CBSE Class 10 Coordinate geometry Sure Shot Questions Set 01

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1. Find the distance between the following points:

(i) A(9, 3) and (15, 11)            (ii) A(7, – 4) and b(–5, 1).

(iii) A(–6, –4) and B(9, –12)     (iv) A(1, –3) and B(4, –6)

(v) P(a + b, a – b) and Q(a – b, a + b)

(vi) P(a sin a, a cos a) and Q(a cos a, –a sin a)

2. If A(6, –1), B(1, 3) and C(k, 8) are three points such that AB= BC, find the value of k.

3. Find all the possible value of a for which the distance between the points A(a, –1) and B(5, 3) is 5 units.

4. Determine, whether each of the given points (–2, 1), (2, –2) and (5, 2) are the vertices of right angle.

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

6. By distance formula, show that the points (1, –1), (5, 2) and (9, 5) are collinear.

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

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

9. Find the point on x-axis which is equidistant from (–2, 5) and (2, –3).

10. Find the point on x-axis which is equidistant from (7, 6) and (–3, 4).

11. Find the point on the x-axis which is equidistant from (2, –5) and (–2, 9).

12. Find a point on the y-axis which is equidistant from the points A(6, 5) and B(– 4, 3).

13. Find a point on the y-axis which is equidistant from the points A(5, 2) and B(– 4, 3).

14. Find a point on the y-axis which is equidistant from the points A(5, – 2) and B(– 3, 2).

15. Find the point on y-axis, each of which is at a distance of 13 units from the point (–5, 7).

16. Find the point on x-axis, each of which is at a distance of 10 units from the point (11, –8).

17. Find the values of k for which the distance between the points A(k, –5) and B(2, 7) is 13 units.

18. Prove that the points A(–3, 0), B(1, –3) and C(4, 1) are the vertices of an isosceles right-angled triangle. Find the area of this triangle.

19. Prove that the points A(a, a), B(–a, –a) and C(– 3 a, 3 a) are the vertices of an equilateral triangle. Calculate the area of this triangle.

Multiple Choice Questions

Question. The distance of the point P(2, 3) from the x-axis is
(a) 2
(b) 3
(c) 1
(d) 5
Answer: (b)
Explanation: The distance from x-axis is equal to its ordinate i.e., 3

Question. The distance between the point P(1, 4) and Q(4, 0) is
(a) 4
(b) 5
(c) 6
(d) \( 3\sqrt{3} \)
Answer: (b)
Explanation: The required distance = \( \sqrt{(4 - 1)^2 + (0 - 4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \)

Question. The points (– 5, 1), (1, p) and (4, – 2) are collinear if the value of p is
(a) 3
(b) 2
(c) 1
(d) –1
Answer: (d)
Explanation: The points are collinear if area of \( \Delta = 0 \)
\( \Rightarrow \frac{1}{2} [-5(p + 2) + 1(-2 - 1) + 4(1 - p)] = 0 \)
\( \Rightarrow -5p - 10 - 3 + 4 - 4p = 0 \Rightarrow -9p = +9 \)
\( \therefore p = -1 \)

Question. The area of the triangle ABC with the vertices A(–5, 7), B(–4, –5) and C(4, 5) is
(a) 63
(b) 35
(c) 53
(d) 36
Answer: (c)
Explanation: Area of \( \Delta ABC \)
\( = \frac{1}{2} [x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)] \)
\( = \frac{1}{2} [-5(-5 - 5) - 4(5 - 7) + 4(7 - (-5))] = \frac{1}{2} [-5(-10) - 4(-2) + 4(12)] \)
\( = \frac{1}{2} [50 + 8 + 48] = \frac{1}{2} \times 106 = 53 \) sq. units

Question. The distance of the point \( (\alpha, \beta) \) from the origin is
(a) \( \alpha + \beta \)
(b) \( \alpha^2 + \beta^2 \)
(c) \( |\alpha| + |\beta| \)
(d) \( \sqrt{\alpha^2 + \beta^2} \)
Answer: (d)
Explanation: Distance of \( (\alpha, \beta) \) from origin (0, 0) \( = \sqrt{(\alpha - 0)^2 + (\beta - 0)^2} = \sqrt{\alpha^2 + \beta^2} \)

Question. The area of the triangle whose vertices are A(1, 2), B(–2, 3) and C(–3, –4) is
(a) 11
(b) 22
(c) 33
(d) 21
Answer: (a)
Explanation: Required Area = \( \frac{1}{2} [1(3 + 4) - 2(-4 - 2) - 3(2 - 3)] \)
\( = \frac{1}{2} [7 + 12 + 3] = \frac{1}{2} \times 22 = 11 \)

Question. The line segment joining the points (3, –1) and (–6, 5) is trisected. The coordinates of point of trisection are
(a) (3, 3)
(b) (– 3, 3)
(c) (3, – 3)
(d) (– 3, – 3)
Answer: (b)
Explanation: Since the line segment AB is trisected \( \therefore PB : BQ = 2 : 1 \). Coordinates of B are \( = \left( \frac{2(-6)+1(3)}{2+1}, \frac{2(5)+1(-1)}{2+1} \right) = \left( \frac{-12 + 3}{3}, \frac{10 - 1}{3} \right) = \left( -\frac{9}{3}, \frac{9}{3} \right) = (-3, 3) \)

Question. The line 3x + y – 9 = 0 divides the line joining the points (1, 3) and (2, 7) internally in the ratio
(a) 3 : 4
(b) 3 : 2
(c) 2 : 3
(d) 4 : 3
Answer: (a)
Explanation: Let the line 3x + y – 9 = 0 divide the line segment joining A(1, 3) and B(2, 7) in the ratio K : 1 at point C. The coordinates of C are \( \left( \frac{2K + 1}{K + 1}, \frac{7K + 3}{K + 1} \right) \). But the point C lies on the line 3x + y – 9 = 0. \( \therefore 3\left(\frac{2K + 1}{K + 1}\right) + \frac{7K + 3}{K + 1} - 9 = 0 \Rightarrow 6K + 3 + 7K + 3 - 9 - 9K = 0 \Rightarrow 4K - 3 = 0 \Rightarrow K = \frac{3}{4} \). Required ratio = 3 : 4

Question. The distance between A(a + b, a – b) and B(a – b, – a – b) is
(a) \( \sqrt{a^2 + b^2} \)
(b) \( \sqrt{a^2 - b^2} \)
(c) \( 2\sqrt{a^2 + b^2} \)
(d) \( 4\sqrt{a^2 + b^2} \)
Answer: (c)
Answer: Explanation: AB \( = \sqrt{((a - b) - (a + b))^2 + ((-a - b) - (a - b))^2} = \sqrt{(-2b)^2 + (-2a)^2} = \sqrt{4b^2 + 4a^2} = 2\sqrt{a^2 + b^2} \)

Question. If Q(\( \frac{a}{3} \), 4) is the mid-point of the segment joining the points P(–6, 5) and R(–2, 3), then the value of ‘a’ is
(a) 12
(b) –6
(c) –12
(d) –4
Answer: (c)
Explanation: Mid-point = \( \left( \frac{-6 - 2}{2}, \frac{5 + 3}{2} \right) \Rightarrow \left( \frac{a}{3}, 4 \right) = (-4, 4) \therefore \frac{a}{3} = -4 \Rightarrow a = -12 \)

Question. If the distance between the points (x, –1) and (3, 2) is 5, then the value of x is
(a) –7 or –1
(b) –7 or 1
(c) 7 or 1
(d) 7 or –1
Answer: (d)
Explanation: We have \( \sqrt{(x - 3)^2 + (2 - (-1))^2} = 5 \Rightarrow (x - 3)^2 + 9 = 25 \Rightarrow x^2 - 6x + 9 + 9 = 25 \Rightarrow x^2 - 6x - 7 = 0 \Rightarrow (x - 7)(x + 1) = 0 \Rightarrow x = 7 \) or \( x = -1 \)

Question. The points (1, 1), (–2, 7) and (3, –3) are
(a) vertices of an equilateral triangle
(b) collinear
(c) vertices of an isosceles triangle
(d) none of these
Answer: (b)
Explanation: Let A(1, 1), B(–2, 7) and C(3, –3) are the given points. Then, we have AB \( = \sqrt{(-2 - 1)^2 + (7 - 1)^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5} \); BC \( = \sqrt{(3 - (-2))^2 + (-3 - 7)^2} = \sqrt{25 + 100} = \sqrt{125} = 5\sqrt{5} \); and AC \( = \sqrt{(3 - 1)^2 + (-3 - 1)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5} \). Clearly BC = AB + AC. \( \therefore \) A, B, C are collinear.

Question. The coordinates of the centroid of a triangle whose vertices are (0, 6), (8, 12) and (8, 0) is
(a) (4, 6)
(b) (16, 6)
(c) (8, 6)
(d) (16/3, 6)
Answer: (d)
Explanation: The co-ordinates of the centroid of the triangle is \( \left( \frac{0 + 8 + 8}{3}, \frac{6 + 12 + 0}{3} \right) = \left( \frac{16}{3}, \frac{18}{3} \right) = \left( \frac{16}{3}, 6 \right) \)

Question. Two vertices of a triangle are (3, – 5) and (– 7, 4). If its centroid is (2, – 1), then the third vertex is
(a) (10, 2)
(b) (– 10, 2)
(c) (10, – 2)
(d) (– 10, – 2)
Answer: (c)

Question. The area of the triangle formed by the points A(–1.5, 3), B(6, –2) and C(–3, 4) is
(a) 0
(b) 1
(c) 2
(d) 3/2
Answer: (a)
Explanation: Area of \( \Delta ABC = \frac{1}{2} [-1.5(-2 - 4) + 6(4 - 3) + (-3) (3 + 2)] = \frac{1}{2} [9 + 6 - 15] = 0 \). It is a straight line.

Question. If the points P(1, 2), B(0, 0) and C(a, b) are collinear, then
(a) 2a = b
(b) a = –b
(c) a = 2b
(d) a = b
Answer: (a)
Explanation: Area of \( \Delta PBC = 0 \Rightarrow \frac{1}{2} [1(0 - b) + 0(b - 1) + a(2 - 0)] = 0 \Rightarrow \frac{1}{2} [-b + 2a] = 0 \Rightarrow -b + 2a = 0 \therefore 2a = b \)

Assertion-Reason Questions

Question. Assertion: The point (0, 4) lies on y-axis.
Reason: The x-coordinate on the point on y-axis is zero.
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
(c) Assertion (A) is true but Reason (R) is false.
(d) Assertion (A) is false but Reason (R) is true.
Answer: (a)
Explanation: The x co-ordinate of the point (0, 4) is zero. Point (0, 4) lies on y-axis.

Question. Assertion: Mid-points of a line segment divides line in the ratio 1 : 1.
Reason: If area of triangle is zero that means points are collinear.
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
(c) Assertion (A) is true but Reason (R) is false.
(d) Assertion (A) is false but Reason (R) is true.
Answer: (b)

Question. Assertion: Centroid of a triangle formed by the points (a, b), (b, c) and (c, a) is at origin, Then a + b + c = 0.
Reason: Centroid of a \( \Delta ABC \) with vertices A(\( x_1, y_1 \)), B(\( x_2, y_2 \)) and C(\( x_3, y_3 \)) is given by \( \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) \).
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
(c) Assertion (A) is true but Reason (R) is false.
(d) Assertion (A) is false but Reason (R) is true.
Answer: (a)
Explanation: As we know, Centroid of a triangle \( = \left( \frac{a+b+c}{3}, \frac{b+c+a}{3} \right) = (0, 0) \therefore a + b + c = 0 \)

Question. Assertion: The value of y is 6, for which the distance between the points P(2, –3) and Q(10, y) is 10.
Reason: Distance between two given 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 Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
(c) Assertion (A) is true but Reason (R) is false.
(d) Assertion (A) is false but Reason (R) is true.
Answer: (d)
Explanation: Given. PQ = 10. Using distance formula, \( PQ^2 = (10)^2 = 100 \). \( (10 - 2)^2 + (y + 3)^2 = 100 \Rightarrow (y + 3)^2 = 100 - 64 = 36 \Rightarrow y + 3 = \pm 6 \therefore y = 6 - 3 = 3 \) or \( y = - 6 - 3 = -9 \)

Question. Assertion: The area of the triangle with vertices (–5, –1), (3, –5), (5, 2), is 32 square units.
Reason: The point (x, y) divides the line segment joining the points (\( x_1, y_1 \)) and (\( x_2, y_2 \)) in the ratio k : 1 externally then \( x = \frac{kx_2 + x_1}{k+1}, y = \frac{ky_2 + y_1}{k+1} \)
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).
(c) Assertion (A) is true but Reason (R) is false.
(d) Assertion (A) is false but Reason (R) is true.
Answer: (c)
 

Short Answer Type Questions

Question. Name the type of triangle formed by the points A(–5, 6), B(–4, –2) and C(7, 5).
Answer: Sol. By using distance formula, distance between points A and B is,
\( AB = \sqrt{(–4 + 5)^2 + (–2 – 6)^2} = \sqrt{1 + 64} = \sqrt{65} \) units
Distance between points B and C is,
\( BC = \sqrt{(7 + 4)^2 + (5 + 2)^2} = \sqrt{(11)^2 + (7)^2} = \sqrt{121 + 49} = \sqrt{170} \) units
Distance between points C and A is,
\( CA = \sqrt{(–5 – 7)^2 + (6 – 5)^2} = \sqrt{(–12)^2 + (1)^2} = \sqrt{144 + 1} = \sqrt{145} \) units
As, \( AB \neq BC \neq CA \).
Also, AB, BC and CA does not fulfill the condition of Pythagoras i.e.,
\( (\text{Hypotenuse})^2 = (\text{Base})^2 + (\text{Perpendicular})^2 \)
Therefore, the required triangle is scalene triangle.

Question. Find the value of \( a \), if the distance between the points A(–3, –14) and B(a, –5) is 9 units.
Answer: Sol. Given: Distance between A(–3, –14) and B(a, –5) is, \( AB = 9 \) units
By distance formula,
\( AB = \sqrt{(a + 3)^2 + (–5 + 14)^2} \)
\( 9 = \sqrt{(a + 3)^2 + (9)^2} \) ...(i)
On squaring both sides, we get \( (a + 3)^2 + 81 = 81 \)
\( \Rightarrow (a + 3)^2 = 0 \therefore a = –3 \)

Question. Find the value of \( m \), if the points (5, 1), (–2, –3) and (8, 2m) are collinear.
Answer: Sol. Points are collinear ...[Given
\( \therefore \) Area of \( \Delta PQR = 0 \)
\( \Rightarrow \frac{1}{2} [x_1(y_2 – y_3) + x_2(y_3 – y_1) + x_3(y_1 – y_2)] = 0 \)
\( \Rightarrow \frac{1}{2} [5(–3 – 2m) + (–2)(2m – 1) + 8(1 – (–3))] = 0 \)
\( \Rightarrow \frac{1}{2} [–15 – 10m – 4m + 2 + 32] = 0 \)
\( \Rightarrow \frac{1}{2} [–14m + 19] = 0 \)
\( \Rightarrow m = \frac{19}{14} \)
Therefore, the required value of \( m \) is \( \frac{19}{14} \).

Question. If the point A(2, –4) is equidistant from P(3, 8) and Q(–10, y), then find the values of y. Also, find distance PQ.
Answer: Sol. Given: A(2, –4) is equidistant from P(3, 8) and Q(–10, y).
\( PA = QA \) ...[Given
\( (PA)^2 = (QA)^2 \) ...[Squaring both side
\( (2 – 3)^2 + (–4 – 8)^2 = (2 + 10)^2 + (–4 – y)^2 \)...[Using distance formula]
\( (–1)^2 + (–12)^2 = (12)^2 + (4 + y)^2 \)
\( 1 + 144 = 144 + 16 + y^2 + 8y \)
\( y^2 + 8y + 15 = 0 \)
\( (y^2 + 5y) + (3y + 15) = 0 \)
\( (y + 5) (y + 3) = 0 \)
\( y + 5 = 0 \) or \( y + 3 = 0 \)
\( \therefore y = –5 \) or \( y = –3 \)
Now, distance between P(3, 8) and Q(–10, y), when \( y = –3 \), we have
\( PQ = \sqrt{(–10 – 3)^2 + (–3 – 8)^2} = \sqrt{(–13)^2 + (–11)^2} = \sqrt{169 + 121} = \sqrt{290} \) units
Again, distance between P(3, 8) and Q(–10, y), when \( y = –5 \), we have
\( PQ = \sqrt{(–10 – 3)^2 + (–5 – 8)^2} = \sqrt{(–13)^2 + (–13)^2} = \sqrt{169 + 169} = \sqrt{338} = 13\sqrt{2} \) units

Question. In what ratio does the X-axis divide the line segment joining the points (–4, –6) and (–1, 7)? Find the coordinates of the points of division.
Answer: Sol. Let C(x, 0) be any point on x-axis.
Let \( AC : CB = K : 1 \)
Coordinates of C = \( \left( \frac{-K - 4}{K + 1}, \frac{7K - 6}{K + 1} \right) = (x, 0) \)
\( \therefore \frac{7K - 6}{K + 1} = 0 \Rightarrow 7K – 6 = 0 \Rightarrow K = \frac{6}{7} \)
Now, \( x = \frac{-6/7 - 4}{6/7 + 1} = \frac{-6 - 28}{6 + 7} = -\frac{34}{13} \)
Hence, \( \left(-\frac{34}{13}, 0\right) \) be point on x-axis.

Question. Find the ratio in which the point \( P \left(\frac{3}{4}, \frac{5}{12}\right) \) divides the line segment joining the points \( A \left(\frac{1}{2}, \frac{3}{2}\right) \) and B(2, –5).
Answer: Sol. Suppose, \( P \left(\frac{3}{4}, \frac{5}{12}\right) \) divides AB internally in the ratio \( m : n \).
Using section formula,
\( \frac{2m + \frac{1}{2}n}{m + n} = \frac{3}{4} \Rightarrow 4(2m + \frac{1}{2}n) = 3(m + n) \Rightarrow 8m + 2n = 3m + 3n \)
\( \Rightarrow 5m = n \Rightarrow \frac{m}{n} = \frac{1}{5} \)
Therefore, the required ratio is 1 : 5.

Question. If P(9a – 2, –b) divides line segment joining A(3a + 1, –3) and B(8a, 5) in the ratio 3 : 1, then find the values of a and b.
Answer: Sol. Given: P(9a – 2, –b) divides AB internally in the ratio 3 : 1.
Using section formula,
Coordinates of P = \( \left( \frac{3(8a) + 1(3a + 1)}{3+1}, \frac{3(5) + 1(–3)}{3+1} \right) = (9a – 2, –b) \)
\( 9a – 2 = \frac{24a + 3a + 1}{4} \Rightarrow 36a – 8 = 27a + 1 \Rightarrow 9a = 9 \Rightarrow a = 1 \)
\( –b = \frac{15 - 3}{4} = \frac{12}{4} = 3 \Rightarrow b = –3 \)
\( \therefore a = 1 \) and \( b = –3 \)

Question. If (a, b) is the mid-point of the line segment joining the points A(10, –6), B(k, 4) and \( a – 2b = 18 \), then find the value of k and the distance AB.
Answer: Sol. Given: (a, b) is the mid-point of line segment AB.
\( a – 2b = 18 \) ...(i) ...[Given]
By mid-point formula \( (a, b) = \left( \frac{10 + k}{2}, \frac{-6 + 4}{2} \right) \)
\( a = \frac{10 + k}{2} \) ...(ii) and \( b = –1 \) ...(iii)
From equations (i) and (iii) we get, \( a – 2(–1) = 18 \Rightarrow a = 16 \)
Putting the value of a in equation (ii), we get \( 16 = \frac{10 + k}{2} \Rightarrow 32 = 10 + k \Rightarrow k = 22 \)
Now, coordinates of A & B are (10, –6) & (22, 4) respectively.
Now, \( AB = \sqrt{(22 - 10)^2 + (4 + 6)^2} = \sqrt{144 + 100} = \sqrt{244} = 2\sqrt{61} \)
Therefore, the required distance of AB is \( 2\sqrt{61} \).

Question. The line segment joining the points A(3, 2) and B(5, 1) is divided at the point P in the ratio 1 : 2 and it lies on the line \( 3x – 18y + k = 0 \). Find the value of k.
Answer: Sol. By section formula, coordinates of point P are:
\( P = \left( \frac{1(5) + 2(3)}{1+2}, \frac{1(1) + 2(2)}{1+2} \right) = \left( \frac{11}{3}, \frac{5}{3} \right) \)
But given that, the point P lies on the line \( 3x – 18y + k = 0 \)
\( \therefore 3 \left(\frac{11}{3}\right) – 18 \left(\frac{5}{3}\right) + k = 0 \)
\( 11 – 30 + k = 0 \Rightarrow k = 19 \)
Therefore, the required value of k is 19.

Question. If the points A(2, 9), B(a, 5) and C(5, 5) are the vertices of a \( \Delta ABC \) right angled at B. Find the values of a and hence the area of \( \Delta ABC \).
Answer: Sol. By distance formula,
\( AB = \sqrt{(a - 2)^2 + (5 - 9)^2} = \sqrt{a^2 - 4a + 20} \)
\( BC = \sqrt{(5 - a)^2 + (5 - 5)^2} = 5 - a \)
\( CA = \sqrt{(2 - 5)^2 + (9 - 5)^2} = \sqrt{9 + 16} = 5 \)
Using pythagoras theorem, \( CA^2 = AB^2 + BC^2 \)
\( 25 = (a^2 - 4a + 20) + (5 - a)^2 \)
\( 25 = a^2 - 4a + 20 + 25 + a^2 - 10a \)
\( 2a^2 – 14a + 20 = 0 \Rightarrow a^2 – 7a + 10 = 0 \Rightarrow (a - 2)(a - 5) = 0 \)
\( a = 2 \) (Taking \( a=5 \) makes BC length 0).
Vertices are A(2, 9), B(2, 5), C(5, 5).
Area of \( \Delta ABC = \frac{1}{2} [2(5 – 5) + 2(5 – 9) + 5(9 – 5)] = 6 \) sq. units.

Question. Find the coordinates of the point R on the line segment joining the points P(–1, 3) and Q(2, 5) such that \( PR = \frac{3}{5} PQ \).
Answer: Sol. \( PR = \frac{3}{5} PQ \Rightarrow \frac{PR}{PQ} = \frac{3}{5} \).
Let \( PR = 3K, PQ = 5K \Rightarrow RQ = 2K \).
So, \( PR : RQ = 3 : 2 \).
Using section formula, \( R = \left( \frac{3(2) + 2(-1)}{3+2}, \frac{3(5) + 2(3)}{3+2} \right) = \left( \frac{4}{5}, \frac{21}{5} \right) \).
Required coordinates are \( \left(\frac{4}{5}, \frac{21}{5}\right) \).

Question. Find the values of k, if the points A(k + 1, 2k), B(3k, 2k + 3) and C(5k – 1, 5k) are collinear.
Answer: Sol. Area of \( \Delta ABC = 0 \)
\( \frac{1}{2} [(k + 1)(2k + 3 – 5k) + 3k(5k – 2k) + (5k – 1)(2k – (2k + 3))] = 0 \)
\( \frac{1}{2} [(k + 1)(3 – 3k) + 9k^2 – 3(5k – 1)] = 0 \)
\( 6k^2 – 15k + 6 = 0 \Rightarrow 2k^2 – 5k + 2 = 0 \)
\( (k - 2)(2k - 1) = 0 \Rightarrow k = 2, 1/2 \).

Question. Find the ratio in which the line \( 2x + 3y – 5 = 0 \) divides the line segment joining the points (8, –9) and (2, 1). Also, find the coordinates of the point of division.
Answer: Sol. Let the ratio be \( K : 1 \). Coordinates of P are \( \left( \frac{2K + 8}{K + 1}, \frac{K - 9}{K + 1} \right) \).
P lies on \( 2x + 3y – 5 = 0 \), so:
\( 2 \left(\frac{2K + 8}{K + 1}\right) + 3 \left(\frac{K - 9}{K + 1}\right) - 5 = 0 \Rightarrow 4K + 16 + 3K - 27 - 5K - 5 = 0 \)
\( 2K – 16 = 0 \Rightarrow K = 8 \).
Ratio is 8 : 1. Coordinates of P = \( \left( \frac{16 + 8}{9}, \frac{8 - 9}{9} \right) = \left( \frac{8}{3}, -\frac{1}{9} \right) \).

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