CBSE Class 10 Mathematics Coordinate Geometry Assignment Set 25

Type I: Very Short Answer Type Questions (1 Mark Each)

Question 1. Find the distance of the point P(−2, −3) from the x-axis.
Answer:
The distance from the x-axis is the absolute value of the ordinate: \( |-3| = \) 3 units.
In simple words: Distance from the x-axis = |y|.

Teacher's Note:
1. The distance from the y-axis would be |x|.
2. Distances are never negative.

Question 2. Find the distance of the point P(−6, 8) from the origin.
Answer:
\( OP = \sqrt{(-6)^2 + 8^2} = \sqrt{100} = \) 10 units.
In simple words: Use OP = √(x² + y²).

Teacher's Note:
1. This is a 6-8-10 right triangle.

Question 3. Find the perimeter of a triangle with vertices (0, 9), (0, 0) and (12, 0).
Answer:
Let A(0, 9), O(0, 0), B(12, 0). \( AO = 9 \), \( OB = 12 \), \( AB = \sqrt{12^2 + 9^2} = 15 \).
Perimeter \( = 9 + 12 + 15 = \) 36 units.
In simple words: Find all three side lengths and add.

Teacher's Note:
1. The triangle is right-angled at the origin.

Question 4. Find the point where the perpendicular bisector of the line segment joining A(−2, −5) and B(2, 5) meets AB.
Answer:
The perpendicular bisector meets AB at its midpoint: \( \left(\dfrac{-2 + 2}{2}, \dfrac{-5 + 5}{2}\right) = \) (0, 0).
In simple words: The bisector passes through the midpoint of AB.

Teacher's Note:
1. Use the midpoint formula.

Question 5. If \( P\left(\dfrac{a}{3}, 4\right) \) is the mid-point of the line segment joining the points Q(−6, 5) and R(−2, 3), then find the value of a.
Answer:
Midpoint \( = \left(\dfrac{-6 - 2}{2}, \dfrac{5 + 3}{2}\right) = (-4, 4) \). So \( \dfrac{a}{3} = -4 \Rightarrow \) a = −12.
In simple words: Equate the x-coordinates of the midpoint.

Teacher's Note:
1. The y-coordinate 4 already matches.

Type II: Short Answer Type Questions (2 Marks Each)

Question 6. Find the points on the x-axis which are at a distance of \( 2\sqrt{5} \) from the point (7, −4). How many such points are there?
Answer:
Let the point be (x, 0). \( (x - 7)^2 + (0 + 4)^2 = 20 \Rightarrow (x - 7)^2 = 4 \Rightarrow x = 9 \) or 5.
The points are (5, 0) and (9, 0), so there are two such points.
In simple words: Use the distance formula with y = 0.

Teacher's Note:
1. Square both sides to remove the root.
2. Take both signs.

Question 7. What type of a quadrilateral do the points A(2, −2), B(7, 3), C(11, −1) and D(6, −6), taken in that order, form?
Answer:
\( AB = \sqrt{25 + 25} = 5\sqrt{2} \), \( BC = \sqrt{16 + 16} = 4\sqrt{2} \), \( CD = \sqrt{25 + 25} = 5\sqrt{2} \), \( DA = \sqrt{16 + 16} = 4\sqrt{2} \).
Diagonals: \( AC = \sqrt{81 + 1} = \sqrt{82} \), \( BD = \sqrt{1 + 81} = \sqrt{82} \).
Opposite sides are equal and the diagonals are equal, so ABCD is a rectangle (not a square, since adjacent sides are unequal).
In simple words: Compare sides and diagonals.

Teacher's Note:
1. Equal diagonals separate a rectangle from a general parallelogram.
2. Unequal adjacent sides rule out a square.

Question 8. Find the value of a, if the distance between the points A(−3, −14) and B(a, −5) is 9 units.
Answer:
\( (a + 3)^2 + (-5 + 14)^2 = 81 \Rightarrow (a + 3)^2 + 81 = 81 \Rightarrow a + 3 = 0 \).
So a = −3.
In simple words: Use the distance formula and solve for a.

Teacher's Note:
1. The vertical gap is already 9, so the horizontal gap is zero.

Question 9. 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:
Let the ratio be k : 1. The point of division is \( \left(\dfrac{2k + 8}{k + 1}, \dfrac{k - 9}{k + 1}\right) \).
It lies on the line: \( 2(2k + 8) + 3(k - 9) - 5(k + 1) = 0 \Rightarrow 2k - 16 = 0 \Rightarrow k = 8 \).
The ratio is 8 : 1 and the point is \( \left(\dfrac{24}{9}, \dfrac{-1}{9}\right) = \) \( \left(\dfrac{8}{3}, -\dfrac{1}{9}\right) \).
In simple words: Put the section-formula point in the equation of the line.

Teacher's Note:
1. Take the ratio as k : 1.
2. Substitute and clear the denominator.

Question 10. The centre of a circle is (2a, a − 7). Find the values of a if the circle passes through the point (11, −9) and has diameter \( 10\sqrt{2} \) units.
Answer:
The radius is \( 5\sqrt{2} \), so the squared distance of the centre from (11, −9) is 50: \( (11 - 2a)^2 + (-9 - a + 7)^2 = 50 \).
\( 121 - 44a + 4a^2 + a^2 + 4a + 4 = 50 \Rightarrow 5a^2 - 40a + 75 = 0 \Rightarrow a^2 - 8a + 15 = 0 \).
\( (a - 3)(a - 5) = 0 \), so a = 3 or a = 5.
In simple words: Radius = distance from the centre to a point on the circle.

Teacher's Note:
1. Radius is half the diameter.
2. Solve the quadratic in a.

Type III: Long Answer Type Questions (4 Marks Each)

Question 11. The mid-points D, E, F of the sides of a triangle ABC are (3, 4), (8, 9) and (6, 7). Find the coordinates of the vertices of the triangle.
Answer:
Take D as the midpoint of BC, E of CA and F of AB. Then each vertex is the sum of the two midpoints adjacent to it minus the opposite midpoint.
\( A = E + F - D = (8 + 6 - 3, 9 + 7 - 4) = (11, 12) \)
\( B = D + F - E = (3 + 6 - 8, 4 + 7 - 9) = (1, 2) \)
\( C = D + E - F = (3 + 8 - 6, 4 + 9 - 7) = (5, 6) \)
The vertices are (11, 12), (1, 2), (5, 6). (Check: midpoint of BC is (3, 4), of CA is (8, 9), of AB is (6, 7).)
In simple words: Use the midpoint formula for each side.

Teacher's Note:
1. Write the three midpoint equations.
2. Always check by recomputing the midpoints.

Question 12. Ayush starts walking from his house to office. Instead of going to the office directly, he goes to a bank first, from there to his daughter's school and then reaches the office. What is the extra distance travelled by Ayush in reaching his office? (Assume that all distances covered are in straight lines.) If the house is situated at (2, 4), bank at (5, 8), school at (13, 14) and office at (13, 26) and coordinates are in km.
Answer:
House H(2, 4), bank B(5, 8), school S(13, 14), office O(13, 26).
\( HB = \sqrt{9 + 16} = 5 \), \( BS = \sqrt{64 + 36} = 10 \), \( SO = \sqrt{0 + 144} = 12 \). Distance travelled \( = 27 \) km.
Direct distance \( HO = \sqrt{11^2 + 22^2} = \sqrt{605} = 11\sqrt{5} \approx 24.6 \) km.
Extra distance \( = 27 - 11\sqrt{5} \approx \) 2.4 km.
In simple words: Add the three legs and subtract the straight-line distance.

Teacher's Note:
1. Find each leg with the distance formula.
2. 11√5 is about 24.6.

Type IV: Very Long Answer Type Questions (4 Marks Each)

Question 13. If (−4, 3) and (4, 3) are two vertices of an equilateral triangle, find the coordinates of the third vertex, given that the origin lies in the interior of the triangle.
Answer:
Let A(−4, 3), B(4, 3) and C(x, y). \( AB = 8 \). From \( CA = CB \): \( (x + 4)^2 = (x - 4)^2 \Rightarrow x = 0 \).
From \( CA = 8 \): \( 16 + (y - 3)^2 = 64 \Rightarrow (y - 3)^2 = 48 \Rightarrow y = 3 \pm 4\sqrt{3} \).
The origin must lie inside the triangle. Since AB is on the line y = 3 (above the origin), the third vertex must be below the origin: \( y = 3 - 4\sqrt{3} \approx -3.93 \). The third vertex is \( (0, 3 - 4\sqrt{3}) \).
In simple words: The third vertex lies on the perpendicular bisector of AB.

Teacher's Note:
1. Use CA = CB and CA = AB.
2. Choose the sign using the condition on the origin.

Question 14. The points A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) are the vertices of \( \triangle ABC \).
(i) The median from A meets BC at D. Find the coordinates of the point D.
(ii) Find the coordinates of the point P on AD such that AP : PD = 2 : 1.
(iii) Find the coordinates of points Q and R on medians BE and CF respectively such that BQ : QE = 2 : 1 and CR : RF = 2 : 1.
(iv) What are the coordinates of the centroid of the triangle ABC?
Answer:
(i) D is the midpoint of BC: \( D = \left(\dfrac{x_2 + x_3}{2}, \dfrac{y_2 + y_3}{2}\right) \).
(ii) \( P = \left(\dfrac{2 \cdot \frac{x_2 + x_3}{2} + 1 \cdot x_1}{3}, \dfrac{2 \cdot \frac{y_2 + y_3}{2} + 1 \cdot y_1}{3}\right) = \left(\dfrac{x_1 + x_2 + x_3}{3}, \dfrac{y_1 + y_2 + y_3}{3}\right) \).
(iii) In the same way, with E the midpoint of CA and F the midpoint of AB, both Q and R work out to \( \left(\dfrac{x_1 + x_2 + x_3}{3}, \dfrac{y_1 + y_2 + y_3}{3}\right) \).
(iv) P, Q and R coincide, so the three medians are concurrent. The centroid is \( \left(\dfrac{x_1 + x_2 + x_3}{3}, \dfrac{y_1 + y_2 + y_3}{3}\right) \).
In simple words: The point two-thirds of the way along each median is the same.

Teacher's Note:
1. Use the section formula with ratio 2 : 1 on each median.
2. The common point is the centroid.