OP Malhotra Class 9 Maths Solutions Chapter 20 Coordinates and Graphs of Simultaneous Linear Equations Exercise 20 (D)

Question 1. Find the distance between each of the following pairs of points :
(i) (0, 0) and (2, 3)
(ii) (-3, 0), (0, \( \sqrt{7} \))
(iii) (a, 0), (0, a)
(iv) (7, 9), (4, 5)
(v) (-6,-1), (-6, 11)
(vi) (a + b, a - b), (a - b,-a - b)
(vii) (2, -11), (-4, -3)
(viii) (a, b), (2a, b)
Answer: We use the distance formula to find the length between two points \( P(x_1, y_1) \) and \( Q(x_2, y_2) \), which is given by \( \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \).
(i) Distance between (0, 0) and (2, 3):
\( = \sqrt{(2-0)^2+(3-0)^2} \)
\( = \sqrt{2^2+3^2} \)
\( = \sqrt{4+9} \)
\( = \sqrt{13} \)
(ii) Distance between (-3, 0) and (0, \( \sqrt{7} \)):
\( = \sqrt{[0-(-3)]^2+(\sqrt{7}-0)^2} \)
\( = \sqrt{(3)^2+(\sqrt{7})^2} \)
\( = \sqrt{9+7} \)
\( = \sqrt{16} \)
\( = 4 \)
(iii) Distance between (a, 0) and (0, a):
\( = \sqrt{(0-a)^2+(a-0)^2} \)
\( = \sqrt{a^2+a^2} \)
\( = \sqrt{2a^2} \)
\( = a\sqrt{2} \)
(iv) Distance between (7, 9) and (4, 5):
\( = \sqrt{(4-7)^2+(5-9)^2} \)
\( = \sqrt{(-3)^2+(-4)^2} \)
\( = \sqrt{9+16} \)
\( = \sqrt{25} \)
\( = 5 \)
(v) Distance between (-6, -1) and (-6, 11):
\( = \sqrt{[-6-(-6)]^2+[11-(-1)]^2} \)
\( = \sqrt{(-6+6)^2+(11+1)^2} \)
\( = \sqrt{0^2+12^2} \)
\( = \sqrt{144} \)
\( = 12 \)
(vi) Distance between (a + b, a - b) and (a - b, -a - b):
\( = \sqrt{[(a-b)-(a+b)]^2+[(-a-b)-(a-b)]^2} \)
\( = \sqrt{(a-b-a-b)^2+(-a-b-a+b)^2} \)
\( = \sqrt{(-2b)^2+(-2a)^2} \)
\( = \sqrt{4b^2+4a^2} \)
\( = \sqrt{4(b^2+a^2)} \)
\( = 2\sqrt{a^2+b^2} \)
(vii) Distance between (2, -11) and (-4, -3):
\( = \sqrt{(-4-2)^2+[-3-(-11)]^2} \)
\( = \sqrt{(-6)^2+(-3+11)^2} \)
\( = \sqrt{(-6)^2+(8)^2} \)
\( = \sqrt{36+64} \)
\( = \sqrt{100} \)
\( = 10 \)
(viii) Distance between (a, b) and (2a, b):
\( = \sqrt{(2a-a)^2+(b-b)^2} \)
\( = \sqrt{(a)^2+(0)^2} \)
\( = \sqrt{a^2} \)
\( = a \)
In simple words: For each pair of points, we use the distance formula. This formula finds how far apart two points are by using their x and y coordinates. We subtract the coordinates, square the results, add them, and then take the square root.

🎯 Exam Tip: Remember to apply the distance formula correctly, especially when dealing with negative signs or square roots. Pay attention to simplifying the square root at the end of the calculation.

Question 2. The distance between two points (0, 0) and (x, 3) is 5. Find x.
Answer: We use the distance formula for two points \( (x_1, y_1) \) and \( (x_2, y_2) \), which is \( \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \).
Given points are (0, 0) and (x, 3), and the distance is 5.
\( \implies \sqrt{(x-0)^2+(3-0)^2} = 5 \)
\( \implies \sqrt{x^2+3^2} = 5 \)
\( \implies \sqrt{x^2+9} = 5 \)
To remove the square root, we square both sides of the equation:
\( \implies ( \sqrt{x^2+9} )^2 = 5^2 \)
\( \implies x^2+9 = 25 \)
Now, we solve for \( x^2 \):
\( \implies x^2 = 25-9 \)
\( \implies x^2 = 16 \)
To find \( x \), we take the square root of both sides:
\( \implies x = \pm\sqrt{16} \)
\( \implies x = \pm 4 \)
So, the possible values for \( x \) are 4 or -4.
In simple words: We are given the distance between two points, including one with an unknown 'x' coordinate. We set up the distance formula with the given values and then solve the equation for 'x'. Squaring both sides helps remove the square root to find 'x'.

🎯 Exam Tip: When taking the square root of a number to solve for x, always remember to include both the positive and negative solutions, as both will satisfy the squared equation.

Question 3. Find the radius of the circle whose
(i) centre is at (0, 0) and which passes through (-6, 8);
(ii) centre is at (2, 0) and which passes through (7, -12).
Answer: The radius of a circle is the distance from its center to any point on its circumference. We use the distance formula to find this length.
(i) Center (0, 0) and passing through (-6, 8):
Radius \( = \sqrt{(-6-0)^2+(8-0)^2} \)
\( = \sqrt{(-6)^2+(8)^2} \)
\( = \sqrt{36+64} \)
\( = \sqrt{100} \)
\( = 10 \) units
(ii) Center (2, 0) and passing through (7, -12):
Radius \( = \sqrt{(7-2)^2+(-12-0)^2} \)
\( = \sqrt{(5)^2+(-12)^2} \)
\( = \sqrt{25+144} \)
\( = \sqrt{169} \)
\( = 13 \) units
In simple words: The radius is simply the distance from the center of the circle to any point on its edge. We calculate this distance using the distance formula for each case.

🎯 Exam Tip: The radius calculation is a direct application of the distance formula. Be careful with signs when subtracting coordinates, especially with negative numbers.

Question 4. Find the lengths of the sides of the triangle whose vertices are A (3, 4), B (2, -1) and C (4, -6).
Answer: To find the lengths of the sides of the triangle, we calculate the distance between each pair of vertices using the distance formula \( \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \).
The vertices are A (3, 4), B (2, -1), and C (4, -6).
Length of side AB:
\( = \sqrt{(2-3)^2+(-1-4)^2} \)
\( = \sqrt{(-1)^2+(-5)^2} \)
\( = \sqrt{1+25} \)
\( = \sqrt{26} \)
Length of side BC:
\( = \sqrt{(4-2)^2+(-6-(-1))^2} \)
\( = \sqrt{(2)^2+(-6+1)^2} \)
\( = \sqrt{2^2+(-5)^2} \)
\( = \sqrt{4+25} \)
\( = \sqrt{29} \)
Length of side CA:
\( = \sqrt{(4-3)^2+(-6-4)^2} \)
\( = \sqrt{1^2+(-10)^2} \)
\( = \sqrt{1+100} \)
\( = \sqrt{101} \)
In simple words: We find the length of each side of the triangle by using the distance formula for the two points that make up that side. We do this three times for sides AB, BC, and CA.

🎯 Exam Tip: When calculating lengths, ensure you apply the distance formula for all three sides. Always double-check calculations involving negative numbers to avoid sign errors.

Question 5. Find the co-ordinates of the points on the x- axis which are at a distance of 10 units from the point (-4, 8).
Answer: A point on the x-axis always has a y-coordinate of 0. Let this unknown point be \( (x, 0) \). We are given that the distance between \( (x, 0) \) and \( (-4, 8) \) is 10 units. We use the distance formula to solve for \( x \).
\( \text{Distance} = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \)
\( \implies 10 = \sqrt{(-4-x)^2+(8-0)^2} \)
\( \implies 10 = \sqrt{(-4-x)^2+8^2} \)
\( \implies 10 = \sqrt{(-4-x)^2+64} \)
To eliminate the square root, we square both sides of the equation:
\( \implies 10^2 = (-4-x)^2+64 \)
\( \implies 100 = (x+4)^2+64 \) (Since \( (-4-x)^2 = (-(4+x))^2 = (4+x)^2 \))
\( \implies 100-64 = (x+4)^2 \)
\( \implies 36 = (x+4)^2 \)
Now, we take the square root of both sides to find \( x+4 \):
\( \implies \pm\sqrt{36} = x+4 \)
\( \implies \pm 6 = x+4 \)
This gives us two possible cases for \( x \):
Case 1: \( 6 = x+4 \)
\( \implies x = 6-4 \)
\( \implies x = 2 \)
Case 2: \( -6 = x+4 \)
\( \implies x = -6-4 \)
\( \implies x = -10 \)
So, the points on the x-axis are \( (-10, 0) \) and \( (2, 0) \).
In simple words: First, we imagine the unknown point is on the x-axis, so its y-part is zero. Then, we use the distance formula, setting the distance to 10. We square both sides to get rid of the square root and solve the equation for 'x', which gives two possible answers.

🎯 Exam Tip: When a problem asks for points on the x-axis or y-axis, remember that the other coordinate is zero (e.g., \( (x, 0) \) or \( (0, y) \)). Always remember to consider both positive and negative roots when solving a quadratic equation like \( x^2 = k \).

Question 6. What point on the y-axis is equidistant from P (0, 8) and Q (-4, 4)?
Answer: A point on the y-axis has an x-coordinate of 0. Let the required point be \( A(0, y) \). If point A is equidistant from P(0, 8) and Q(-4, 4), then the distance AP must be equal to the distance AQ.
We use the distance formula \( \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \).
Distance AP:
\( = \sqrt{(0-0)^2+(y-8)^2} \)
\( = \sqrt{0^2+(y-8)^2} \)
\( = \sqrt{(y-8)^2} \)
Distance AQ:
\( = \sqrt{(-4-0)^2+(4-y)^2} \)
\( = \sqrt{(-4)^2+(4-y)^2} \)
\( = \sqrt{16+(4-y)^2} \)
Since AP = AQ, we can write:
\( \implies \sqrt{(y-8)^2} = \sqrt{16+(4-y)^2} \)
To remove the square roots, we square both sides:
\( \implies (y-8)^2 = 16+(4-y)^2 \)
Now, expand both sides:
\( \implies y^2-16y+64 = 16+(16-8y+y^2) \)
\( \implies y^2-16y+64 = 16+16-8y+y^2 \)
\( \implies y^2-16y+64 = 32-8y+y^2 \)
Subtract \( y^2 \) from both sides and rearrange to solve for \( y \):
\( \implies -16y+64 = 32-8y \)
\( \implies 64-32 = 16y-8y \)
\( \implies 32 = 8y \)
\( \implies y = \frac{32}{8} \)
\( \implies y = 4 \)
So, the point on the y-axis equidistant from P and Q is \( (0, 4) \).
In simple words: We know the point is on the y-axis, so its x-value is 0. We then say the distance from this point to P is the same as its distance to Q. We use the distance formula for both, set them equal, and solve for the unknown y-value.

🎯 Exam Tip: When a point is equidistant from two other points, set up an equation where the square of the distance to the first point equals the square of the distance to the second point. This simplifies calculations by removing square roots early.

Question 7. A line is of length 10 and one end is at the point (-3, 2). If the ordinate of the other end be 10, prove that the abscissa will be 3 or -9.
Answer: Let the first end of the line be point A(-3, 2). The other end, point B, has an ordinate (y-coordinate) of 10. Let its abscissa (x-coordinate) be \( x \). So, point B is \( (x, 10) \). The length of the line AB is given as 10 units. We use the distance formula to find \( x \).
Length \( AB = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \)
\( \implies 10 = \sqrt{[x-(-3)]^2+(10-2)^2} \)
\( \implies 10 = \sqrt{(x+3)^2+8^2} \)
\( \implies 10 = \sqrt{(x+3)^2+64} \)
Square both sides to eliminate the square root:
\( \implies 10^2 = (x+3)^2+64 \)
\( \implies 100 = (x+3)^2+64 \)
Rearrange the equation to solve for \( (x+3)^2 \):
\( \implies 100-64 = (x+3)^2 \)
\( \implies 36 = (x+3)^2 \)
Take the square root of both sides to find \( x+3 \):
\( \implies \pm\sqrt{36} = x+3 \)
\( \implies \pm 6 = x+3 \)
This leads to two possible values for \( x \):
Case 1: \( 6 = x+3 \)
\( \implies x = 6-3 \)
\( \implies x = 3 \)
Case 2: \( -6 = x+3 \)
\( \implies x = -6-3 \)
\( \implies x = -9 \)
Thus, the abscissa of the other end will be 3 or -9. This proves the statement.
In simple words: We use the distance formula because we know the length of the line and one end point, plus the y-coordinate of the other end. We set up the equation, square both sides to remove the square root, and then solve for the unknown x-coordinate.

🎯 Exam Tip: Problems that "prove" a result often involve setting up an equation using the distance formula and carefully solving it. Ensure all algebraic steps, especially factoring or quadratic formula application, are accurate.

Question 8.
(i) Show that the points (-5, 1), (1, -1) and (1,-2) are collinear.
(ii) Find the value of p for which the points (-1, 3) (2, p) and (5, -1) are collinear.
Answer:
(i) To check if three points A, B, and C are collinear (lie on the same straight line), we find the distances between each pair of points. If the sum of the lengths of any two segments equals the length of the third segment, the points are collinear.
Let A = (-5, 1), B = (1, -1), and C = (1, -2).
Length of AB:
\( = \sqrt{(1-(-5))^2+(-1-1)^2} \)
\( = \sqrt{(1+5)^2+(-2)^2} \)
\( = \sqrt{6^2+(-2)^2} \)
\( = \sqrt{36+4} \)
\( = \sqrt{40} = 2\sqrt{10} \)
Length of BC:
\( = \sqrt{(1-1)^2+(-2-(-1))^2} \)
\( = \sqrt{0^2+(-2+1)^2} \)
\( = \sqrt{0^2+(-1)^2} \)
\( = \sqrt{1} = 1 \)
Length of CA:
\( = \sqrt{(-5-1)^2+(1-(-2))^2} \)
\( = \sqrt{(-6)^2+(1+2)^2} \)
\( = \sqrt{(-6)^2+3^2} \)
\( = \sqrt{36+9} \)
\( = \sqrt{45} = 3\sqrt{5} \)
Now, we check for collinearity:
\( AB+BC = 2\sqrt{10}+1 \approx 2 \times 3.16 + 1 = 6.32+1 = 7.32 \)
\( CA = 3\sqrt{5} \approx 3 \times 2.24 = 6.72 \)
Since \( AB+BC \neq CA \), the points A, B, and C are not collinear.
(ii) For points A(-1, 3), B(2, p), and C(5, -1) to be collinear, one distance must be the sum of the other two. Let's calculate the distances AB, BC, and AC.
Length of AB:
\( = \sqrt{(2-(-1))^2+(p-3)^2} \)
\( = \sqrt{3^2+(p-3)^2} \)
\( = \sqrt{9+p^2-6p+9} \)
\( = \sqrt{p^2-6p+18} \)
Length of BC:
\( = \sqrt{(5-2)^2+(-1-p)^2} \)
\( = \sqrt{3^2+(p+1)^2} \)
\( = \sqrt{9+p^2+2p+1} \)
\( = \sqrt{p^2+2p+10} \)
Length of AC:
\( = \sqrt{(5-(-1))^2+(-1-3)^2} \)
\( = \sqrt{6^2+(-4)^2} \)
\( = \sqrt{36+16} \)
\( = \sqrt{52} \)
If A, B, C are collinear, then \( AB+BC = AC \) (or another combination). Assuming \( B \) is between \( A \) and \( C \), we have \( AC = AB+BC \).
So, \( \sqrt{p^2+2p+10} = \sqrt{p^2-6p+18} + \sqrt{52} \)
Squaring both sides:
\( (p^2+2p+10) = (p^2-6p+18) + 52 + 2\sqrt{52}\sqrt{p^2-6p+18} \)
\( p^2+2p+10 = p^2-6p+70 + 2\sqrt{52}\sqrt{p^2-6p+18} \)
\( 2p+10+6p-70 = 2\sqrt{52}\sqrt{p^2-6p+18} \)
\( 8p-60 = 2\sqrt{52}\sqrt{p^2-6p+18} \)
Divide both sides by 2:
\( 4p-30 = \sqrt{52}\sqrt{p^2-6p+18} \)
Since \( \sqrt{52} = \sqrt{4 \times 13} = 2\sqrt{13} \):
\( 4p-30 = 2\sqrt{13}\sqrt{p^2-6p+18} \)
Divide both sides by 2:
\( 2p-15 = \sqrt{13}\sqrt{p^2-6p+18} \)
Squaring both sides again:
\( (2p-15)^2 = 13(p^2-6p+18) \)
\( 4p^2-60p+225 = 13p^2-78p+234 \)
Move all terms to one side to form a quadratic equation:
\( 0 = 13p^2-4p^2-78p+60p+234-225 \)
\( 0 = 9p^2-18p+9 \)
Divide the entire equation by 9:
\( 0 = p^2-2p+1 \)
This is a perfect square trinomial:
\( \implies (p-1)^2 = 0 \)
Therefore, \( p-1 = 0 \)
\( \implies p = 1 \)
In simple words: (i) To see if points are in a line, we measure the distance between each pair. If adding the two shorter distances gives the longest distance, they are in a line. Here, they were not. (ii) For the points to be in a line, we set up an equation where the sum of two distances equals the third. We solve this equation by squaring both sides to find the unknown 'p'.

🎯 Exam Tip: For collinearity, calculate all three pairwise distances. The sum of the two smaller distances must exactly equal the largest distance. If solving for an unknown, remember to simplify and square equations carefully to find the correct value.

Question 9. Find the value of x such that AB = BC, where the coordinates of A, B and C are (-5, 2), (1, -2) and (x, 4) respectively.
Answer: We are given three points A(-5, 2), B(1, -2), and C(x, 4). We are told that the distance AB is equal to the distance BC. We will use the distance formula \( \sqrt{(x_2-x_1)^2+(y_2-y_1)^2} \) to set up an equation and solve for \( x \).
First, calculate the length of AB:
\( AB = \sqrt{(1-(-5))^2+(-2-2)^2} \)
\( AB = \sqrt{(1+5)^2+(-4)^2} \)
\( AB = \sqrt{6^2+(-4)^2} \)
\( AB = \sqrt{36+16} \)
\( AB = \sqrt{52} \)
Next, calculate the length of BC:
\( BC = \sqrt{(x-1)^2+(4-(-2))^2} \)
\( BC = \sqrt{(x-1)^2+(4+2)^2} \)
\( BC = \sqrt{(x-1)^2+6^2} \)
\( BC = \sqrt{(x-1)^2+36} \)
Since AB = BC, we can set their expressions equal:
\( \implies \sqrt{52} = \sqrt{(x-1)^2+36} \)
Square both sides to eliminate the square roots:
\( \implies 52 = (x-1)^2+36 \)
Now, rearrange the equation to solve for \( (x-1)^2 \):
\( \implies 52-36 = (x-1)^2 \)
\( \implies 16 = (x-1)^2 \)
Take the square root of both sides:
\( \implies \pm\sqrt{16} = x-1 \)
\( \implies \pm 4 = x-1 \)
This gives two possible cases for \( x \):
Case 1: \( 4 = x-1 \)
\( \implies x = 4+1 \)
\( \implies x = 5 \)
Case 2: \( -4 = x-1 \)
\( \implies x = -4+1 \)
\( \implies x = -3 \)
Thus, the possible values for \( x \) are 5 or -3.
In simple words: We are given that two line segments, AB and BC, have the same length. We use the distance formula to write down equations for AB and BC. We set these two equations equal to each other and then solve for the unknown 'x' value.

🎯 Exam Tip: When setting two distances equal to each other, squaring both sides is usually the first step to simplify the equation. Remember to solve for both positive and negative roots in the final step of a quadratic equation.

Question 10. Prove that the triangles whose vertices are P (2, 0), Q (6, 0) and R (4, 4) is an isosceles triangles.
Answer: A triangle is considered isosceles if at least two of its sides have the same length. To prove that triangle PQR is isosceles, we need to calculate the lengths of its three sides using the distance formula.
The vertices are P (2, 0), Q (6, 0), and R (4, 4).
Length of side PQ:
\( = \sqrt{(6-2)^2+(0-0)^2} \)
\( = \sqrt{4^2+0^2} \)
\( = \sqrt{16+0} \)
\( = \sqrt{16} \)
\( = 4 \)
Length of side QR:
\( = \sqrt{(4-6)^2+(4-0)^2} \)
\( = \sqrt{(-2)^2+4^2} \)
\( = \sqrt{4+16} \)
\( = \sqrt{20} \)
Length of side RP:
\( = \sqrt{(4-2)^2+(4-0)^2} \)
\( = \sqrt{2^2+4^2} \)
\( = \sqrt{4+16} \)
\( = \sqrt{20} \)
We observe that the lengths of sides QR and RP are both \( \sqrt{20} \). Since two sides of the triangle (QR and RP) have equal lengths, triangle PQR is an isosceles triangle. This proves the statement.
In simple words: We find the length of all three sides of the triangle. If two of these sides turn out to be the same length, then the triangle is called an isosceles triangle. In this case, two sides were indeed equal.

🎯 Exam Tip: To prove a triangle is isosceles, calculate all three side lengths. Showing that any two sides are equal is sufficient. No need to check angle properties for this specific proof.

Question 11. Which of the triangles, having the following vertices, are right-angled triangles ?
(a) A (7, 0), B (6, 3) and C (12, 5)
(b) D (2, 0), E (5, 2) and F (1, 8)
(c) P (-4, 0), Q (-2, 5) and R (4, -1).
Answer: To determine if a triangle is a right-angled triangle, we use the Pythagorean theorem. This theorem states that in a right-angled triangle, the square of the longest side (hypotenuse) is equal to the sum of the squares of the other two sides ( \( a^2+b^2=c^2 \) ). We first calculate the length of each side for all triangles.
(a) Vertices A (7, 0), B (6, 3), and C (12, 5)
Length of AB:
\( = \sqrt{(6-7)^2+(3-0)^2} = \sqrt{(-1)^2+3^2} = \sqrt{1+9} = \sqrt{10} \)
Length of BC:
\( = \sqrt{(12-6)^2+(5-3)^2} = \sqrt{6^2+2^2} = \sqrt{36+4} = \sqrt{40} \)
Length of CA:
\( = \sqrt{(12-7)^2+(5-0)^2} = \sqrt{5^2+5^2} = \sqrt{25+25} = \sqrt{50} \)
Now, check the Pythagorean condition:
\( AB^2+BC^2 = (\sqrt{10})^2+(\sqrt{40})^2 = 10+40 = 50 \)
\( CA^2 = (\sqrt{50})^2 = 50 \)
Since \( AB^2+BC^2 = CA^2 \), triangle ABC is a right-angled triangle.
(b) Vertices D (2, 0), E (5, 2), and F (1, 8)
Length of DE:
\( = \sqrt{(5-2)^2+(2-0)^2} = \sqrt{3^2+2^2} = \sqrt{9+4} = \sqrt{13} \)
Length of EF:
\( = \sqrt{(1-5)^2+(8-2)^2} = \sqrt{(-4)^2+6^2} = \sqrt{16+36} = \sqrt{52} \)
Length of FD:
\( = \sqrt{(2-1)^2+(0-8)^2} = \sqrt{1^2+(-8)^2} = \sqrt{1+64} = \sqrt{65} \)
Now, check the Pythagorean condition:
\( DE^2+EF^2 = (\sqrt{13})^2+(\sqrt{52})^2 = 13+52 = 65 \)
\( FD^2 = (\sqrt{65})^2 = 65 \)
Since \( DE^2+EF^2 = FD^2 \), triangle DEF is a right-angled triangle.
(c) Vertices P (-4, 0), Q (-2, 5), and R (4, -1)
Length of PQ:
\( = \sqrt{(-2-(-4))^2+(5-0)^2} = \sqrt{2^2+5^2} = \sqrt{4+25} = \sqrt{29} \)
Length of QR:
\( = \sqrt{(4-(-2))^2+(-1-5)^2} = \sqrt{6^2+(-6)^2} = \sqrt{36+36} = \sqrt{72} \)
Length of RP:
\( = \sqrt{(-4-4)^2+(0-(-1))^2} = \sqrt{(-8)^2+1^2} = \sqrt{64+1} = \sqrt{65} \)
Now, check the Pythagorean condition:
\( PQ^2+RP^2 = (\sqrt{29})^2+(\sqrt{65})^2 = 29+65 = 94 \)
\( QR^2 = (\sqrt{72})^2 = 72 \)
Since \( PQ^2+RP^2 \neq QR^2 \), triangle PQR is not a right-angled triangle.
In simple words: To find out if a triangle is right-angled, we first calculate the length of all three sides. Then we check if the square of the longest side is equal to the sum of the squares of the other two sides. If it is, then it's a right-angled triangle.

🎯 Exam Tip: The key to identifying a right-angled triangle is to apply the converse of the Pythagorean theorem: calculate all three side lengths, square them, and check if the sum of the squares of the two shorter sides equals the square of the longest side.

Question 12.
(i) The coordinates of the points A, B, C are (0, 4),(2, 5) and (3, 3) respectively. Prove that it is an isosceles right angled triangle. Also find its area.
(ii) Show that A (3, 5), B (1, 1), C (5, 3) and D (7, 7) are the vertices of a rhombus, is it a square ? Find its area.
Answer:
(i) Vertices are A(0, 4), B(2, 5), C(3, 3). To prove it's an isosceles right-angled triangle, we need to show two sides are equal and it satisfies the Pythagorean theorem.
Length of AB:
\( = \sqrt{(2-0)^2+(5-4)^2} = \sqrt{2^2+1^2} = \sqrt{4+1} = \sqrt{5} \)
Length of BC:
\( = \sqrt{(3-2)^2+(3-5)^2} = \sqrt{1^2+(-2)^2} = \sqrt{1+4} = \sqrt{5} \)
Length of CA:
\( = \sqrt{(3-0)^2+(3-4)^2} = \sqrt{3^2+(-1)^2} = \sqrt{9+1} = \sqrt{10} \)
Since \( AB = BC = \sqrt{5} \), the triangle is isosceles.
Now, check for a right angle using the Pythagorean theorem:
\( AB^2+BC^2 = (\sqrt{5})^2+(\sqrt{5})^2 = 5+5 = 10 \)
\( CA^2 = (\sqrt{10})^2 = 10 \)
Since \( AB^2+BC^2 = CA^2 \), the triangle is a right-angled triangle at B.
Therefore, \( \triangle ABC \) is an isosceles right-angled triangle.
Area of \( \triangle ABC \):
For a right-angled triangle, Area \( = \frac{1}{2} \times \text{base} \times \text{height} \). Here, base = AB and height = BC.
Area \( = \frac{1}{2} \times AB \times BC = \frac{1}{2} \times \sqrt{5} \times \sqrt{5} = \frac{1}{2} \times 5 = 2.5 \) square units.
(ii) Vertices are A(3, 5), B(1, 1), C(5, 3), and D(7, 7). To show it's a rhombus, all four sides must be equal. To check if it's a square, its diagonals must also be equal.
Length of AB:
\( = \sqrt{(1-3)^2+(1-5)^2} = \sqrt{(-2)^2+(-4)^2} = \sqrt{4+16} = \sqrt{20} \)
Length of BC:
\( = \sqrt{(5-1)^2+(3-1)^2} = \sqrt{4^2+2^2} = \sqrt{16+4} = \sqrt{20} \)
Length of CD:
\( = \sqrt{(7-5)^2+(7-3)^2} = \sqrt{2^2+4^2} = \sqrt{4+16} = \sqrt{20} \)
Length of DA:
\( = \sqrt{(3-7)^2+(5-7)^2} = \sqrt{(-4)^2+(-2)^2} = \sqrt{16+4} = \sqrt{20} \)
Since all four sides \( AB=BC=CD=DA=\sqrt{20} \), the quadrilateral ABCD is a rhombus.
Now, check the lengths of the diagonals to see if it's a square:
Length of diagonal AC:
\( = \sqrt{(5-3)^2+(3-5)^2} = \sqrt{2^2+(-2)^2} = \sqrt{4+4} = \sqrt{8} \)
Length of diagonal BD:
\( = \sqrt{(7-1)^2+(7-1)^2} = \sqrt{6^2+6^2} = \sqrt{36+36} = \sqrt{72} \)
Since \( AC \neq BD \) (\( \sqrt{8} \neq \sqrt{72} \)), the diagonals are not equal.
Therefore, ABCD is a rhombus but not a square.
Area of rhombus ABCD:
Area \( = \frac{1}{2} \times d_1 \times d_2 \) (where \( d_1 \) and \( d_2 \) are the lengths of the diagonals)
Area \( = \frac{1}{2} \times \sqrt{8} \times \sqrt{72} = \frac{1}{2} \times \sqrt{8 \times 72} = \frac{1}{2} \times \sqrt{576} = \frac{1}{2} \times 24 = 12 \) square units.
In simple words: (i) We found the lengths of all sides of the triangle. Two sides were equal, making it isosceles. Then we checked if the square of the longest side matched the sum of the squares of the other two, confirming it's right-angled. We used the base and height (the two equal sides) to find the area. (ii) We calculated all four side lengths of the quadrilateral. Since all were equal, it's a rhombus. Next, we checked the lengths of its diagonals. Because the diagonals were not equal, it is not a square. Finally, we used the lengths of the diagonals to calculate the area.

🎯 Exam Tip: For quadrilateral proofs, calculating all four side lengths and both diagonal lengths is essential. Equal sides indicate a rhombus; equal diagonals in a rhombus indicate a square. Use the appropriate area formulas for each shape.

Question 13. Show that ABC is an equilateral triangle if A, B, C, have the following coordinates:
(i) A (1, \( \sqrt{3} \)), B (3, \( \sqrt{3} \)), C (2, \( 2\sqrt{3} \))
(ii) A (1, 1), B (-1, -1), C (- \( \sqrt{3} \), \( \sqrt{3} \)).
Answer: An equilateral triangle has all three sides of equal length. We will use the distance formula to calculate the length of each side for both sets of given coordinates.
(i) Vertices A (1, \( \sqrt{3} \)), B (3, \( \sqrt{3} \)), C (2, \( 2\sqrt{3} \))
Length of AB:
\( = \sqrt{(3-1)^2+(\sqrt{3}-\sqrt{3})^2} \)
\( = \sqrt{2^2+0^2} \)
\( = \sqrt{4} = 2 \)
Length of BC:
\( = \sqrt{(2-3)^2+(2\sqrt{3}-\sqrt{3})^2} \)
\( = \sqrt{(-1)^2+(\sqrt{3})^2} \)
\( = \sqrt{1+3} = \sqrt{4} = 2 \)
Length of CA:
\( = \sqrt{(2-1)^2+(2\sqrt{3}-\sqrt{3})^2} \)
\( = \sqrt{1^2+(\sqrt{3})^2} \)
\( = \sqrt{1+3} = \sqrt{4} = 2 \)
Since \( AB = BC = CA = 2 \), triangle ABC is an equilateral triangle.
(ii) Vertices A (1, 1), B (-1, -1), C (- \( \sqrt{3} \), \( \sqrt{3} \))
Length of AB:
\( = \sqrt{(-1-1)^2+(-1-1)^2} \)
\( = \sqrt{(-2)^2+(-2)^2} \)
\( = \sqrt{4+4} = \sqrt{8} \)
Length of BC:
\( = \sqrt{(-\sqrt{3}-(-1))^2+(\sqrt{3}-(-1))^2} \)
\( = \sqrt{(1-\sqrt{3})^2+(1+\sqrt{3})^2} \)
\( = \sqrt{(1-2\sqrt{3}+3)+(1+2\sqrt{3}+3)} \)
\( = \sqrt{4-2\sqrt{3}+4+2\sqrt{3}} \)
\( = \sqrt{8} \)
Length of CA:
\( = \sqrt{(1-(-\sqrt{3}))^2+(1-\sqrt{3})^2} \)
\( = \sqrt{(1+\sqrt{3})^2+(1-\sqrt{3})^2} \)
\( = \sqrt{(1+2\sqrt{3}+3)+(1-2\sqrt{3}+3)} \)
\( = \sqrt{4+2\sqrt{3}+4-2\sqrt{3}} \)
\( = \sqrt{8} \)
Since \( AB = BC = CA = \sqrt{8} \), triangle ABC is an equilateral triangle.
In simple words: To show a triangle is equilateral, we must check that all three of its sides have the exact same length. We use the distance formula three times, once for each side. If all calculated lengths are equal, then it is an equilateral triangle.

🎯 Exam Tip: When dealing with square roots in coordinates (like \( \sqrt{3} \)), be meticulous with algebraic expansion, especially when squaring terms like \( (a \pm \sqrt{b})^2 \). These often lead to terms that cancel out, simplifying the final result.

Question 14. Show that the points (-2, 6),(5, 3), (-1, -11) and (-8, -8) are the vertices of a rectangle.
Answer: To prove that the given points form a rectangle, we need to show two properties: first, that opposite sides are equal in length (which means it's a parallelogram), and second, that its diagonals are also equal in length. This extra condition differentiates a rectangle from other parallelograms.
Let the points be A(-2, 6), B(5, 3), C(-1, -11), and D(-8, -8).
First, calculate the lengths of all four sides:
Length of AB:
\( = \sqrt{(5-(-2))^2+(3-6)^2} \)
\( = \sqrt{(5+2)^2+(-3)^2} \)
\( = \sqrt{7^2+(-3)^2} \)
\( = \sqrt{49+9} \)
\( = \sqrt{58} \)
Length of BC:
\( = \sqrt{(-1-5)^2+(-11-3)^2} \)
\( = \sqrt{(-6)^2+(-14)^2} \)
\( = \sqrt{36+196} \)
\( = \sqrt{232} \)
Length of CD:
\( = \sqrt{(-8-(-1))^2+(-8-(-11))^2} \)
\( = \sqrt{(-8+1)^2+(-8+11)^2} \)
\( = \sqrt{(-7)^2+3^2} \)
\( = \sqrt{49+9} \)
\( = \sqrt{58} \)
Length of DA:
\( = \sqrt{(-2-(-8))^2+(6-(-8))^2} \)
\( = \sqrt{(-2+8)^2+(6+8)^2} \)
\( = \sqrt{6^2+14^2} \)
\( = \sqrt{36+196} \)
\( = \sqrt{232} \)
We observe that \( AB = CD = \sqrt{58} \) and \( BC = DA = \sqrt{232} \). Since opposite sides are equal, ABCD is a parallelogram.
Next, calculate the lengths of the diagonals:
Length of diagonal AC:
\( = \sqrt{(-1-(-2))^2+(-11-6)^2} \)
\( = \sqrt{(-1+2)^2+(-17)^2} \)
\( = \sqrt{1^2+(-17)^2} \)
\( = \sqrt{1+289} \)
\( = \sqrt{290} \)
Length of diagonal BD:
\( = \sqrt{(-8-5)^2+(-8-3)^2} \)
\( = \sqrt{(-13)^2+(-11)^2} \)
\( = \sqrt{169+121} \)
\( = \sqrt{290} \)
Since the diagonals \( AC = BD = \sqrt{290} \), and it is a parallelogram, ABCD is a rectangle.
In simple words: To show that the points form a rectangle, we first calculate the length of all four sides. If opposite sides are equal, it's a parallelogram. Then, we calculate the lengths of the two diagonals. If the diagonals are also equal, then it is a rectangle.

🎯 Exam Tip: For coordinate geometry proofs of quadrilaterals, always compute all four side lengths and both diagonal lengths. This methodical approach helps distinguish between different types of parallelograms and ensures all conditions are met for the specific shape requested.

Question 14. Show that the points (-2, 6),(5, 3), (-1, -11) and (-8, -8) are the vertices of a rectangle.
Answer: Let the given points be A(-2, 6), B(5, 3), C(-1, -11) and D(-8, -8). We find the distance between each pair of points to determine the lengths of the sides: Length of AB \( = \sqrt{[5-(-2)]^2+(3-6)^2} = \sqrt{(7)^2+(-3)^2} = \sqrt{49+9} = \sqrt{58} \) Length of BC \( = \sqrt{(-1-5)^2+(-11-3)^2} = \sqrt{(-6)^2+(-14)^2} = \sqrt{36+196} = \sqrt{232} \) Length of CD \( = \sqrt{[-8-(-1)]^2+[-8-(-11)]^2} = \sqrt{(-7)^2+(3)^2} = \sqrt{49+9} = \sqrt{58} \) Length of DA \( = \sqrt{[-2-(-8)]^2+[6-(-8)]^2} = \sqrt{(6)^2+(14)^2} = \sqrt{36+196} = \sqrt{232} \) We can see that AB = CD and BC = DA. This means the opposite sides are equal in length, which confirms that ABCD is a parallelogram. Now, we find the lengths of the diagonals to check if it's a rectangle: Length of diagonal AC \( = \sqrt{[-1-(-2)]^2+(-11-6)^2} = \sqrt{(1)^2+(-17)^2} = \sqrt{1+289} = \sqrt{290} \) Length of diagonal BD \( = \sqrt{[-8-5]^2+[-8-3]^2} = \sqrt{(-13)^2+(-11)^2} = \sqrt{169+121} = \sqrt{290} \) Since the diagonals AC and BD are equal in length, the parallelogram ABCD is indeed a rectangle. A parallelogram with equal diagonals always forms a rectangle.In simple words: First, we find the length of all four sides of the shape. We see that the opposite sides are equal, which tells us it's a parallelogram. Then, we find the length of the diagonal lines inside the shape. Because these diagonal lines are also equal in length, the shape must be a rectangle.

🎯 Exam Tip: To prove a quadrilateral is a rectangle, first show it's a parallelogram (opposite sides equal), and then show either that its diagonals are equal or that one angle is 90 degrees (using Pythagorean theorem for adjacent sides).

Question 15. The centre of a circle is at the origin and its radius is 10. Tell whether the following points lie (i) on (ii) inside or (iii) outside the circle.
(i) (6, 8)
(ii) (0, 11)
(ii) (-10, 0)
(iv) (7, 7)
(v) (-9, 4)
Answer: The circle has its center at the origin O(0, 0) and a radius (R) of 10 units. We need to find the distance (d) of each given point from the origin and compare it with the radius (R). - If \( d = R \), the point lies on the circle. - If \( d < R \), the point lies inside the circle. - If \( d > R \), the point lies outside the circle. (i) For point (6, 8): Distance \( d = \sqrt{(6-0)^2+(8-0)^2} = \sqrt{6^2+8^2} = \sqrt{36+64} = \sqrt{100} = 10 \) units. Since \( d = 10 \) and \( R = 10 \), the distance is equal to the radius, so this point lies on the circle. (ii) For point (0, 11): Distance \( d = \sqrt{(0-0)^2+(11-0)^2} = \sqrt{0^2+11^2} = \sqrt{0+121} = \sqrt{121} = 11 \) units. Since \( d = 11 \) and \( R = 10 \), the distance is greater than the radius, so this point lies outside the circle. (iii) For point (-10, 0): Distance \( d = \sqrt{(-10-0)^2+(0-0)^2} = \sqrt{(-10)^2+0^2} = \sqrt{100+0} = \sqrt{100} = 10 \) units. Since \( d = 10 \) and \( R = 10 \), the distance is equal to the radius, so this point lies on the circle. (iv) For point (7, 7): Distance \( d = \sqrt{(7-0)^2+(7-0)^2} = \sqrt{7^2+7^2} = \sqrt{49+49} = \sqrt{98} \) units. Since \( \sqrt{98} \approx 9.9 \) and \( R = 10 \), the distance is less than the radius, so this point lies inside the circle. (v) For point (-9, 4): Distance \( d = \sqrt{(-9-0)^2+(4-0)^2} = \sqrt{(-9)^2+4^2} = \sqrt{81+16} = \sqrt{97} \) units. Since \( \sqrt{97} \approx 9.85 \) and \( R = 10 \), the distance is less than the radius, so this point lies inside the circle.In simple words: To check where a point is, we find how far it is from the center of the circle. If this distance is the same as the circle's radius, the point is on the circle. If it's shorter, the point is inside. If it's longer, it's outside. This is a basic use of the distance formula.

🎯 Exam Tip: Remember the three conditions: distance = radius (on circle), distance < radius (inside circle), distance > radius (outside circle). Always compare the calculated distance to the given radius clearly.

Question 16. The point P (2, -5) is mapped onto point P' on reflection in the x-axis and Q(3, 7) is mapped onto the point Q' on reflection in the origin. Find the length PQ and P' Q'.
Answer: First, we find the coordinates of the reflected points. When point P(2, -5) is reflected in the x-axis, its x-coordinate stays the same and its y-coordinate changes its sign. So, P' becomes (2, 5). When point Q(3, 7) is reflected in the origin, both x and y coordinates change their signs. So, Q' becomes (-3, -7). Next, we calculate the length of the line segment PQ using the distance formula for P(2, -5) and Q(3, 7): Length of PQ \( = \sqrt{(3-2)^2+[7-(-5)]^2} \) \( = \sqrt{1^2+(7+5)^2} \) \( = \sqrt{1^2+12^2} \) \( = \sqrt{1+144} \) \( = \sqrt{145} \) units. Finally, we calculate the length of the line segment P'Q' using the distance formula for P'(2, 5) and Q'(-3, -7): Length of P'Q' \( = \sqrt{(-3-2)^2+(-7-5)^2} \) \( = \sqrt{(-5)^2+(-12)^2} \) \( = \sqrt{25+144} \) \( = \sqrt{169} \) \( = 13 \) units.In simple words: We find the new locations of the points after they are reflected (flipped). For reflection in the x-axis, only the y-part changes sign. For reflection in the origin, both x and y parts change signs. Then, we use the distance formula to find the gap between the original points and the gap between the new, reflected points.

🎯 Exam Tip: Remember the rules for reflections: for x-axis, \( (x, y) \to (x, -y) \); for y-axis, \( (x, y) \to (-x, y) \); for origin, \( (x, y) \to (-x, -y) \). Always apply these correctly before using the distance formula.

Question 17. P and Q have coordinates (4, 1) and (2, 0). Find
(i) the image P' of P under reflection in the y- axis.
(ii) the image Q' of Q under reflection in the line PP'.
(iii) the length of P'Q'.
Answer: The coordinates of point P are (4, 1) and Q are (2, 0). (i) To find the image P' of P(4, 1) under reflection in the y-axis: When a point \( (x, y) \) is reflected in the y-axis, its new coordinates are \( (-x, y) \). So, P(4, 1) becomes P'(-4, 1). (ii) To find the image Q' of Q(2, 0) under reflection in the line PP': First, we need to find the equation of the line PP'. P is (4, 1) and P' is (-4, 1). Since both points P and P' have the same y-coordinate (1), the line joining them is a horizontal line with the equation \( y = 1 \). When a point \( (x, y) \) is reflected in the horizontal line \( y = k \), its new coordinates are \( (x, 2k-y) \). Here, Q is (2, 0) and the line of reflection is \( y = 1 \), so \( k=1 \). Therefore, Q' becomes \( (2, 2 \times 1 - 0) = (2, 2) \). (iii) To find the length of P'Q': P' is (-4, 1) and Q' is (2, 2). Length of P'Q' \( = \sqrt{[2-(-4)]^2+(2-1)^2} \) \( = \sqrt{(2+4)^2+1^2} \) \( = \sqrt{6^2+1^2} \) \( = \sqrt{36+1} \) \( = \sqrt{37} \) units.In simple words: First, we find the new spot for point P when it mirrors across the y-axis (P'). Then, we find the line that connects P and P'. This line (y=1) becomes the new mirror for point Q. After finding Q' (the mirrored Q), we use the distance formula to measure the length between P' and Q'.

🎯 Exam Tip: When reflecting in a line like \( y=k \) or \( x=k \), use the formulas \( (x, 2k-y) \) for \( y=k \) or \( (2k-x, y) \) for \( x=k \). Pay close attention to identifying the correct mirror line first.

Question 18. Point A(5, 1) is the centre of a circle with radius 13 units. AB is perpendicular to the chord PQ. The coordinates of B are (2, -3). Calculate the length of (i) AB (ii) PB (iii) PQ.
Answer: We are given the center of the circle A(5, 1) and its radius, which is 13 units. We are also given point B(2, -3) on the chord PQ, such that AB is perpendicular to PQ.
(i) To find the length of AB: We use the distance formula for the center A(5, 1) and point B(2, -3): Length of AB \( = \sqrt{(2-5)^2+(-3-1)^2} \) \( = \sqrt{(-3)^2+(-4)^2} \) \( = \sqrt{9+16} \) \( = \sqrt{25} \) \( = 5 \) units. (ii) To find the length of PB: Since AB is perpendicular to the chord PQ, triangle APB is a right-angled triangle, with the right angle at B. AP is the radius of the circle, so AP = 13 units. We already found AB = 5 units. Using the Pythagorean theorem in right-angled \( \triangle APB \): \( AP^2 = AB^2 + PB^2 \) \( 13^2 = 5^2 + PB^2 \) \( 169 = 25 + PB^2 \) \( PB^2 = 169 - 25 \) \( PB^2 = 144 \) \( PB = \sqrt{144} \) \( PB = 12 \) units. (iii) To find the length of PQ: A line drawn from the center of a circle perpendicular to a chord bisects the chord. This means that point B is the midpoint of the chord PQ. Therefore, the length of PQ is twice the length of PB: PQ \( = 2 \times PB \) PQ \( = 2 \times 12 \) PQ \( = 24 \) units.In simple words: First, we use the distance formula to find the length of the line segment AB. Then, because AB is perpendicular to the chord, it forms a right-angled triangle with the radius (AP). We use the Pythagoras theorem to find the length of PB. Finally, since the center line cuts the chord exactly in half, the total length of the chord PQ is simply double the length of PB.

🎯 Exam Tip: Remember the key circle theorems: a perpendicular from the center to a chord bisects the chord. This creates a right-angled triangle (center-endpoint of chord-midpoint of chord) where you can apply the Pythagorean theorem.