CBSE Class 10 Mathematics Coordinate Geometry Assignment Set 24

Very Short Answer Type Questions

Question 1. Find the value of x, if the distance between the points \( A(x, -1) \) and \( B(3, 2) \) is 5.
Answer:
\( AB = 5 \Rightarrow \sqrt{(3 - x)^2 + (2 + 1)^2} = 5 \Rightarrow (3 - x)^2 + 9 = 25 \Rightarrow (3 - x)^2 = 16 \).
\( 3 - x = \pm 4 \). Hence \( x = -1 \) or \( x = 7 \).
In simple words: Square both sides of the distance equation and solve the quadratic.

Teacher's Note:
1. Do not forget both signs of the square root.

Question 2. Find the distance between the points \( A(a + b,\ a - b) \) and \( B(a,\ -b) \).
Answer:
\( AB = \sqrt{(a + b - a)^2 + (a - b + b)^2} = \sqrt{b^2 + a^2} \).
Hence \( AB = \sqrt{a^2 + b^2} \) units.
In simple words: Subtract the coordinates carefully: the differences simplify to b and a.

Teacher's Note:
1. Write the two points as (x₁, y₁) and (x₂, y₂), substitute in the distance formula and simplify the root.
2. Leave the answer in the simplest form.

Question 3. Find the point whose abscissa is equal to its ordinate and which is equidistant from \( P(-1, 0) \) and \( B(0, 5) \).
Answer:
Let the point be \( A(a, a) \). \( AP = AB \Rightarrow (a + 1)^2 + a^2 = a^2 + (a - 5)^2 \Rightarrow a^2 + 2a + 1 = a^2 - 10a + 25 \Rightarrow 12a = 24 \Rightarrow a = 2 \).
Hence the point is \( (2, 2) \).
In simple words: A point with equal abscissa and ordinate has the form (a, a).

Teacher's Note:
1. Points that are equidistant from two points satisfy PA² = PB²; squaring removes the square roots.
2. Expand carefully and cancel the common terms.

Question 4. Find the distance between the points \( P(at_1^2,\ 2at_1) \) and \( Q(at_2^2,\ 2at_2) \).
Answer:
\( PQ = \sqrt{(at_2^2 - at_1^2)^2 + (2at_2 - 2at_1)^2} = \sqrt{a^2(t_2 - t_1)^2(t_2 + t_1)^2 + 4a^2(t_2 - t_1)^2} \).
\( = a(t_2 - t_1)\sqrt{(t_2 + t_1)^2 + 4} \), taking \( t_2 > t_1 \). Hence \( PQ = a(t_2 - t_1)\sqrt{(t_1 + t_2)^2 + 4} \).
In simple words: Take out the common factors a and (t₂ − t₁) from under the root.

Teacher's Note:
1. Write the two points as (x₁, y₁) and (x₂, y₂), substitute in the distance formula and simplify the root.
2. Leave the answer in the simplest form.

Question 5. If the three vertices of a parallelogram taken in order are \( (-1, 0),\ (3, 1) \) and \( (2, 2) \) respectively, then find the coordinates of the fourth vertex.
Answer:
Let \( A(-1, 0),\ B(3, 1),\ C(2, 2) \) and the fourth vertex \( D(x, y) \). The diagonals AC and BD bisect each other.
Mid-point of \( AC = \left( \frac{1}{2}, 1 \right) \) and mid-point of \( BD = \left( \frac{3 + x}{2}, \frac{1 + y}{2} \right) \). So \( \frac{3 + x}{2} = \frac{1}{2} \Rightarrow x = -2 \) and \( \frac{1 + y}{2} = 1 \Rightarrow y = 1 \).
Hence the fourth vertex is \( D(-2, 1) \).
In simple words: The diagonals of a parallelogram bisect each other, so they share a mid-point.

Teacher's Note:
1. Join the vertices in the given order to find which pair forms a diagonal.

Short Answer Type Questions

Question 6. If the point \( A(x, y) \) is equidistant from \( B(a + b,\ b - a) \) and \( C(a - b,\ a + b) \), then prove that \( bx = ay \).
Answer:
\( AB = AC \Rightarrow AB^2 = AC^2 \Rightarrow \{x - (a + b)\}^2 + \{y - (b - a)\}^2 = \{x - (a - b)\}^2 + \{y - (a + b)\}^2 \).
Expanding and cancelling \( x^2,\ y^2 \) and \( a^2 + b^2 \)-type terms: \( -2x(a + b) - 2y(b - a) = -2x(a - b) - 2y(a + b) \).
\( \Rightarrow -2x(a + b) + 2x(a - b) = 2y(b - a) - 2y(a + b) \Rightarrow -4bx = -4ay \Rightarrow bx = ay \). Hence proved.
In simple words: Equate the squared distances AB² and AC² and simplify.

Teacher's Note:
1. Points that are equidistant from two points satisfy PA² = PB²; squaring removes the square roots.
2. Expand carefully and cancel the common terms.

Question 7. If A and B are two points whose coordinates are \( (at^2,\ 2at) \) and \( \left( \frac{a}{t^2},\ \frac{2a}{t} \right) \) respectively and C is the point \( (a, 0) \). Show that \( \frac{1}{CA} + \frac{1}{CB} \) is independent of t.
Answer:
\( CA = \sqrt{(at^2 - a)^2 + (2at)^2} = a\sqrt{(t^2 - 1)^2 + 4t^2} = a(t^2 + 1) \).
\( CB = \sqrt{\left(\frac{a}{t^2} - a\right)^2 + \left(\frac{2a}{t}\right)^2} = \frac{a}{t^2}\sqrt{(1 - t^2)^2 + 4t^2} = \frac{a(1 + t^2)}{t^2} \).
\( \frac{1}{CA} + \frac{1}{CB} = \frac{1}{a(t^2 + 1)} + \frac{t^2}{a(1 + t^2)} = \frac{1 + t^2}{a(1 + t^2)} = \frac{1}{a} \), which does not contain t. Hence proved.
In simple words: Both distances simplify to expressions with the factor (1 + t²), which cancels.

Teacher's Note:
1. Use (t² − 1)² + 4t² = (t² + 1)² to remove the root.

Question 8. Find the coordinates of the circumcentre of the triangle whose vertices are \( (8, 6),\ (8, -2) \) and \( (2, -2) \). Also, find its circum-radius.

Answer:
The circumcentre is equidistant from the three vertices. Let \( P(8, 6),\ Q(8, -2),\ R(2, -2) \) and the circumcentre \( M(x, y) \).
\( MP^2 = MQ^2 \): \( (x - 8)^2 + (y - 6)^2 = (x - 8)^2 + (y + 2)^2 \Rightarrow -12y + 36 = 4y + 4 \Rightarrow y = 2 \).
\( MQ^2 = MR^2 \): \( (x - 8)^2 + (y + 2)^2 = (x - 2)^2 + (y + 2)^2 \Rightarrow -16x + 64 = -4x + 4 \Rightarrow x = 5 \).
Circum-radius \( = MP = \sqrt{(5 - 8)^2 + (2 - 6)^2} = \sqrt{9 + 16} = 5 \).
Hence the circumcentre is \( (5, 2) \) and the circum-radius is 5 units.
In simple words: The circumcentre is equidistant from all three vertices, so equate two pairs of distances.

Teacher's Note:
1. Use two equations from MP = MQ = MR to find x and y.
2. The radius is the distance from the centre to any vertex.

Question 9. Find the coordinates of the remaining vertices, if the opposite vertices of a square are \( (1, -1) \) and \( (3, 4) \).
Answer:
Let \( P(1, -1) \) and \( R(3, 4) \) be the opposite vertices and \( Q(x, y) \) one of the other vertices. In a square, \( PQ = QR \): \( (x - 1)^2 + (y + 1)^2 = (x - 3)^2 + (y - 4)^2 \Rightarrow 4x + 10y = 23 \Rightarrow x = \frac{23 - 10y}{4} \). (i)
Since \( \angle PQR = 90^\circ \), \( PQ^2 + QR^2 = PR^2 = 4 + 25 = 29 \): \( (x - 1)^2 + (y + 1)^2 + (x - 3)^2 + (y - 4)^2 = 29 \Rightarrow x^2 + y^2 - 4x - 3y - 1 = 0 \). (ii)
Substituting (i) in (ii): \( 4y^2 - 12y + 5 = 0 \Rightarrow (2y - 1)(2y - 5) = 0 \Rightarrow y = \frac{1}{2} \) or \( y = \frac{5}{2} \).
For \( y = \frac{1}{2} \): \( x = \frac{23 - 5}{4} = \frac{9}{2} \). For \( y = \frac{5}{2} \): \( x = \frac{23 - 25}{4} = -\frac{1}{2} \).
Hence the remaining vertices are \( \left( \frac{9}{2}, \frac{1}{2} \right) \) and \( \left( -\frac{1}{2}, \frac{5}{2} \right) \).
In simple words: Use PQ = QR and the right angle at Q (Pythagoras on PQR) to get two equations in x and y.

Teacher's Note:
1. The diagonal PR is also the hypotenuse of the right triangle PQR.
2. Two values of y give the two remaining vertices.

Question 10. If a circle passes through the points \( (0, 0),\ (-2, 1) \) and \( (-3, 2) \), find the coordinates of the centre of the circle and also find its radius.
Answer:
Let the centre be \( O(x, y) \) and the points \( A(0, 0),\ B(-2, 1),\ C(-3, 2) \). Then \( OA = OB = OC \).
\( OA^2 = OB^2 \Rightarrow x^2 + y^2 = (x + 2)^2 + (y - 1)^2 \Rightarrow 4x - 2y + 5 = 0 \). (i)
\( OA^2 = OC^2 \Rightarrow x^2 + y^2 = (x + 3)^2 + (y - 2)^2 \Rightarrow 6x - 4y + 13 = 0 \). (ii)
Solving (i) and (ii): \( x = \frac{3}{2},\ y = \frac{11}{2} \). Radius \( = OA = \sqrt{\frac{9}{4} + \frac{121}{4}} = \frac{\sqrt{130}}{2} \).
Hence the centre is \( \left( \frac{3}{2}, \frac{11}{2} \right) \) and the radius is \( \frac{1}{2}\sqrt{130} \) units.
In simple words: The centre of a circle is equidistant from every point on it, so OA = OB = OC.

Teacher's Note:
1. Form two linear equations by equating pairs of squared distances.
2. The radius is the distance from the centre to any of the points.

Question 11. Prove that \( (4, -1),\ (6, 0),\ (7, 2) \) and \( (5, 1) \) are the vertices of a rhombus. Is it a square?
Answer:
Let \( P(4, -1),\ Q(6, 0),\ R(7, 2),\ S(5, 1) \). \( PQ = \sqrt{4 + 1} = \sqrt{5} \), \( QR = \sqrt{1 + 4} = \sqrt{5} \), \( RS = \sqrt{4 + 1} = \sqrt{5} \), \( SP = \sqrt{1 + 4} = \sqrt{5} \).
All four sides are equal, so PQRS is a rhombus. The diagonals are \( PR = \sqrt{9 + 9} = 3\sqrt{2} \) and \( QS = \sqrt{1 + 1} = \sqrt{2} \), which are not equal.
Hence PQRS is a rhombus but not a square.
In simple words: Four equal sides give a rhombus; it is a square only if the diagonals are also equal.

Teacher's Note:
1. Check the diagonals to separate a square from a rhombus.

Question 12. If the coordinates of two points P and Q are \( (3, 4) \) and \( (5, -2) \) respectively, find the coordinates of the point M, if \( MP = MQ \). Also, the area of \( \triangle MPQ \) is 10 sq. units.
Answer:
M is equidistant from P and Q: \( (x - 3)^2 + (y - 4)^2 = (x - 5)^2 + (y + 2)^2 \Rightarrow 4x - 12y - 4 = 0 \Rightarrow x - 3y - 1 = 0 \). (i)
\( PQ = \sqrt{4 + 36} = 2\sqrt{10} \). The mid-point of PQ is \( D(4, 1) \), and MD is perpendicular to PQ, so \( \text{area} = \frac{1}{2} \times PQ \times MD = 10 \Rightarrow MD = \frac{20}{2\sqrt{10}} = \sqrt{10} \).
\( MD^2 = 10 \Rightarrow (x - 4)^2 + (y - 1)^2 = 10 \). From (i), \( x = 3y + 1 \): \( (3y - 3)^2 + (y - 1)^2 = 10 \Rightarrow 10(y - 1)^2 = 10 \Rightarrow y = 2 \) or \( 0 \).
For \( y = 2 \): \( x = 7 \). For \( y = 0 \): \( x = 1 \).
Hence M is \( (7, 2) \) or \( (1, 0) \).
In simple words: M lies on the perpendicular bisector of PQ at a distance from PQ given by the area (base times height).

Teacher's Note:
1. Area = half the base times the height, and the height is MD.
2. Two positions of M, one on each side of PQ.

Question 13. If the point \( C(-1, 2) \) divides internally the line segment joining \( A(2, 5) \) and B in the ratio \( 3 : 4 \), find the coordinates of B.
Answer:
Let \( B(\alpha, \beta) \). By the section formula, \( C = \left( \frac{3\alpha + 4(2)}{7},\ \frac{3\beta + 4(5)}{7} \right) = (-1, 2) \).
\( \frac{3\alpha + 8}{7} = -1 \Rightarrow \alpha = -5 \) and \( \frac{3\beta + 20}{7} = 2 \Rightarrow \beta = -2 \).
Hence \( B(-5, -2) \).
In simple words: Write the coordinates of C by the section formula in terms of B and equate them to the given point.

Teacher's Note:
1. Here AC : CB = 3 : 4, so the weights are 3 for B and 4 for A.

Question 14. Find the ratio in which the point \( P(-3, p) \) divides the line segment joining \( A(-5, -4) \) and \( B(-2, 3) \). Hence, find the value of p.
Answer:
Let P divide AB in the ratio \( k : 1 \). Then \( P = \left( \frac{-2k - 5}{k + 1},\ \frac{3k - 4}{k + 1} \right) \).
\( \frac{-2k - 5}{k + 1} = -3 \Rightarrow -2k - 5 = -3k - 3 \Rightarrow k = 2 \). Then \( p = \frac{3(2) - 4}{2 + 1} = \frac{2}{3} \).
Hence the ratio is \( 2 : 1 \) and \( p = \frac{2}{3} \).
In simple words: Use the known x-coordinate to find the ratio, then the y-coordinate to find p.

Teacher's Note:
1. The ratio comes from the coordinate that has no unknown.

Long Answer Type Questions

Question 15. Find the third vertex of an equilateral triangle, if two of its vertices are \( (0, 0) \) and \( (3, \sqrt{3}) \).

Answer:
Let \( O(0, 0),\ P(3, \sqrt{3}) \) and the third vertex \( Q(x, y) \). \( OP = \sqrt{9 + 3} = \sqrt{12} \). Since \( OQ = PQ = OP \): \( x^2 + y^2 = 12 \) and \( (x - 3)^2 + (y - \sqrt{3})^2 = 12 \).
The second equation gives \( x^2 + y^2 - 6x - 2\sqrt{3}y + 12 = 12 \), i.e. \( 12 - 6x - 2\sqrt{3}y = 0 \Rightarrow 3x + \sqrt{3}y = 6 \Rightarrow y = \frac{6 - 3x}{\sqrt{3}} \).
Putting in \( x^2 + y^2 = 12 \): \( x^2 + \frac{(6 - 3x)^2}{3} = 12 \Rightarrow 12x^2 - 36x = 0 \Rightarrow x = 0 \) or \( 3 \).
For \( x = 0 \): \( y = 2\sqrt{3} \). For \( x = 3 \): \( y = -\sqrt{3} \).
Hence the third vertex is \( (0, 2\sqrt{3}) \) or \( (3, -\sqrt{3}) \).
In simple words: All three sides are equal, so the third vertex is at distance OP from both given vertices.

Teacher's Note:
1. Two such triangles exist, one on each side of OP.
2. Substitute the linear equation into the circle-type equation to solve.

Question 16. If the coordinates of the mid-points of the sides of a triangle are \( (1, 2),\ (0, -1),\ (2, -1) \). Find the coordinates of its vertices.

Answer:
Let \( P(x_1, y_1),\ Q(x_2, y_2),\ R(x_3, y_3) \) be the vertices and \( L(1, 2),\ M(0, -1),\ N(2, -1) \) the mid-points of QR, RP and PQ respectively.
\( x_2 + x_3 = 2,\ y_2 + y_3 = 4 \); \( x_1 + x_3 = 0,\ y_1 + y_3 = -2 \); \( x_1 + x_2 = 4,\ y_1 + y_2 = -2 \).
Adding: \( x_1 + x_2 + x_3 = 3 \) and \( y_1 + y_2 + y_3 = 0 \). Hence \( x_1 = 3 - 2 = 1,\ y_1 = 0 - 4 = -4 \); \( x_2 = 3 - 0 = 3,\ y_2 = 0 + 2 = 2 \); \( x_3 = 3 - 4 = -1,\ y_3 = 0 + 2 = 2 \).
Hence the vertices are \( P(1, -4),\ Q(3, 2),\ R(-1, 2) \).
In simple words: Write the mid-point formula for each side, add the equations and subtract to get every vertex.

Teacher's Note:
1. Each vertex equals the sum of all three x (or y) values minus the sum for the opposite side.
2. Verify by finding a mid-point of two of the vertices.

Question 17. The vertices of a \( \triangle ABC \) are \( A(5, 5),\ B(1, 1) \) and \( C(9, 1) \). A line is drawn to intersect sides AB and AC at P and Q respectively, such that \( \frac{AP}{AB} = \frac{AQ}{AC} = \frac{3}{4} \). Find the length of the line segment PQ.

Answer:
\( \frac{AP}{AB} = \frac{3}{4} \Rightarrow 4AP = 3AP + 3PB \Rightarrow AP : PB = 3 : 1 \), and similarly \( AQ : QC = 3 : 1 \).
\( P = \left( \frac{3(1) + 1(5)}{4},\ \frac{3(5) + 1(5)}{4} \right) = (2, 5) \) and \( Q = \left( \frac{3(9) + 1(5)}{4},\ \frac{3(1) + 1(5)}{4} \right) = (8, 2) \).
\( PQ = \sqrt{(8 - 2)^2 + (2 - 5)^2} = \sqrt{36 + 9} = \sqrt{45} = 3\sqrt{5} \).
Hence the length of PQ is \( 3\sqrt{5} \) units.
In simple words: AP : PB = 3 : 1 gives the coordinates of P and Q by the section formula.

Teacher's Note:
1. Convert AP/AB = 3/4 into the ratio AP : PB first.
2. Then use the distance formula for PQ.