Class 9 Distance Formula Mock Test (ICSE)

Topics in Distance Formula

• Distance between two points
• Proving collinearity
• Proving shapes of triangles and quadrilaterals
• Finding an unknown coordinate

Distance Formula practice questions with answers

1. Q1Multiple choice·1 mark
   If the distance between the points $(x, 2)$ and $(3, 6)$ is $5$ units, then the positive value of $x$ is:

   1. (a)$5$
   2. (b)$3$
   3. (c)$8$
   4. (d)$6$
   Show answer and step marks

   1. Correct option: (d) $6$+1

   Answer: (d)
2. Q2Multiple choice·1 mark
   If the points $(1, 2)$, $(3, 4)$, and $(t, 6)$ are collinear, the value of $t$ is:

   1. (a)$4$
   2. (b)$5$
   3. (c)$6$
   4. (d)$3$
   Show answer and step marks

   1. Correct option: (b) $5$+1

   Answer: (b)
3. Q3Multiple choice·1 mark
   The points $A(0, 3)$, $B(3, 0)$ and $C(0, 0)$ are the vertices of $\triangle ABC$. Which of the following describes the triangle most completely?

   1. (a)Right-angled triangle
   2. (b)Scalene triangle
   3. (c)Equilateral triangle
   4. (d)Isosceles right-angled triangle
   Show answer and step marks

   1. Correct option: (d) Isosceles right-angled triangle. $AC = BC = 3$ and $AB = 3\sqrt{2}$; $AC^2 + BC^2 = 18 = AB^2$, so it is right-angled at $C$ with two equal sides.+1

   Answer: (d)
4. Q4Multiple choice·1 mark
   The distance of the point $P(-5, 12)$ from the origin is:

   1. (a)$13$ units
   2. (b)$5$ units
   3. (c)$17$ units
   4. (d)$12$ units
   Show answer and step marks

   1. Correct option: (a) $13$ units+1

   Answer: (a)
5. Q5Short answer·2 marks
   Prove that the points $A(-2, -1)$, $B(1, 0)$, $C(4, 3)$ and $D(1, 2)$ are the vertices of a parallelogram.
   Show answer and step marks

   1. $AB = \sqrt{(1 - (-2))^2 + (0 - (-1))^2} = \sqrt{3^2 + 1^2} = \sqrt{10}$ units.+0.5
   2. $CD = \sqrt{(1 - 4)^2 + (2 - 3)^2} = \sqrt{(-3)^2 + (-1)^2} = \sqrt{10}$ units, so $AB = CD$.+0.5
   3. $BC = \sqrt{(4 - 1)^2 + (3 - 0)^2} = \sqrt{3^2 + 3^2} = \sqrt{18}$ units.+0.5
   4. $DA = \sqrt{(-2 - 1)^2 + (-1 - 2)^2} = \sqrt{(-3)^2 + (-3)^2} = \sqrt{18}$ units, so $BC = DA$. Both pairs of opposite sides are equal, hence $ABCD$ is a parallelogram.+0.5

   Answer: $AB = CD = \sqrt{10}$ units and $BC = DA = \sqrt{18}$ units; both pairs of opposite sides are equal, so $ABCD$ is a parallelogram.
6. Q6Short answer·2 marks
   Show that the points $A(1, 7)$, $B(4, 2)$, $C(-1, -1)$ and $D(-4, 4)$ are the vertices of a rhombus.
   Show answer and step marks

   1. $AB = \sqrt{(4-1)^2 + (2-7)^2} = \sqrt{9 + 25} = \sqrt{34}$ units.+0.5
   2. $BC = \sqrt{(-1-4)^2 + (-1-2)^2} = \sqrt{25 + 9} = \sqrt{34}$ units.+0.5
   3. $CD = \sqrt{(-4-(-1))^2 + (4-(-1))^2} = \sqrt{9 + 25} = \sqrt{34}$ units.+0.5
   4. $DA = \sqrt{(1-(-4))^2 + (7-4)^2} = \sqrt{25 + 9} = \sqrt{34}$ units. Since $AB = BC = CD = DA$, all four sides are equal, so $ABCD$ is a rhombus.+0.5

   Answer: $AB = BC = CD = DA = \sqrt{34}$ units; all four sides are equal, so $ABCD$ is a rhombus.

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