is 10 units.
Answer: The given points are \(P(2, -3)\) and \(Q(10, y)\).
\(\therefore PQ = \sqrt{(10 - 2)^2 + [y - (-3)]^2} \)
\( = \sqrt{8^2 + (y + 3)^2} \)
\( = \sqrt{64 + y^2 + 6y + 9} \)
\( = \sqrt{y^2 + 6y + 73} \)
But \(PQ = 10 \Rightarrow \sqrt{y^2 + 6y + 73} = 10 \)
Squaring both sides,
\(y^2 + 6y + 73 = 100 \)
\(\Rightarrow y^2 + 6y - 27 = 0 \)
\(\Rightarrow y^2 - 3y + 9y - 27 = 0 \)
\(\Rightarrow (y - 3)(y + 9) = 0 \)
\(\Rightarrow\) Either \(y - 3 = 0 \Rightarrow y = 3\)
or \(y + 9 = 0 \Rightarrow y = -9\)
\(\therefore\) The required value of \(y\) is \(3\) or \(-9\).

Question. If Q (0, 1) is equidistant from P (5, − 3) and R (x, 6), find the values of x. Also find the distances QR and PR.
Answer: Here, \(QP = \sqrt{(5 - 0)^2 + [-3 - 1]^2} = \sqrt{5^2 + (-4)^2} = \sqrt{25 + 16} = \sqrt{41} \)
\(QR = \sqrt{(x - 0)^2 + (6 - 1)^2} = \sqrt{x^2 + 5^2} = \sqrt{x^2 + 25} \)
\(\because QP = QR \Rightarrow \sqrt{41} = \sqrt{x^2 + 25} \)
Squaring both sides, we have: \(x^2 + 25 = 41 \)
\(\Rightarrow x^2 = 41 - 25 = 16 \Rightarrow x = \pm 4\)
Thus, the point \(R\) is \((4, 6)\) or \((-4, 6)\)
Now, \(QR = \sqrt{(\pm 4 - 0)^2 + (6 - 1)^2} = \sqrt{16 + 25} = \sqrt{41} \)
and \(PR = \sqrt{(\pm 4 - 5)^2 + (6 + 3)^2} \)
\(\Rightarrow PR = \sqrt{(4 - 5)^2 + (6 + 3)^2}\) or \(\sqrt{(-4 - 5)^2 + (6 + 3)^2} \)
\(\Rightarrow PR = \sqrt{1 + 81}\) or \(\sqrt{(-9)^2 + 9^2} \)
\(\Rightarrow PR = \sqrt{82}\) or \(\sqrt{2 \times 9^2} = 9\sqrt{2} \)

Question. Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (− 3, 4).
Answer: Let the points be \(A(x, y), B(3, 6)\) and \(C(-3, 4)\).
\(\therefore AB = \sqrt{(x - 3)^2 + (y - 6)^2} \)
And \(AC = \sqrt{[x - (-3)]^2 + (y - 4)^2} \)
Since, the point \((x, y)\) is equidistant from \((3, 6)\) and \((-3, 4)\).
\(\therefore AB = AC \)
\(\Rightarrow \sqrt{(x - 3)^2 + (y - 6)^2} = \sqrt{(x + 3)^2 + (y - 4)^2} \)
Squaring both sides,
\((3 - x)^2 + (6 - y)^2 = (-3 - x)^2 + (4 - y)^2 \)
\(\Rightarrow (9 + x^2 - 6x) + (36 + y^2 - 12y) = (9 + x^2 + 6x) + (16 + y^2 - 8y) \)
\(\Rightarrow 9 + x^2 - 6x + 36 + y^2 - 12y - 9 - x^2 - 6x - 16 - y^2 + 8y = 0 \)
\(\Rightarrow -12x - 4y + 20 = 0 \)
\(\Rightarrow -3x - y + 5 = 0\) [Dividing by 4]
\(\Rightarrow 3x + y - 5 = 0\)
which is the required relation between \(x\) and \(y\).

Question. Find the coordinates of the point which divides the join of (− 1, 7) and (4, − 3) in the ratio 2 : 3.
Answer: Let the required point be \(P(x, y)\).
Here, the end points are: \((-1, 7)\) and \((4, -3)\)
\(\because\) Ratio \( = 2 : 3 = m_1 : m_2 \)
\(\therefore x = \frac{m_1x_2 + m_2x_1}{m_1 + m_2} = \frac{2 \times 4 + 3 \times (-1)}{2 + 3} = \frac{8 - 3}{5} = \frac{5}{5} = 1 \)
And \(y = \frac{m_1y_2 + m_2y_1}{m_1 + m_2} = \frac{2 \times (-3) + 3 \times 7}{2 + 3} = \frac{-6 + 21}{5} = \frac{15}{5} = 3 \)
Thus, the required point is \((1, 3)\).

Question. Find the coordinates of the points of trisection of the line segment joining (4, − 1) and (− 2, − 3).
Answer: Let the given points be \(A(4, -1)\) and \(B(-2, -3)\).
Let the points \(P\) and \(Q\) trisect \(AB\).
i.e., \(AP = PQ = QB\)
i.e., \(P\) divides \(AB\) in the ratio of \(1 : 2\)
\(Q\) divides \(AB\) in the ratio of \(2 : 1\)
Let the coordinates of \(P\) be \((x, y)\).
\(\therefore x = \frac{m_1x_2 + m_2x_1}{m_1 + m_2} = \frac{1(-2) + 2(4)}{1 + 2} = \frac{-2 + 8}{3} = 2 \)
\(y = \frac{m_1y_2 + m_2y_1}{m_1 + m_2} = \frac{1(-3) + 2(-1)}{1 + 2} = \frac{-3 - 2}{3} = \frac{-5}{3} \)
\(\therefore\) The required co-ordinates of \(P\) are \((2, -\frac{5}{3})\).
Let the co-ordinates of \(Q\) be \((X, Y)\).
\(\therefore X = \frac{2(-2) + 1(4)}{2 + 1} = \frac{-4 + 4}{3} = 0 \)
\(Y = \frac{2(-3) + 1(-1)}{2 + 1} = \frac{-6 - 1}{3} = \frac{-7}{3} \)
\(\therefore\) The required coordinates of \(Q\) are \((0, -\frac{7}{3})\).

Question. Find the ratio in which the line segment joining the points (− 3, 10) and (6, − 8) is divided by (− 1, 6).
Answer: Let the given points are: \(A(-3, 10)\) and \(B(6, -8)\).
Let the point \(P(-1, 6)\) divides \(AB\) in the ratio \(m_1 : m_2\).
\(\therefore\) Using the section formula, we have:
\((-1, 6) = (\frac{x_2m_1 + m_2x_1}{m_1 + m_2}, \frac{m_1y_2 + m_2y_1}{m_1 + m_2})\)
\(\Rightarrow (-1, 6) = (\frac{6 \times m_1 + m_2 \times (-3)}{m_1 + m_2}, \frac{m_1 \times (-8) + m_2 \times 10}{m_1 + m_2})\)
\(\Rightarrow -1 = \frac{6m_1 - 3m_2}{m_1 + m_2}\) and \(6 = \frac{-8m_1 + 10m_2}{m_1 + m_2}\)
\(\Rightarrow -1(m_1 + m_2) = 6m_1 - 3m_2\) and \(6(m_1 + m_2) = -8m_1 + 10m_2\)
\(\Rightarrow -m_1 - m_2 - 6m_1 + 3m_2 = 0\) and \(6m_1 + 6m_2 + 8m_1 - 10m_2 = 0\)
\(\Rightarrow -7m_1 + 2m_2 = 0\) and \(14m_1 - 4m_2 = 0 \Rightarrow 7m_1 - 2m_2 = 0\)
\(\Rightarrow 2m_2 = 7m_1 \Rightarrow \frac{m_1}{m_2} = \frac{2}{7}\)
Thus, the required ratio is \(2 : 7\).

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: Part-I: To find the ratio
The given points are: \(A(1, -5)\) and \(B(-4, 5)\).
Let the required ratio \( = k : 1\) and the required point be \(P(x, y)\).
Since the point \(P\) lies on x-axis, \(\therefore\) Its y-coordinate is \(0\).
\(0 = \frac{m_1y_2 + m_2y_1}{m_1 + m_2} = \frac{k(5) + 1(-5)}{k + 1}\)
\(\Rightarrow 0 = \frac{5k - 5}{k + 1} \Rightarrow 5k - 5 = 0 \Rightarrow k = 1\)
Part II : To find coordinates of \(P\):
\(x = \frac{k(-4) + 1(1)}{k + 1} = \frac{1(-4) + 1}{1 + 1} = -\frac{3}{2}\) [\(\because k = 1\)]
\(\therefore\) The required ratio \(k : 1 = 1 : 1\).
Coordinates of \(P\) are \((x, 0) = (-\frac{3}{2}, 0)\).

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: Here, the given points are:
A \( (- 2, 2) \) and B \( (2, 8) \)
Let \( P_1, P_2 \) and \( P_3 \) divide AB in four equal parts.
\(\therefore AP_1 = P_1 P_2 = P_2 P_3 = P_3 B \)
Obviously, \( P_2 \) is the mid point of AB
\(\therefore\) Coordinates of \( P_2 \) are:
\( \left( \frac{- 2 + 2}{2}, \frac{2 + 8}{2} \right) \) or \( (0, 5) \)
Again, \( P_1 \) is the mid point of \( AP_2 \).
\(\therefore\) Coordinates of \( P_1 \) are:
\( \left( \frac{- 2 + 0}{2}, \frac{2 + 5}{2} \right) \) or \( \left( - 1, \frac{7}{2} \right) \)
Also \( P_3 \) is the mid point of \( P_2 B \).
\(\therefore\) Coordinates of \( P_3 \) are:
\( \left( \frac{0 + 2}{2}, \frac{5 + 8}{2} \right) \) or \( \left( 1, \frac{13}{2} \right) \)
Thus, the coordinates of \( P_1, P_2 \) and \( P_3 \) are:
\( (0, 5), \left( - 1, \frac{7}{2} \right) \) and \( \left( 1, \frac{13}{2} \right) \) respectively.

Area of Triangle

• I. If \( A (x_1, y_1); B (x_2, y_2) \) and \( C (x_3, y_3) \) are the vertices of \( \Delta ABC \), then the area of \( \Delta ABC = \frac{1}{2} [x_1 (y_2 - y_3) + x_2 (y_3 - y_1) + x_3 (y_1 - y_2)] \).
• II. The three points A, B and C are collinear if and only if area of \( \Delta ABC = 0 \).

NCERT TEXTBOOK QUESTIONS SOLVED

EXERCISE 7.3

Question. Find the area of the triangle whose vertices are:
(i) \( (2, 3), (- 1, 0), (2, - 4) \) (ii) \( (- 5, - 1), (3, - 5), (5, 2) \)
Answer: (i) Let the vertices of the triangle be
A \( (2, 3), B (- 1, 0) \) and C \( (2, - 4) \)
Here, \( x_1 = 2, y_1 = 3 \)
\( x_2 = - 1, y_2 = 0 \)
\( x_3 = 2, y_3 = - 4 \)
\(\because\) Area of a \( \Delta = \frac{1}{2} [x_1 (y_2 - y_3) + x_2 (y_3 - y_1) + x_3 (y_1 - y_2)] \)
\(\therefore\) Area of \( \Delta ABC = \frac{1}{2} [2 \{0 - (- 4)\} + (- 1) \{- 4 - (3)\} + 2 \{3 - 0\}] \)
\( = \frac{1}{2} [2 (0 + 4) + (- 1) (- 4 - 3) + 2 (3)] \)
\( = \frac{1}{2} [8 + 7 + 6] \)
\( = \frac{1}{2} [21] = \frac{21}{2} \) sq. units.
(ii) Let the vertices of the triangle be
A \( (- 5, - 1), B (3, - 5) \) and C \( (5, 2) \)
i.e., \( x_1 = - 5, y_1 = - 1 \)
\( x_2 = 3, y_2 = - 5 \)
\( x_3 = 5, y_3 = 2 \)
\(\because\) Area of a \( \Delta = \frac{1}{2} [x_1 (y_2 - y_3) + x_2 (y_3 - y_1) + x_3 (y_1 - y_2)] \)
\(\therefore\) Area of \( \Delta ABC = \frac{1}{2} [- 5 \{ - 5 - 2 \} + 3 \{ 2 - (- 1) \} + 5 \{ - 1 - (- 5) \} ] \)
\( = \frac{1}{2} [ - 5 \{ - 7 \} + 3 \{ 2 + 1 \} + 5 \{ - 1 + 5 \} ] \)
\( = \frac{1}{2} [ ( - 5) ( - 7) + 3 (3) + 5 (4) ] \)
\( = \frac{1}{2} [ 35 + 9 + 20 ] \)
\( = \frac{1}{2} \times 64 = 32 \) sq. units.

Question. In each of the following find the value of ‘k’, for which the points are collinear.
(i) \( (7, - 2), (5, 1), (3, k) \) (ii) \( (8, 1), (k, - 4), (2, - 5) \)
Answer: The given three points will be collinear if the \( \Delta \) formed by them has zero area.
(i) Let A \( (7, - 2), B (5, 1) \) and C \( (3, k) \) be the vertices of a triangle.
\(\therefore\) The given points will be collinear, if \( ar (\Delta ABC) = 0 \)
or \( 7 (1 - k) + 5 (k + 2) + 3 (- 2 - 1) = 0 \)
\(\Rightarrow 7 - 7k + 5k + 10 + (- 6) - 3 = 0 \)
\(\Rightarrow 17 - 9 + 5k - 7k = 0 \)
\(\Rightarrow 8 - 2k = 0 \)
\(\Rightarrow 2k = 8 \)
\(\Rightarrow k = \frac{8}{2} = 4 \)
The required value of \( k = 4 \).
(ii) Let \( (8, 1), (k, - 4) \) and \( (2, - 5) \) be the verticles of a triangle.
\(\therefore\) For the above points being collinear, \( ar (\Delta ABC) = 0 \)
i.e., \( 8 (- 4 + 5) + k (- 5 - 1) + 2 [1 - (- 4)] = 0 \)
\(\Rightarrow 8 (+ 1) + k (- 6) + 2 (5) = 0 \)
\(\Rightarrow 8 + (- 6k) + 10 = 0 \)
\(\Rightarrow - 6k + 18 = 0 \)
\(\Rightarrow k = (- 18) \div (- 6) = 3 \)
Thus, \( k = 3 \).

Question. Determine the ratio in which the line \( 2x + y - 4 = 0 \) divides the line segment joining the points A \( (2, - 2) \) and B \( (3, 7) \).
Answer: Let the required ratio be \( k : 1 \) and the point C divides them in the above ratio.
\(\therefore\) Coordinates of C are:
\( \left( \frac{3k + 2}{k + 1}, \frac{7k - 2}{k + 1} \right) \),
Since the point C lies on the given line \( 2x + y - 4 = 0 \),
\(\therefore\) We have:
\( 2 \left( \frac{3k + 2}{k + 1} \right) + \left( \frac{7k - 2}{k + 1} \right) - 4 = 0 \)
\(\Rightarrow 2 (3k + 2) + (7k - 2) = 4 \times (k + 1) \)
\(\Rightarrow 6k + 4 + 7k - 2 - 4k - 4 = 0 \)
\(\Rightarrow (6 + 7 - 4) k + (4 - 2 - 4) = 0 \)
\(\Rightarrow 9k + (- 2) = 0 \)
\(\Rightarrow 9k - 2 = 0 \)
\(\Rightarrow k = \frac{2}{9} \)
\(\therefore\) The required ratio
\( = k : 1 \)
\( = \frac{2}{9} : 1 \)
\( = 2 : 9 \)

Question. Find a relation between x and y if the points \( (x, y), (1, 2) \) and \( (7, 0) \) are collinear.
Answer: The given points are:
A \( (x, y), B (1, 2) \) and C \( (7, 0) \)
The points A, B and C will be collinear if
\( x (2 - 0) + 1 (0 - y) + 7 (y - 2) = 0 \)
or if \( 2x - y + 7y - 14 = 0 \)
or if \( 2x + 6y - 14 = 0 \)
or if \( x + 3y - 7 = 0 \)
which is the required relation between x and y.

Question. Find the centre of a circle passing through the points \( (6, - 6), (3, - 7) \) and \( (3, 3) \).
Answer: Let \( P (x, y) \) be the centre of the circle passing through
A \( (6, - 6), B (3, - 7) \) and C \( (3, 3) \)
\(\therefore AP = BP = CP \)
Taking \( AP = BP \), we have \( AP^2 = BP^2 \)
\(\Rightarrow (x - 6)^2 + (y + 6)^2 = (x - 3)^2 + (y + 7)^2 \)
\(\Rightarrow x^2 - 12x + 36 + y^2 + 12y + 36 = x^2 - 6x + 9 + y^2 + 14y + 49 \)
\(\Rightarrow - 12x + 6x + 12y - 14y + 72 - 58 = 0 \)
\(\Rightarrow - 6x - 2y + 14 = 0 \)
\(\Rightarrow 3x + y - 7 = 0 \) ...(1) [Dividing by \( (- 2) \)]
Taking \( BP = CP \), we have \( BP^2 = CP^2 \)
\(\Rightarrow (x - 3)^2 + (y + 7)^2 = (x - 3)^2 + (y - 3)^2 \)
\(\Rightarrow x^2 - 6x + 9 + y^2 + 14y + 49 = x^2 - 6x + 9 + y^2 - 6y + 9 \)
\(\Rightarrow - 6x + 6x + 14y + 6y + 58 - 18 = 0 \)
\(\Rightarrow 20y + 40 = 0 \)
\(\Rightarrow y = \frac{- 40}{20} = - 2 \) ...(2)
From (1) and (2),
\( 3x - 2 - 7 = 0 \)
\(\Rightarrow 3x = 9 \Rightarrow x = 3 \)
i.e., \( x = 3 \) and \( y = - 2 \)
\(\therefore\) The required centre is \( (3, - 2) \).

VERY SHORT ANSWER TYPE QUESTIONS

Question. Find a point on the y-axis equidistant from \( (-5, 2) \) and \( (9, -2) \). 
Answer: Let the required point on the y-axis be \( P(0, y) \)
\(\therefore PA = PB \)
\(\Rightarrow \sqrt{(0 + 5)^2 + (y - 2)^2} = \sqrt{(0 - 9)^2 + (y + 2)^2} \)
\(\Rightarrow \sqrt{5^2 + y^2 + 4 - 4y} = \sqrt{(-9)^2 + y^2 + 4 + 4y} \)
\(\Rightarrow 25 + y^2 + 4 - 4y = 81 + y^2 + 4 + 4y \)
\(\Rightarrow y^2 - y^2 - 4y - 4y = 81 + 4 - 4 - 25 \)
\(\Rightarrow -8y = 85 - 29 \)
\(\Rightarrow -8y = 56 \)
\(\Rightarrow y = \frac{56}{-8} = -7 \)
\(\therefore\) The required point is \( (0, -7) \).

Question. Find a point on x-axis at a distance of 4 units from the point \( A(2, 1) \).
Answer: Let the required point on x-axis be \( P(x, 0) \).
\(\therefore PA = 4 \)
\(\Rightarrow \sqrt{(x - 2)^2 + (0 - 1)^2} = 4 \)
\(\Rightarrow x^2 - 4x + 4 + 1 = 4^2 = 16 \)
\(\Rightarrow x^2 - 4x + 5 - 16 = 0 \)
\(\Rightarrow x^2 - 4x - 11 = 0 \)
\(\Rightarrow x = 2 \pm \sqrt{15} \)
Thus, the coordinates of P are: \( (2 \pm \sqrt{15}, 0) \).

Question. Find the distance of the point \( (3, -4) \) from the origin.
Answer: The coordinates of origin \( (0, 0) \).
\(\therefore\) Distance of \( (3, -4) \) from the origin
\( = \sqrt{(3 - 0)^2 + (-4 - 0)^2} \)
\( = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \).

Question. For what value of x is the distance between the points \( A(-3, 2) \) and \( B(x, 10) \) 10 units?
Answer: The distance between \( A(-3, 2) \) and \( B(x, 10) \)
\( = \sqrt{(x + 3)^2 + (10 - 2)^2} \)
\(\therefore \sqrt{(x + 3)^2 + (8)^2} = 10 \)
\(\Rightarrow (x + 3)^2 + 64 = 10^2 \)
\(\Rightarrow (x + 3)^2 = 100 - 64 \)
\(\Rightarrow (x + 3)^2 = 36 \)
\(\Rightarrow x + 3 = \pm \sqrt{36} = \pm 6 \)
For +ve sign, \( x = 6 - 3 = 3 \)
For -ve sign, \( x = -6 - 3 = -9 \)

Question. Find a point on the x-axis which is equidistant from the points \( A(5, 2) \) and \( B(1, -2) \).
Answer: The given points are:
\( A(5, 2) \) and \( B(1, -2) \)
Let the required point on the x-axis be \( C(x, 0) \).
Since, C is equidistant from A and B.
\(\therefore AC = BC \)
\(\therefore \sqrt{(x - 5)^2 + (0 - 2)^2} = \sqrt{(x - 1)^2 + (0 + 2)^2} \)
\(\Rightarrow (x - 5)^2 + (-2)^2 = (x - 1)^2 + (2)^2 \)
\(\Rightarrow x^2 + 25 - 10x + 4 = x^2 + 1 - 2x + 4 \)
\(\Rightarrow -10x + 2x = 5 - 29 \)
\(\Rightarrow -8x = -24 \)
\(\Rightarrow x = \frac{-24}{-8} = 3 \)
\(\therefore\) The required point is \( (0, 3) \).

Question. Establish the relation between x and y when \( P(x, y) \) is equidistant from the points \( A(-1, 2) \) and \( B(2, -1) \).
Answer: \(\because P \) is equidistant from A and B
\(\therefore PA = PB \)
\( \sqrt{(x + 1)^2 + (y - 2)^2} = \sqrt{(x - 2)^2 + (y + 1)^2} \)
\(\Rightarrow (x + 1)^2 + (y - 2)^2 = (x - 2)^2 + (y + 1)^2 \)
\(\Rightarrow x^2 + 1 + 2x + y^2 - 4y + 4 = x^2 + 4 - 4x + y^2 + 1 + 2y \)
\(\Rightarrow 2x - 4y + 5 = -4x + 2y + 5 \)
\(\Rightarrow 2x + 4x + 5 = 2y + 4y + 5 \)
\(\Rightarrow 6x = 6y \)
\(\Rightarrow x = y \)
which is the required relation.

Question. Show that the points \( (7, -2), (2, 3) \) and \( (-1, 6) \) are collinear.
Answer: Here, the vertices of a triangle are \( (7, -2), (2, 3) \) and \( (-1, 6) \)
\(\therefore\) Area of the triangle
\( = \frac{1}{2} [x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)] \)
\( = \frac{1}{2} [7(3 - 6) + 2(6 + 2) + (-1)(-2 - 3)] \)
\( = \frac{1}{2} [7 \times (-3) + 2 \times 8 + (-1)(-5)] \)
\( = \frac{1}{2} [-21 + 16 + 5] \)
\( = \frac{1}{2} [0] = 0 \)
Since area of triangle = 0
\(\therefore\) The vertices of the triangle are collinear. Thus, the given points are collinear.

Question. Find the distance between the points \( \left( -\frac{8}{5}, 2 \right) \) and \( \left( \frac{2}{5}, 2 \right) \). 
Answer: Distance between \( \left( -\frac{8}{5}, 2 \right) \) and \( \left( \frac{2}{5}, 2 \right) \) is given by
\( \sqrt{\left( \frac{2}{5} + \frac{8}{5} \right)^2 + (2 - 2)^2} = \sqrt{2^2 - 0^2} = 2 \text{ units.} \)

Question. If the mid point of the line joining the points \( P(6, b - 2) \) and \( Q(-2, 4) \) is \( (2, -3) \), find the value of b. 
Answer: Here, \( P(6, b - 2) \) and \( Q(-2, 4) \) are the given points.
\(\therefore\) Mid point of PQ is given by:
\( \left[ \frac{6 + (-2)}{2}, \frac{4 + b - 2}{2} \right] \)
or \( \left[ \frac{6 - 2}{2}, \frac{4 - 2 + b}{2} \right] \)
or \( \left[ 2, \frac{2 + b}{2} \right] \)
\(\therefore \frac{2 + b}{2} = -3 \Rightarrow 2 + b = -6 \)
\(\Rightarrow b = -6 - 2 \)
\(\Rightarrow b = -8 \)

SHORT ANSWER TYPE QUESTIONS

Question. Point \( P(5, -3) \) is one of the two points of trisection of the line segment joining the points \( A(7, -2) \) and \( B(1, -5) \) near to A. Find the coordinates of the other point of trisection. 
Answer: Since P is near to A \(\therefore\) other point Q is the mid point of PB
\(\Rightarrow x = \frac{5 + 1}{2} = 3 \)
\(\Rightarrow y = \frac{-3 - 5}{2} = \frac{-8}{2} = -4 \)
Thus, the point Q is \( (3, -4) \)

Question. Points P, Q, R and S, in this order, divide a line segment joining \( A(2, 6), B(7, -4) \) in five equal parts. Find the coordinates of P and R. 
Answer: \(\because P, Q, R \) and S divide AB in five equal parts.
\(\therefore AP = PQ = QR = RS = SB \)
Now, P divides AB in the ratio \( 1 : 4 \)
\(\therefore\) Coordinates of P are:
\( \left[ \frac{1 \times 7 + 4 \times 2}{1 + 4}, \frac{1 \times (-4) + 4 \times 6}{1 + 4} \right] \)
or \( \left[ \frac{7 + 8}{5}, \frac{-4 + 24}{5} \right] \) or \( (3, 4) \)
Again, R divides AB in the ratio \( 3 : 2 \)
\(\therefore\) Coordinates of R are:
\( \left[ \frac{3 \times 7 + 2 \times 2}{2 + 3}, \frac{3 \times (-4) + 2 \times 6}{2 + 3} \right] \)
or \( \left[ \frac{21 + 4}{5}, \frac{0}{5} \right] \) or \( (5, 0) \)

Question. Find the point on y-axis which is equidistant from the points \( (5, -2) \) and \( (-3, 2) \).
Answer: Let the required point be \( P(0, y) \)
\(\because\) The given points are \( A(5, -2) \) and \( B(-3, 2) \)
\(\therefore PA = PB \Rightarrow PA^2 = PB^2 \)
\(\therefore (5 - 0)^2 + (-2 - y)^2 = (-3 - 0)^2 + (2 - y)^2 \)
\(\Rightarrow 5^2 + (-2 - y)^2 = (-3)^2 + (2 - y)^2 \)
\(\Rightarrow 25 + 4 + y^2 + 4y = 9 + 4 + y^2 - 4y \)
\(\Rightarrow 25 + 4y = 9 - 4y \)
\(\Rightarrow 8y = -16 \Rightarrow y = -2 \)
Thus, the required point is \( (0, -2) \)

Question. Find the point on y-axis which is equidistant from \( (-5, 2) \) and \( (9, -2) \). 
Answer: Let the required point on Y-axis be \( P(0, y) \).
The given points are \( A(-5, 2) \) and \( B(9, -2) \)
\(\because AP = BP \)
\(\therefore \sqrt{(0 + 5)^2 + (y - 2)^2} = \sqrt{(0 - 9)^2 + (y + 2)^2} \)
\(\Rightarrow 5^2 + (y - 2)^2 = 9^2 + (y + 2)^2 \)
\(\Rightarrow 25 + y^2 - 4y + 4 = 81 + y^2 + 4 + 4y \)
\(\Rightarrow -4y - 4y = 81 + 4 - 4 - 25 \)
\(\Rightarrow -8y = 56 \)
\(\Rightarrow y = \frac{56}{-8} = -7 \)
\(\therefore\) The required point \( = (0, -7) \)

Question. Find the value of x for which the distance between the points \( P(4, -5) \) and \( Q(12, x) \) is 10 units
Answer: The given points are \( P(4, -5) \) and \( Q(12, x) \) such that \( PQ = 10 \)
\(\therefore \sqrt{(12 - 4)^2 + (x + 5)^2} = 10 \)
\(\Rightarrow (12 - 4)^2 + (x + 5)^2 = 10^2 \)
\(\Rightarrow 8^2 + (x + 5)^2 = 100 \)
\(\Rightarrow 64 + x^2 + 25 + 10x = 100 \)
\(\Rightarrow x^2 + 10x - 11 = 0 \)
\(\Rightarrow (x - 1)(x + 11) = 0 \)
\(\Rightarrow x = 1 \) or \( x = -11 \)

Question. Find the relation between x and y if the points \( (2, 1), (x, y) \) and \( (7, 5) \) are collinear. 
Answer: Here, \( x_1 = 2, y_1 = 1, x_2 = x, y_2 = y, x_3 = 7, y_3 = 5 \)
\(\therefore\) Area of triangle \( = \frac{1}{2} [x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)] \)
\( = \frac{1}{2} [2(y - 5) + x(5 - 1) + 7(1 - y)] \)
\( = \frac{1}{2} [2y - 10 + 4x + 7 - 7y] \)
\( = \frac{1}{2} [-5y + 4x - 3] \)
\(\because ar(\Delta) = 0 \)
\(\therefore \frac{1}{2} [-5y + 4x - 3] = 0 \)
\(\Rightarrow 4x - 5y - 3 = 0 \)
which is the required relation.

Question. If the points \( A(4, 3) \) and \( B(x, 5) \) are on the circle with the centre \( O(2, 3) \), find the value of x. 
Answer: Let \( O(2, 3) \) be the centre of the circle.
\(\therefore OA = OB \Rightarrow OA^2 = OB^2 \)
\(\Rightarrow (4 - 2)^2 + (3 - 3)^2 = (x - 2)^2 + (5 - 3)^2 \)
\(\Rightarrow 2^2 = (x - 2)^2 + 2^2 \)
\(\Rightarrow (x - 2)^2 = 0 \)
\(\Rightarrow x - 2 = 0 \)
\(\Rightarrow x = 2 \)
Thus, the required value of x is 2.

Question. If the vertices of a \( \Delta \) are \( (2, 4), (5, k) \) and \( (3, 10) \) and its area is 15 sq. units, then find the value of ‘k’. 
Answer: The area of the given \( \Delta \)
\( = \frac{1}{2} [2(k - 10) + 5(10 - 4) + 3(4 - k)] \)
\( = \frac{1}{2} [2k - 20 + 30 + 12 - 3k] \)
\( = \frac{1}{2} [-k + 22] \)
But \( ar(\Delta) = 15 \) [given]
\(\therefore \frac{1}{2} [-k + 22] = 15 \)
\(\Rightarrow -k + 22 = 30 \)
\(\Rightarrow -k = 30 - 22 = 8 \)
\(\Rightarrow k = -8 \)

Question. The vertices of a triangle are: \( (1, k), (4, -3), (-9, 7) \) and its area is 15 sq. units. Find the value of k. 
Answer: Area of the given triangle
\( = \frac{1}{2} [1(-3 - 7) + 4(7 - k) - 9(k + 3)] = 15 \)
\(\Rightarrow \frac{1}{2} [-10 + 28 - 4k - 9k - 27] = 15 \)
\(\Rightarrow -13k - 9 = 30 \)
\(\Rightarrow -13k = 39 \)
\(\Rightarrow k = \frac{39}{-13} \Rightarrow k = -3 \)

Question. Find the area of a \( \Delta ABC \) whose vertices are \( A(-5, 7), B(-4, -5) \) and \( C(4, 5) \). 
Answer: Here, \( x_1 = -5, y_1 = 7, x_2 = -4, y_2 = -5, x_3 = 4, y_3 = 5 \)
Now, \( ar(\Delta) = \frac{1}{2} [x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)] \)
\( = \frac{1}{2} [(-5)(-5 - 5) + (-4)(5 - 7) + 4(7 + 5)] \)
\( = \frac{1}{2} [50 + 8 + 48] \)
\( = \frac{1}{2} \times 106 = 53 \)
\(\therefore\) The required \( ar(\Delta ABC) = 53 \) sq. units.

Question. Find the value of k such that the points \( (1, 1), (3, k) \) and \( (-1, 4) \) are collinear. 
Answer: For the three points to be collinear, the area of triangle formed by them must be zero.
\(\therefore\) Area of triangle = 0
\(\Rightarrow \frac{1}{2} [1(k - 4) + (3)(4 - 1) + (-1)(1 - k)] = 0 \)
\(\Rightarrow \frac{1}{2} [k - 4 + 9 - 1 + k] = 0 \)
\(\Rightarrow \frac{1}{2} [2k + 4] = 0 \)
\(\Rightarrow k + 2 = 0 \)
\(\Rightarrow k = -2 \)

Question. For what value of p, the points \( (-5, 1), (1, p) \) and \( (4, -2) \) are collinear? 
Answer: Since the points are collinear, \(\therefore\) The area of the \( \Delta \) formed by these points must be zero.
i.e., \( \frac{1}{2} [-5(p + 2) + 1(-2 - 1) + 4(1 - p)] = 0 \)
\(\Rightarrow -5p - 10 - 3 + 4 - 4p = 0 \)
\(\Rightarrow -9p - 9 = 0 \)
\(\Rightarrow -9p = 9 \)
\(\Rightarrow p = \frac{9}{-9} = -1 \)

Question. For what value of p, are the points \( (2, 1), (p, -1) \) and \( (-1, 3) \) collinear? 
Answer: \(\because\) The given points are collinear. \(\therefore\) The area of a triangle formed by these points must be zero.
i.e., Area of triangle = 0
\(\Rightarrow \frac{1}{2} [2(-1 - 3) + p(3 - 1) + (-1)(1 + 1)] = 0 \)
\(\Rightarrow \frac{1}{2} [-8 + 2p - 2] = 0 \)
\(\Rightarrow \frac{1}{2} [-10 + 2p] = 0 \)
\(\Rightarrow -5 + p = 0 \)
\(\Rightarrow p = 5 \)

Please click the link below to download CBSE Class 10 Coordinate geometry Sure Shot Questions Set C.