Kerala SSLC · Mathematics · Class 10

Coordinates: Important Questions with Answers

These are 10 important multiple-choice questions from the Kerala SSLC Mathematics chapter “Coordinates” (Kerala SCERT syllabus). Each one shows the correct answer and a short explanation of why it is right. Try answering before you read the answer. ExamSummary has 20 practice questions on this chapter in total.

1. Q1.What is the distance between the points (3, -2) and (-1, 1)?

   • A)3 units
   • B)4 units
   • C)5 units
   • D)6 units

   Answer: C) 5 units

   Using the distance formula d = sqrt((x2-x1)^2 + (y2-y1)^2), we get sqrt((-1-3)^2 + (1-(-2))^2) = sqrt(16 + 9) = sqrt(25) = 5.

2. Q2.Find the coordinates of the midpoint of the line segment joining (-5, 8) and (3, 2).

   • A)(-1, 5)
   • B)(1, 5)
   • C)(-1, -5)
   • D)(1, -5)

   Answer: A) (-1, 5)

   The midpoint formula is ((x1+x2)/2, (y1+y2)/2). Substituting values gives ((-5+3)/2, (8+2)/2) = (-2/2, 10/2) = (-1, 5).

3. Q3.In what ratio does the point (4, 5) divide the line segment joining (2, 3) and (8, 9)?

   • A)1:2
   • B)2:1
   • C)3:1
   • D)1:3

   Answer: A) 1:2

   Using the section formula for x-coordinates: 4 = (m*8 + n*2)/(m+n). Solving 4m + 4n = 8m + 2n leads to 2n = 4m, so m/n = 1/2.

4. Q4.Calculate the area of a triangle with vertices at (0, 0), (4, 0), and (0, 3).

   • A)3 square units
   • B)6 square units
   • C)12 square units
   • D)7 square units

   Answer: B) 6 square units

   Since two vertices lie on the axes, the base is 4 and height is 3. Area = 1/2 * base * height = 1/2 * 4 * 3 = 6 square units.

5. Q5.If the points (1, 2), (3, 4), and (k, 6) are collinear, what is the value of k?

   • A)3
   • B)4
   • C)5
   • D)6

   Answer: C) 5

   For collinear points, slopes must be equal. Slope between first two is (4-2)/(3-1) = 1. Setting slope between second and third to 1 gives (6-4)/(k-3) = 1, so k-3 = 2, thus k=5.

6. Q6.Which of the following points lies on the X-axis?

   • A)(2, 3)
   • B)(3, 2)
   • C)(0, 5)
   • D)(5, 0)

   Answer: D) (5, 0)

   Any point lying on the X-axis has a y-coordinate (ordinate) of zero. Among the options, only (5, 0) satisfies this condition.

7. Q7.What is the distance of the point P(4, 3) from the origin O(0, 0)?

   • A)4 units
   • B)5 units
   • C)7 units
   • D)25 units

   Answer: B) 5 units

   The distance from the origin is calculated as sqrt(x^2 + y^2). For (4, 3), this is sqrt(4^2 + 3^2) = sqrt(16 + 9) = sqrt(25) = 5.

8. Q8.In what ratio does the Y-axis divide the line segment joining (-2, 3) and (4, -1)?

   • A)1:2
   • B)2:1
   • C)3:1
   • D)1:3

   Answer: A) 1:2

   On the Y-axis, the x-coordinate is 0. Using the section formula x = (mx2 + nx1)/(m+n), we set 0 = (4m - 2n). This implies 4m = 2n, giving the ratio m:n = 1:2.

9. Q9.If the area of the triangle formed by points (2, 1), (3, -2), and (-4, -1) is A, what is the value of A?

   • A)10
   • B)15
   • C)20
   • D)25

   Answer: A) 10

   Using the area formula 1/2 |x1(y2-y3) + x2(y3-y1) + x3(y1-y2)|, we substitute values to get 1/2 |2(-2+1) + 3(-1-1) - 4(1+2)| = 1/2 |-2 - 6 - 12| = 10.

10. Q10.Find the value of k if the point (k, 0) is equidistant from (2, 1) and (-1, 2).

    • A)0
    • B)1
    • C)2
    • D)-1

    Answer: A) 0

    Equating distances squared: (k-2)^2 + (0-1)^2 = (k+1)^2 + (0-2)^2. Simplifying yields k^2 - 4k + 5 = k^2 + 2k + 5, which reduces to 6k = 0, so k=0.

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