CBSE Class 10 Mathematics Coordinate Geometry Assignment Set 27

Exercise 1 Subjective Questions: Section A (Introduction and Distance Formula)

Question 1. Find the distance between the points: (i) P(−6, 7) and Q(−1, −5); (ii) A(at₁², 2at₁) and B(at₂², 2at₂).
Answer:
(i) \( PQ = \sqrt{(-1 + 6)^2 + (-5 - 7)^2} = \sqrt{25 + 144} = \) 13 units.
(ii) \( AB = \sqrt{a^2(t_2^2 - t_1^2)^2 + 4a^2(t_2 - t_1)^2} = |a(t_2 - t_1)|\sqrt{(t_2 + t_1)^2 + 4} \) units. So \( AB = |a(t_2 - t_1)|\sqrt{(t_1 + t_2)^2 + 4} \).
In simple words: Apply the distance formula and factor out common terms.

Teacher's Note:
1. In (ii) take out a(t₂ − t₁) from both squares.
2. The distance is never negative.

Question 2. Find the value of x, if the distance between the points (x, −1) and (3, 2) is 5.
Answer:
\( (x - 3)^2 + (-1 - 2)^2 = 25 \Rightarrow (x - 3)^2 = 16 \Rightarrow x - 3 = \pm 4 \).
So x = 7 or x = −1.
In simple words: Square both sides.

Teacher's Note:
1. Give both values.

Question 3. If the point P(x, y) is equidistant from the points A(3, 6) and B(−3, 4), prove that \( 3x + y - 5 = 0 \).
Answer:
\( PA^2 = PB^2 \Rightarrow (x - 3)^2 + (y - 6)^2 = (x + 3)^2 + (y - 4)^2 \).
\( -6x - 12y + 45 = 6x - 8y + 25 \Rightarrow 12x + 4y - 20 = 0 \Rightarrow 3x + y - 5 = 0 \). Hence proved.
In simple words: Equal distances give equal squared distances.

Teacher's Note:
1. The locus of such points is the perpendicular bisector of AB.

Question 4. What is the distance between the points A(c, 0) and B(0, −c)?
Answer:
\( AB = \sqrt{(0 - c)^2 + (-c - 0)^2} = \sqrt{2c^2} = \) \( \sqrt{2}\,c \) (taking c > 0).
In simple words: Use the distance formula.

Teacher's Note:
1. For negative c the distance is √2 |c|.

Question 5. Show that the points \( (a, a) \), \( (-a, -a) \) and \( (-\sqrt{3}a, \sqrt{3}a) \) are the vertices of an equilateral triangle.
Answer:
Let A(a, a), B(−a, −a), C(−√3a, √3a). \( AB = \sqrt{4a^2 + 4a^2} = 2\sqrt{2}|a| \).
\( BC = \sqrt{(\sqrt{3} - 1)^2a^2 + (\sqrt{3} + 1)^2a^2} = |a|\sqrt{8} = 2\sqrt{2}|a| \) and \( CA = \sqrt{(1 + \sqrt{3})^2a^2 + (1 - \sqrt{3})^2a^2} = 2\sqrt{2}|a| \).
All three sides are equal, so the triangle is equilateral.
In simple words: Show all three sides are equal.

Teacher's Note:
1. Use (√3 ± 1)² = 4 ± 2√3.
2. The cross terms cancel when added.

Question 6. Show that the points (1, 1), (−2, 7) and (3, −3) are collinear.
Answer:
Let A(1, 1), B(−2, 7), C(3, −3). \( AB = \sqrt{9 + 36} = 3\sqrt{5} \), \( AC = \sqrt{4 + 16} = 2\sqrt{5} \), \( BC = \sqrt{25 + 100} = 5\sqrt{5} \).
Since \( AB + AC = 5\sqrt{5} = BC \), the points are collinear (A lies between B and C).
In simple words: Three points are collinear when two distances add up to the third.

Teacher's Note:
1. Add the two shorter distances.

Question 7. Prove that (2, −2), (−2, 1) and (5, 2) are the vertices of a right-angled triangle. Find the area of the triangle and the length of the hypotenuse.
Answer:
Let A(2, −2), B(−2, 1), C(5, 2). \( AB = \sqrt{16 + 9} = 5 \), \( AC = \sqrt{9 + 16} = 5 \), \( BC = \sqrt{49 + 1} = \sqrt{50} \).
\( AB^2 + AC^2 = 50 = BC^2 \), so the triangle is right-angled at A.
Area \( = \tfrac{1}{2} \times AB \times AC = \tfrac{1}{2} \times 5 \times 5 = \) 12.5 square units; hypotenuse \( BC = \) \( 5\sqrt{2} \) units.
In simple words: Use Pythagoras for the right angle; the legs give the area.

Teacher's Note:
1. Area = ½ × product of the legs.

Question 8. Show that the points A(1, 2), B(5, 4), C(3, 8) and D(−1, 6) are the vertices of a square.
Answer:
\( AB = \sqrt{16 + 4} = 2\sqrt{5} \), \( BC = \sqrt{4 + 16} = 2\sqrt{5} \), \( CD = \sqrt{16 + 4} = 2\sqrt{5} \), \( DA = \sqrt{4 + 16} = 2\sqrt{5} \).
\( AC = \sqrt{4 + 36} = 2\sqrt{10} \), \( BD = \sqrt{36 + 4} = 2\sqrt{10} \).
All sides are equal and the diagonals are equal, so ABCD is a square.
In simple words: Four equal sides and equal diagonals give a square.

Teacher's Note:
1. A rhombus needs only equal sides.
2. Equal diagonals make it a square.

Question 9. Show that the points A(2, −2), B(14, 10), C(11, 13) and D(−1, 1) are the vertices of a rectangle.
Answer:
\( AB = \sqrt{144 + 144} = 12\sqrt{2} \), \( CD = \sqrt{144 + 144} = 12\sqrt{2} \), \( BC = \sqrt{9 + 9} = 3\sqrt{2} \), \( DA = \sqrt{9 + 9} = 3\sqrt{2} \).
\( AC = \sqrt{81 + 225} = \sqrt{306} \), \( BD = \sqrt{225 + 81} = \sqrt{306} \).
Opposite sides are equal and the diagonals are equal, so ABCD is a rectangle.
In simple words: Equal opposite sides and equal diagonals give a rectangle.

Teacher's Note:
1. The adjacent sides are unequal, so it is not a square.

Exercise 1 Subjective Questions: Section B (Section Formula)

Question 10. If two vertices of a triangle are (3, 7) and (−1, 5) and its centroid is (1, 3), find the coordinates of the third vertex.
Answer:
Let the third vertex be (x, y). Centroid: \( \dfrac{3 - 1 + x}{3} = 1 \) and \( \dfrac{7 + 5 + y}{3} = 3 \).
So \( x + 2 = 3 \Rightarrow x = 1 \) and \( 12 + y = 9 \Rightarrow y = -3 \). The third vertex is (1, −3).
In simple words: The centroid is the average of the three vertices.

Teacher's Note:
1. Multiply the centroid coordinates by 3 and subtract the known vertices.

Question 11. If the mid point of the line segment joining the points (−7, 14) and (K, 4) is (a, b), where 2a + 3b = 5, find the value of K.
Answer:
\( a = \dfrac{K - 7}{2} \) and \( b = \dfrac{14 + 4}{2} = 9 \).
\( 2a + 3b = 5 \Rightarrow (K - 7) + 27 = 5 \Rightarrow \) K = −15.
In simple words: Write a and b with the midpoint formula.

Teacher's Note:
1. Substitute them in 2a + 3b = 5.

Question 12. The coordinates of one end point of a diameter of a circle are (4, −1) and the coordinates of the centre of the circle are (1, −3). Find the coordinates of the other end of the diameter.
Answer:
The centre is the midpoint of the diameter. Let the other end be (x, y): \( \dfrac{4 + x}{2} = 1 \) and \( \dfrac{-1 + y}{2} = -3 \).
\( x = -2 \), \( y = -5 \). The other end is (−2, −5).
In simple words: The centre bisects every diameter.

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

Question 13. 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:
The radius \( OA = \sqrt{4 + 0} = 2 \). Since B is on the circle, \( OB = 2 \): \( (x - 2)^2 + (5 - 3)^2 = 4 \Rightarrow (x - 2)^2 = 0 \).
So x = 2.
In simple words: All points on a circle are at the same distance from the centre.

Teacher's Note:
1. Find the radius from A first.

Question 14. If A(−1, 3), B(1, −1) and C(5, 1) are the vertices of a triangle ABC, find the length of the median passing through the vertex A.
Answer:
The midpoint of BC is \( D\left(\dfrac{1 + 5}{2}, \dfrac{-1 + 1}{2}\right) = (3, 0) \).
\( AD = \sqrt{(3 + 1)^2 + (0 - 3)^2} = \sqrt{16 + 9} = \) 5 units.
In simple words: A median joins a vertex to the midpoint of the opposite side.

Teacher's Note:
1. Find the midpoint of BC first.

Question 15. Determine the ratio in which the point (−6, a) divides the join of A(−3, −1) and B(−8, 9). Also find the value of a.
Answer:
Let the ratio be k : 1. \( \dfrac{-8k - 3}{k + 1} = -6 \Rightarrow -8k - 3 = -6k - 6 \Rightarrow k = \dfrac{3}{2} \). The ratio is 3 : 2.
\( a = \dfrac{9k - 1}{k + 1} = \dfrac{\frac{27}{2} - 1}{\frac{5}{2}} = \dfrac{25}{5} = \) 5.
In simple words: Use the known x-coordinate to find the ratio.

Teacher's Note:
1. Then substitute the ratio in the y-formula.

Question 16. If A(−2, −1), B(a, 0), C(4, b) and D(1, 2) are the vertices of a parallelogram, find the values of a and b.
Answer:
The diagonals of a parallelogram bisect each other, so the midpoints of AC and BD coincide.
Midpoint of AC \( = \left(1, \dfrac{b - 1}{2}\right) \); midpoint of BD \( = \left(\dfrac{a + 1}{2}, 1\right) \).
So \( \dfrac{a + 1}{2} = 1 \Rightarrow a = 1 \) and \( \dfrac{b - 1}{2} = 1 \Rightarrow b = 3 \). a = 1, b = 3.
In simple words: Equal midpoints of the diagonals.

Teacher's Note:
1. Take the vertices in order.
2. A and C are opposite, B and D are opposite.

Question 17. The point R divides the line segment AB, where A(−4, 0) and B(0, 6), such that \( AR = \dfrac{3}{4}AB \). Find the co-ordinates of R.
Answer:
\( AR = \dfrac{3}{4}AB \) gives \( AR : RB = 3 : 1 \).
\( R = \left(\dfrac{3(0) + 1(-4)}{4}, \dfrac{3(6) + 1(0)}{4}\right) = \) \( \left(-1, \dfrac{9}{2}\right) \).
In simple words: AR = 3/4 AB means RB = 1/4 AB.

Teacher's Note:
1. The ratio is AR : RB = 3 : 1.

Question 18. The line segment joining the points (3, −4) and (1, 2) is trisected at the points P and Q. If the coordinates of P and Q are (p, −2) and \( \left(\dfrac{5}{3}, q\right) \) respectively, find the values of p and q.
Answer:
P divides the segment in 1 : 2 and Q in 2 : 1. \( P = \left(\dfrac{1(1) + 2(3)}{3}, \dfrac{1(2) + 2(-4)}{3}\right) = \left(\dfrac{7}{3}, -2\right) \).
\( Q = \left(\dfrac{2(1) + 1(3)}{3}, \dfrac{2(2) + 1(-4)}{3}\right) = \left(\dfrac{5}{3}, 0\right) \).
So p = 7/3 and q = 0.
In simple words: Find P and Q by the section formula and compare.

Teacher's Note:
1. Check that the given y = −2 and x = 5/3 agree.

Question 19. If A(4, −8), B(3, 6) and C(5, −4) are the vertices of a \( \triangle ABC \), D is the mid-point of BC and P is a point on AD joined such that \( \dfrac{AP}{PD} = 2 \), find the coordinates of P.
Answer:
\( D = \left(\dfrac{3 + 5}{2}, \dfrac{6 - 4}{2}\right) = (4, 1) \).
P divides AD in 2 : 1: \( P = \left(\dfrac{2(4) + 1(4)}{3}, \dfrac{2(1) + 1(-8)}{3}\right) = \) (4, −2).
In simple words: First find D, then use the section formula on AD.

Teacher's Note:
1. AP : PD = 2 : 1.

Question 20. Find the ratio in which the point (x, 2) divides the line segment joining the points (−3, −4) and (3, 5). Also find the value of x.
Answer:
Let the ratio be k : 1. \( \dfrac{5k - 4}{k + 1} = 2 \Rightarrow 3k = 6 \Rightarrow k = 2 \). The ratio is 2 : 1.
\( x = \dfrac{3k - 3}{k + 1} = \dfrac{3}{3} = \) 1.
In simple words: The y-coordinate 2 is known, so find k from it.

Teacher's Note:
1. Then find x.

Question 21. Find the ratio in which the point (2, y) divides the line segment joining the points A(−2, 2) and B(3, 7). Also find the value of y.
Answer:
Let the ratio be k : 1. \( \dfrac{3k - 2}{k + 1} = 2 \Rightarrow k = 4 \). The ratio is 4 : 1.
\( y = \dfrac{7(4) + 2}{5} = \) 6.
In simple words: Use the x-coordinate to find the ratio.

Teacher's Note:
1. Then find y.

Question 22. Point P divides the line segment joining the points A(−1, 3) and B(9, 8) such that \( \dfrac{AP}{PB} = \dfrac{k}{1} \). If P lies on the line \( x - y + 2 = 0 \), find the value of k.
Answer:
\( P = \left(\dfrac{9k - 1}{k + 1}, \dfrac{8k + 3}{k + 1}\right) \).
P lies on \( x - y + 2 = 0 \): \( (9k - 1) - (8k + 3) + 2(k + 1) = 0 \Rightarrow 3k - 2 = 0 \Rightarrow \) k = 2/3.
In simple words: Substitute P into the line equation.

Teacher's Note:
1. Multiply through by k + 1 to clear the denominator.