GSEB Class 10 Maths Solutions Chapter 7 Coordinate Geometry Exercise 7.1

Question 1. Find the distance between the following pairs of points:
(i) (2, 3), (4, 1)
(ii) (-5, 7), (-1, 3)
(iii) (a, b), (-a, -b)
Answer:
(i) Let the provided points be A(2, 3) and B(4, 1). We understand that the distance between two points \( A(x_1, y_1) \) and \( B(x_2, y_2) \) is found using the formula:
\( AB = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \)
\( \implies AB = \sqrt{(4-2)^2 + (1-3)^2} \)
\( = \sqrt{2^2 + (-2)^2} \)
\( = \sqrt{4+4} \)
\( = \sqrt{8} \)
\( = 2\sqrt{2} \)
(ii) Let the provided points be A(-5, 7) and B(-1, 3). We understand that the distance between two points \( A(x_1, y_1) \) and \( B(x_2, y_2) \) is found using the formula:
\( AB = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \)
\( \implies AB = \sqrt{(-1 - (-5))^2 + (3 - 7)^2} \)
\( = \sqrt{(4)^2 + (-4)^2} \)
\( = \sqrt{16 + 16} \)
\( = \sqrt{32} \)
\( = 4\sqrt{2} \)
(iii) Let the provided points be A(a, b) and B(-a, -b). We understand that the distance between two points \( A(x_1, y_1) \) and \( B(x_2, y_2) \) is found using the formula:
\( AB = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \)
\( \implies AB = \sqrt{(-a - a)^2 + (-b - b)^2} \)
\( = \sqrt{(-2a)^2 + (-2b)^2} \)
\( = \sqrt{4a^2 + 4b^2} \)
\( = \sqrt{4(a^2 + b^2)} \)
\( = 2\sqrt{a^2 + b^2} \)
In simple words: To find the distance between two points, use the distance formula. Plug in the x and y values for each point, calculate the squared differences, add them, and then take the square root. Remember to simplify the square root if possible.

Exam Tip: Always write down the distance formula first to avoid errors, especially with negative coordinates. Be careful with double negatives when subtracting coordinates.

Question 2. Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B discussed in Section 7.2
Answer: Let the provided points be A(0, 0) and B(36, 15). We calculate the distance AB using the formula: \( AB = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \).
\( \implies AB = \sqrt{(36-0)^2 + (15-0)^2} \)
\( = \sqrt{(36)^2 + (15)^2} \)
\( = \sqrt{1296 + 225} \)
\( = \sqrt{1521} \)
\( = 39 \)
Yes, we are able to determine the distance between the two towns A and B, which were mentioned in section 7.2, and this distance is 39 km.
In simple words: To find the distance between the two points, use the distance formula. Square the differences in the x and y coordinates, add them, and find the square root. The distance comes out to 39, so the towns are 39 km apart.

Exam Tip: Remember that the distance formula is derived from the Pythagorean theorem, relating horizontal and vertical displacement. For problems involving real-world scenarios like towns, include units if specified.

Question 3. Determine if the points (1, 5), (2, 3) and (-2, -11) are collinear.
Answer: Let the provided points be A(1, 5), B(2, 3), and C(-2, -11). To check for collinearity, we must determine the distances between each pair of points.
First, calculate AB:
\( AB = \sqrt{(2-1)^2 + (3-5)^2} \)
\( \implies AB = \sqrt{(1)^2 + (-2)^2} \)
\( \implies AB = \sqrt{1 + 4} \)
\( \implies AB = \sqrt{5} \)
Next, calculate AC:
\( AC = \sqrt{(-2-1)^2 + (-11-5)^2} \)
\( \implies AC = \sqrt{(-3)^2 + (-16)^2} \)
\( \implies AC = \sqrt{9 + 256} \)
\( \implies AC = \sqrt{265} \)
Lastly, calculate BC:
\( BC = \sqrt{(-2-2)^2 + (-11-3)^2} \)
\( \implies BC = \sqrt{(-4)^2 + (-14)^2} \)
\( \implies BC = \sqrt{16 + 196} \)
\( \implies BC = \sqrt{212} \)
Now, we check if the sum of any two distances equals the third distance. We observe that \( AB + BC \neq AC \), \( BC + AC \neq AB \), and \( AB + AC \neq BC \). Therefore, the points A, B, and C are not collinear.
In simple words: To see if points are in a straight line, calculate the distance between all three pairs. If two of the distances add up to the third, then the points are collinear. In this case, they don't, so the points are not on the same line.

Exam Tip: Points are collinear if the sum of the distances of the two shorter segments equals the length of the longest segment. This is the primary condition to check for collinearity using the distance formula.

Question 4. Check whether (5, -2), (6, 4) and (7, -2) are the vertices of an isosceles triangle.
Answer: Let the provided points be A(5, -2), B(6, 4), and C(7, -2). To determine if they form an isosceles triangle, we need to calculate the lengths of its sides.
First, calculate the length of side AB:
\( AB = \sqrt{(6-5)^2 + (4 - (-2))^2} \)
\( = \sqrt{(1)^2 + (6)^2} \)
\( = \sqrt{1 + 36} \)
\( = \sqrt{37} \)
Next, calculate the length of side BC:
\( BC = \sqrt{(7-6)^2 + (-2-4)^2} \)
\( = \sqrt{(1)^2 + (-6)^2} \)
\( = \sqrt{1 + 36} \)
\( = \sqrt{37} \)
Since the length of AB is equal to the length of BC (\( AB = BC = \sqrt{37} \)), the triangle ABC is an isosceles triangle.
In simple words: An isosceles triangle has two sides of equal length. Calculate the lengths of all three sides using the distance formula. If two sides have the same length, then it is an isosceles triangle.

Exam Tip: Remember that an isosceles triangle only requires two sides to be equal, not all three. Make sure to clearly state your conclusion after showing the side lengths.

Question 5. In a classroom, 4 friends are seated at the points A, B, C and D as shown in figure. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don't you think ABCD is a square"? Chameli disagrees, using distance formula, find which of them is correct.
Answer: First, let's identify the coordinates of the friends from the figure: A(3, 4), B(6, 7), C(9, 4), and D(6, 1). To check if ABCD forms a square, we must calculate the lengths of all four sides and both diagonals.
**Calculate side lengths:**
\( AB = \sqrt{(6-3)^2 + (7-4)^2} \)
\( = \sqrt{3^2 + 3^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt{2} \)
\( BC = \sqrt{(9-6)^2 + (4-7)^2} \)
\( = \sqrt{3^2 + (-3)^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt{2} \)
\( CD = \sqrt{(6-9)^2 + (1-4)^2} \)
\( = \sqrt{(-3)^2 + (-3)^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt{2} \)
\( DA = \sqrt{(3-6)^2 + (4-1)^2} \)
\( = \sqrt{(-3)^2 + 3^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt{2} \)
Since all four sides are equal (\( AB = BC = CD = DA = 3\sqrt{2} \)), the figure could be a rhombus or a square.
**Calculate diagonal lengths:**
\( AC = \sqrt{(9-3)^2 + (4-4)^2} \)
\( = \sqrt{6^2 + 0^2} = \sqrt{36} = 6 \)
\( BD = \sqrt{(6-6)^2 + (1-7)^2} \)
\( = \sqrt{0^2 + (-6)^2} = \sqrt{36} = 6 \)
Since both diagonals are also equal (\( AC = BD = 6 \)), and all sides are equal, the quadrilateral ABCD is indeed a square.
Therefore, Champa is correct.
In simple words: To know if it's a square, measure all four sides and both diagonals using the distance formula. If all sides are the same length AND both diagonals are the same length, then it's a square. Champa was right because all sides and diagonals were equal.

Exam Tip: For quadrilaterals, calculate all four side lengths and both diagonal lengths using the distance formula. A square has all sides equal and both diagonals equal.

Question 6. Name the type of quadrilateral formed, if any, by the following points and give reasons for your answer:
(i) (-1, -2), (1, 0), (-1, 2), (-3, 0)
(ii) (3, 5), (3, 1), (0, 3), (-1, -4)
(iii) (4, 5), (7, 6), (4, 3), (1, 2)
Answer:
(i) Let the provided points be A(-1, -2), B(1, 0), C(-1, 2), and D(-3, 0). We will calculate the lengths of the four sides and the two diagonals to identify the quadrilateral.
**Calculate side lengths:**
\( AB = \sqrt{(1 - (-1))^2 + (0 - (-2))^2} \)
\( = \sqrt{(1+1)^2 + (0+2)^2} = \sqrt{2^2 + 2^2} = \sqrt{4+4} = \sqrt{8} \)
\( BC = \sqrt{(-1-1)^2 + (2-0)^2} \)
\( = \sqrt{(-2)^2 + 2^2} = \sqrt{4+4} = \sqrt{8} \)
\( CD = \sqrt{(-3 - (-1))^2 + (0 - 2)^2} \)
\( = \sqrt{(-3+1)^2 + (-2)^2} = \sqrt{(-2)^2 + (-2)^2} = \sqrt{4+4} = \sqrt{8} \)
\( DA = \sqrt{(-1 - (-3))^2 + (-2 - 0)^2} \)
\( = \sqrt{(-1+3)^2 + (-2)^2} = \sqrt{2^2 + (-2)^2} = \sqrt{4+4} = \sqrt{8} \)
All four sides are equal: \( AB = BC = CD = DA = \sqrt{8} \).
**Calculate diagonal lengths:**
\( AC = \sqrt{(-1 - (-1))^2 + (2 - (-2))^2} \)
\( = \sqrt{(-1+1)^2 + (2+2)^2} = \sqrt{0^2 + 4^2} = \sqrt{0+16} = \sqrt{16} = 4 \)
\( BD = \sqrt{(-3-1)^2 + (0-0)^2} \)
\( = \sqrt{(-4)^2 + 0^2} = \sqrt{16+0} = \sqrt{16} = 4 \)
Both diagonals are also equal: \( AC = BD = 4 \).
Since all four sides of the quadrilateral are equal and its diagonals are also equal, the given points form a square.
(ii) Let the provided points be A(3, 5), B(3, 1), C(0, 3), and D(-1, -4). We must calculate the lengths of all four sides and the two diagonals to classify the quadrilateral.
**Calculate side lengths:**
\( AB = \sqrt{(3-3)^2 + (1-5)^2} = \sqrt{0^2 + (-4)^2} = \sqrt{0+16} = \sqrt{16} = 4 \)
\( BC = \sqrt{(0-3)^2 + (3-1)^2} = \sqrt{(-3)^2 + 2^2} = \sqrt{9+4} = \sqrt{13} \)
\( CD = \sqrt{(-1-0)^2 + (-4-3)^2} = \sqrt{(-1)^2 + (-7)^2} = \sqrt{1+49} = \sqrt{50} = 5\sqrt{2} \)
\( DA = \sqrt{(3-(-1))^2 + (5-(-4))^2} = \sqrt{(3+1)^2 + (5+4)^2} = \sqrt{4^2 + 9^2} = \sqrt{16+81} = \sqrt{97} \)
**Calculate diagonal lengths:**
\( AC = \sqrt{(0-3)^2 + (3-5)^2} = \sqrt{(-3)^2 + (-2)^2} = \sqrt{9+4} = \sqrt{13} \)
\( BD = \sqrt{(-1-3)^2 + (-4-1)^2} = \sqrt{(-4)^2 + (-5)^2} = \sqrt{16+25} = \sqrt{41} \)
Observing the side and diagonal lengths, we find that no pairs of opposite sides are equal (\( AB \neq CD \), \( BC \neq DA \)), and the diagonals are not equal (\( AC \neq BD \)). Therefore, the given points form a general quadrilateral.
(iii) Let the provided points be A(4, 5), B(7, 6), C(4, 3), and D(1, 2). We will calculate the lengths of the four sides and the two diagonals to identify the quadrilateral.
**Calculate side lengths:**
\( AB = \sqrt{(7-4)^2 + (6-5)^2} \)
\( \implies AB = \sqrt{3^2 + 1^2} = \sqrt{9+1} = \sqrt{10} \)
\( BC = \sqrt{(4-7)^2 + (3-6)^2} \)
\( = \sqrt{(-3)^2 + (-3)^2} = \sqrt{9+9} = \sqrt{18} \)
\( CD = \sqrt{(1-4)^2 + (2-3)^2} \)
\( = \sqrt{(-3)^2 + (-1)^2} = \sqrt{9+1} = \sqrt{10} \)
\( DA = \sqrt{(4-1)^2 + (5-2)^2} \)
\( = \sqrt{3^2 + 3^2} = \sqrt{9+9} = \sqrt{18} \)
We see that opposite sides are equal: \( AB = CD = \sqrt{10} \) and \( BC = DA = \sqrt{18} \).
**Calculate diagonal lengths:**
\( AC = \sqrt{(4-4)^2 + (3-5)^2} \)
\( = \sqrt{0^2 + (-2)^2} = \sqrt{0+4} = \sqrt{4} = 2 \)
\( BD = \sqrt{(1-7)^2 + (2-6)^2} \)
\( = \sqrt{(-6)^2 + (-4)^2} = \sqrt{36+16} = \sqrt{52} \)
Since opposite sides are equal (\( AB=CD \) and \( BC=DA \)) and the diagonals are not equal (\( AC \neq BD \)), the quadrilateral formed by the given points is a parallelogram.
In simple words: To classify any quadrilateral, you need to find the length of all four sides and both diagonals. Use the distance formula for each. Then, compare these lengths. Equal opposite sides and equal diagonals mean it's a square or rectangle. Equal opposite sides and unequal diagonals mean it's a parallelogram. All sides equal and equal diagonals mean square. All sides equal and unequal diagonals mean rhombus. If none of these specific patterns match, it's a general quadrilateral.

Exam Tip: To classify a quadrilateral, calculate all four side lengths and both diagonal lengths using the distance formula. Then compare them to determine if it's a square, rhombus, rectangle, parallelogram, or general quadrilateral.

Question 7. Find the point on the x-axis which is equidistant from (2, -5) and (-2, 9).
Answer: Let the desired point be 'P', located on the x-axis. This means its y-coordinate is 0, and its x-coordinate is, for instance, x. So, the coordinates of point P are (x, 0). Let the two provided points be A(2, -5) and B(-2, 9).
We are informed that P is equidistant from A and B, which means \( AP = BP \).
Squaring both sides for easier calculation, we get \( AP^2 = BP^2 \).
Using the distance formula:
\( (x-2)^2 + (0 - (-5))^2 = (x - (-2))^2 + (0-9)^2 \)
\( (x-2)^2 + 5^2 = (x+2)^2 + (-9)^2 \)
\( x^2 - 4x + 4 + 25 = x^2 + 4x + 4 + 81 \)
\( x^2 - 4x + 29 = x^2 + 4x + 85 \)
Subtract \( x^2 \) from both sides:
\( -4x + 29 = 4x + 85 \)
Move x terms to one side and constants to the other:
\( 29 - 85 = 4x + 4x \)
\( -56 = 8x \)
\( x = \frac{-56}{8} \)
\( x = -7 \)
Therefore, the required point on the x-axis is (-7, 0).
In simple words: A point on the x-axis has a y-coordinate of zero. Set the distance from this point to each of the given points equal. Use the distance formula and then solve the resulting equation for x. This gives you the x-coordinate of the equidistant point.

Exam Tip: When a point is on the x-axis, its y-coordinate is 0 (x,0). If it's on the y-axis, its x-coordinate is 0 (0,y). This simplifies the distance formula setup considerably.

Question 8. Find the values of y for which the distance between the points P(2, -3) and Q(10, y) is 10 units.
Answer: We are given two points, P(2, -3) and Q(10, y), and the distance between them, PQ, is 10 units.
Using the distance formula, \( PQ^2 = (x_2-x_1)^2 + (y_2-y_1)^2 \).
We can square the given distance, so \( PQ^2 = 10^2 = 100 \).
Substitute the coordinates into the formula:
\( (10 - 2)^2 + (y - (-3))^2 = 100 \)
\( (8)^2 + (y + 3)^2 = 100 \)
\( 64 + (y^2 + 6y + 9) = 100 \)
\( y^2 + 6y + 73 = 100 \)
\( y^2 + 6y + 73 - 100 = 0 \)
\( y^2 + 6y - 27 = 0 \)
To solve this quadratic equation, we can factor it:
\( y^2 + 9y - 3y - 27 = 0 \)
\( y(y + 9) - 3(y + 9) = 0 \)
\( (y + 9)(y - 3) = 0 \)
This gives us two possible values for y:
\( y + 9 = 0 \implies y = -9 \)
or
\( y - 3 = 0 \implies y = 3 \)
Therefore, the required values for y are -9 or 3.
In simple words: Use the distance formula and set it equal to the given distance (10). Square both sides to remove the square root. Expand and simplify the equation to get a quadratic equation. Factor the quadratic equation to find the two possible values for y.

Exam Tip: When solving for an unknown coordinate in the distance formula, remember to square both sides to eliminate the square root. This often leads to a quadratic equation, which may yield two possible solutions.

Question 9. If Q(0, 1) is equidistant from P(5, -3) and R(x, 6), find the values of x. Also find the distances QR and PR.
Answer: We are given point Q(0, 1) which is equidistant from P(5, -3) and R(x, 6).
This means \( PQ = RQ \), so \( PQ^2 = RQ^2 \).
Using the distance formula:
\( (0-5)^2 + (1 - (-3))^2 = (0-x)^2 + (1-6)^2 \)
\( (-5)^2 + (1+3)^2 = (-x)^2 + (-5)^2 \)
\( 25 + 4^2 = x^2 + 25 \)
\( 25 + 16 = x^2 + 25 \)
\( 41 = x^2 + 25 \)
\( x^2 = 41 - 25 \)
\( x^2 = 16 \)
\( x = \pm 4 \)
Thus, there are two possible coordinates for R: R(4, 6) or R(-4, 6).
**Now, we find the distances QR and PR for both cases:**
**Case 1: R(4, 6)**
Calculate QR:
\( QR = \sqrt{(0-4)^2 + (1-6)^2} \)
\( = \sqrt{(-4)^2 + (-5)^2} = \sqrt{16+25} = \sqrt{41} \)
Calculate PR:
\( PR = \sqrt{(4-5)^2 + (6-(-3))^2} \)
\( = \sqrt{(-1)^2 + (6+3)^2} = \sqrt{(-1)^2 + 9^2} = \sqrt{1+81} = \sqrt{82} \)
**Case 2: R(-4, 6)**
Calculate QR:
\( QR = \sqrt{(0-(-4))^2 + (1-6)^2} \)
\( = \sqrt{4^2 + (-5)^2} = \sqrt{16+25} = \sqrt{41} \)
Calculate PR:
\( PR = \sqrt{(-4-5)^2 + (6-(-3))^2} \)
\( = \sqrt{(-9)^2 + (6+3)^2} = \sqrt{(-9)^2 + 9^2} = \sqrt{81+81} = \sqrt{162} = 9\sqrt{2} \)
In summary, the values for x are \( \pm 4 \). If \( x=4 \), then \( QR = \sqrt{41} \) and \( PR = \sqrt{82} \). If \( x=-4 \), then \( QR = \sqrt{41} \) and \( PR = 9\sqrt{2} \).
In simple words: First, use the distance formula to set the distance from Q to P equal to the distance from Q to R. Solve this equation to find the possible values for x. Then, for each x-value, calculate the distances QR and PR using the distance formula again.

Exam Tip: When you have \( x^2 = k \), remember that x can be both positive and negative \( (\pm \sqrt{k}) \). Make sure to find distances for both possible values of the coordinate.

Question 10. Find the relation between x andy such that the point (x, y) is equidistant from the point (3, 6) and (-3, 4).
Answer: Let P(x, y) be the point that is equidistant from A(3, 6) and B(-3, 4).
Given that \( AP = BP \), we can square both sides to get \( AP^2 = BP^2 \).
Using the distance formula:
\( (x-3)^2 + (y-6)^2 = (x - (-3))^2 + (y-4)^2 \)
\( (x-3)^2 + (y-6)^2 = (x+3)^2 + (y-4)^2 \)
Expand both sides:
\( (x^2 - 6x + 9) + (y^2 - 12y + 36) = (x^2 + 6x + 9) + (y^2 - 8y + 16) \)
Combine terms:
\( x^2 + y^2 - 6x - 12y + 45 = x^2 + y^2 + 6x - 8y + 25 \)
Subtract \( x^2 \) and \( y^2 \) from both sides:
\( -6x - 12y + 45 = 6x - 8y + 25 \)
Move all terms to one side to find the relation:
\( 0 = 6x + 6x - 8y + 12y + 25 - 45 \)
\( 0 = 12x + 4y - 20 \)
Divide the entire equation by 4 to simplify:
\( 0 = 3x + y - 5 \)
So, the relation between x and y is \( 3x + y - 5 = 0 \).
In simple words: Set up an equation where the distance from (x, y) to the first point is equal to the distance from (x, y) to the second point. Square both sides to remove the square roots, then expand and simplify the equation. Arrange the terms to form a linear relationship between x and y.

Exam Tip: The relation found will always be a linear equation, representing the perpendicular bisector of the line segment connecting the two given points. Simplify the final equation by dividing by a common factor if possible.