CBSE Class 10 Mathematics Coordinate Geometry Assignment Set 20

Question. Find the distance between the following two points:
(i) \( A(1, 2) \) ; \( B(3, 4) \)
(ii) \( (a+b, a-b) \) ; \( B(b-a, b+a) \)
Answer:
(i) Using the distance formula, \( d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \):
\( AB = \sqrt{(3 - 1)^2 + (4 - 2)^2} = \sqrt{2^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \)

(ii) Let the first point be \( P(a+b, a-b) \). Using the distance formula:
\( PB = \sqrt{[(b-a) - (a+b)]^2 + [(b+a) - (a-b)]^2} \)
\( PB = \sqrt{(b-a-a-b)^2 + (b+a-a+b)^2} \)
\( PB = \sqrt{(-2a)^2 + (2b)^2} \)
\( PB = \sqrt{4a^2 + 4b^2} = 2\sqrt{a^2+b^2} \)

Question. Find the value of \( p \) for which the distance between the points \( A(3, -5) \) and \( B(p, 2) \) is \( \sqrt{58} \) units.
Answer:
Using the distance formula:
\( AB = \sqrt{(p - 3)^2 + (2 - (-5))^2} = \sqrt{58} \)
Squaring both sides:
\( (p - 3)^2 + (7)^2 = 58 \)
\( (p - 3)^2 + 49 = 58 \)
\( (p - 3)^2 = 58 - 49 \)
\( (p - 3)^2 = 9 \)
Taking square root on both sides:
\( p - 3 = \pm 3 \)
\( p = 3 \pm 3 \)
So, \( p = 6 \) or \( p = 0 \).
(As per the provided answer key, \( p = 6 \))

Question. Prove that the following points are the vertices of a right angled triangle.
\( A(3, 3) \) ; \( B(9, 0) \) and \( C(12, 21) \)
Answer:
Let us find the squares of the lengths of the sides using the distance formula:
\( AB^2 = (9 - 3)^2 + (0 - 3)^2 = 6^2 + (-3)^2 = 36 + 9 = 45 \)
\( BC^2 = (12 - 9)^2 + (21 - 0)^2 = 3^2 + 21^2 = 9 + 441 = 450 \)
\( AC^2 = (12 - 3)^2 + (21 - 3)^2 = 9^2 + 18^2 = 81 + 324 = 405 \)

We observe that:
\( AB^2 + AC^2 = 45 + 405 = 450 = BC^2 \)
Since the sum of the squares of two sides is equal to the square of the third side, by the converse of Pythagoras' theorem, the triangle \( ABC \) is a right-angled triangle (with the right angle at vertex \( A \)).

Question. Verify \( AB + AC = BC \) where \( A(1, 1) \) ; \( B(-2, 7) \) ; \( C(3, -3) \)
Answer:
Let us calculate the distances \( AB \), \( AC \), and \( BC \) using the distance formula:
\( AB = \sqrt{(-2 - 1)^2 + (7 - 1)^2} = \sqrt{(-3)^2 + 6^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5} \)
\( AC = \sqrt{(3 - 1)^2 + (-3 - 1)^2} = \sqrt{2^2 + (-4)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5} \)
\( BC = \sqrt{(3 - (-2))^2 + (-3 - 7)^2} = \sqrt{5^2 + (-10)^2} = \sqrt{25 + 100} = \sqrt{125} = 5\sqrt{5} \)

Now, checking the given relation:
\( AB + AC = 3\sqrt{5} + 2\sqrt{5} = 5\sqrt{5} = BC \)
Hence, \( AB + AC = BC \) is verified.

Question. Show that the points \( A(a, b+c) \), \( B(b, c+a) \) ; \( C(c, a+b) \) lie on a straight line.
Answer:
Three points are collinear (lie on a straight line) if the area of the triangle formed by them is zero.
The area of a triangle with vertices \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) is given by:
\( \text{Area} = \frac{1}{2} | x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) | \)

Substituting the coordinates \( A(a, b+c) \), \( B(b, c+a) \), and \( C(c, a+b) \):
\( \text{Area} = \frac{1}{2} | a[(c+a) - (a+b)] + b[(a+b) - (b+c)] + c[(b+c) - (c+a)] | \)
\( \text{Area} = \frac{1}{2} | a(c - b) + b(a - c) + c(b - a) | \)
\( \text{Area} = \frac{1}{2} | ac - ab + ab - bc + bc - ac | \)
\( \text{Area} = \frac{1}{2} | 0 | = 0 \)

Since the area of the triangle is \( 0 \), the points \( A \), \( B \), and \( C \) are collinear and lie on a straight line.

Question. Prove that the points \( A(1, 1) \), \( B(4, 4) \), \( C(4, 8) \) and \( D(1, 5) \) are the vertices of a parallelogram.
Answer:
A quadrilateral is a parallelogram if its opposite sides are equal in length.
Let us find the lengths of the sides \( AB \), \( BC \), \( CD \), and \( DA \) using the distance formula:
\( AB = \sqrt{(4 - 1)^2 + (4 - 1)^2} = \sqrt{3^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2} \)
\( CD = \sqrt{(1 - 4)^2 + (5 - 8)^2} = \sqrt{(-3)^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2} \)
Here, \( AB = CD = 3\sqrt{2} \).

\( BC = \sqrt{(4 - 4)^2 + (8 - 4)^2} = \sqrt{0^2 + 4^2} = \sqrt{16} = 4 \)
\( DA = \sqrt{(1 - 1)^2 + (1 - 5)^2} = \sqrt{0^2 + (-4)^2} = \sqrt{16} = 4 \)
Here, \( BC = DA = 4 \).

Since the opposite sides of the quadrilateral \( ABCD \) are equal (\( AB = CD \) and \( BC = DA \)), the points \( A \), \( B \), \( C \), and \( D \) are the vertices of a parallelogram.

Question. Prove that the points \( A(0, -1) \) , \( B(2, 1) \) and \( C(-2, 1) \) are the vertices of a Square.
Answer:
Note: A square is a four-sided polygon and thus requires four vertices. The three given points \( A(0, -1) \), \( B(2, 1) \), and \( C(-2, 1) \) represent three of the vertices. Let us find the fourth vertex \( D(x, y) \) and prove that they form a square.

Let \( A(0, -1) \) be one vertex, and \( B(2, 1) \) and \( C(-2, 1) \) be two adjacent or opposite vertices.
Using the distance formula:
\( AB = \sqrt{(2 - 0)^2 + (1 - (-1))^2} = \sqrt{4 + 4} = \sqrt{8} \)
\( AC = \sqrt{(-2 - 0)^2 + (1 - (-1))^2} = \sqrt{4 + 4} = \sqrt{8} \)
\( BC = \sqrt{(-2 - 2)^2 + (1 - 1)^2} = \sqrt{(-4)^2 + 0} = 4 \)

Since \( AB = AC = \sqrt{8} \) and \( BC = 4 = \sqrt{16} \), we have:
\( AB^2 + AC^2 = 8 + 8 = 16 = BC^2 \)
Thus, \( \triangle ABC \) is a right-angled isosceles triangle at \( A \), meaning \( AB \perp AC \) and \( AB = AC \).
In a square \( ABDC \), \( BC \) serves as the diagonal, and \( A \) is one vertex. The fourth vertex \( D(x,y) \) must lie opposite to \( A \).
By symmetry or using the midpoint of the diagonal \( BC \) (which must be the same as the midpoint of \( AD \)):
Midpoint of \( BC = \left(\frac{2 + (-2)}{2}, \frac{1 + 1}{2}\right) = (0, 1) \)
Let \( D(x, y) \) be the fourth vertex. Then the midpoint of \( AD \) is:
\( \left(\frac{0 + x}{2}, \frac{-1 + y}{2}\right) = (0, 1) \)
Solving for \( x \) and \( y \):
\( \frac{x}{2} = 0 \Rightarrow x = 0 \)
\( \frac{y - 1}{2} = 1 \Rightarrow y - 1 = 2 \Rightarrow y = 3 \)

So, the fourth vertex is \( D(0, 3) \).
With the four vertices \( A(0, -1) \), \( B(2, 1) \), \( D(0, 3) \), and \( C(-2, 1) \):
- All four sides are equal: \( AB = BD = DC = CA = \sqrt{8} \)
- Both diagonals are equal: \( BC = AD = 4 \)
Hence, these points form the vertices of a square.

Question. Show that the points \( A(2, 2) \), \( B(-2, 4) \), \( C(2, 6) \) form an isosceles triangle.
Answer:
A triangle is isosceles if any two of its sides are equal in length.
Using the distance formula, let us calculate the lengths of the three sides:
\( AB = \sqrt{(-2 - 2)^2 + (4 - 2)^2} = \sqrt{(-4)^2 + 2^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5} \)
\( BC = \sqrt{(2 - (-2))^2 + (6 - 4)^2} = \sqrt{4^2 + 2^2} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5} \)
\( AC = \sqrt{(2 - 2)^2 + (6 - 2)^2} = \sqrt{0^2 + 4^2} = \sqrt{16} = 4 \)

Since \( AB = BC = 2\sqrt{5} \), the two sides are equal in length. Hence, the points \( A \), \( B \), and \( C \) form an isosceles triangle.

Question. Show that the points \( A(-2, 2) \), \( B(8, -2) \) and \( C(-4, 3) \) are the vertices of a right angled triangle.
Answer:
Let us first check the distances using the given coordinates \( A(-2, 2) \), \( B(8, -2) \), and \( C(-4, 3) \):
\( AB^2 = [8 - (-2)]^2 + [-2 - 2]^2 = 10^2 + (-4)^2 = 100 + 16 = 116 \)
\( BC^2 = [-4 - 8]^2 + [3 - (-2)]^2 = (-12)^2 + 5^2 = 144 + 25 = 169 \)
\( AC^2 = [-4 - (-2)]^2 + (3 - 2)^2 = (-2)^2 + 1^2 = 4 + 1 = 5 \)

Since \( AB^2 + AC^2 = 116 + 5 = 121 \neq BC^2 \), the triangle with the given coordinates is not right-angled.

Note: There is a sign typo in the given coordinate of \( C \). If we use the corrected point \( C(-4, -3) \), then:
\( AC^2 = [-4 - (-2)]^2 + (-3 - 2)^2 = (-2)^2 + (-5)^2 = 4 + 25 = 29 \)
\( BC^2 = [-4 - 8]^2 + [-3 - (-2)]^2 = (-12)^2 + (-1)^2 = 144 + 1 = 145 \)
Now, checking the Pythagoras theorem with these corrected values:
\( AB^2 + AC^2 = 116 + 29 = 145 = BC^2 \)
Since \( AB^2 + AC^2 = BC^2 \), the triangle \( ABC \) with the corrected vertex \( C(-4, -3) \) is a right-angled triangle at vertex \( A \).

Question. Show that the points \( A(0, 5) \); \( B(-2, -2) \); \( C(5, 0) \); \( D(7, 7) \) are the vertices of a rhombus.
Answer:
A quadrilateral is a rhombus if all four of its sides are equal in length, while its diagonals are unequal.
Let us calculate the lengths of the sides using the distance formula:
\( AB = \sqrt{(-2 - 0)^2 + (-2 - 5)^2} = \sqrt{(-2)^2 + (-7)^2} = \sqrt{4 + 49} = \sqrt{53} \)
\( BC = \sqrt{(5 - (-2))^2 + (0 - (-2))^2} = \sqrt{7^2 + 2^2} = \sqrt{49 + 4} = \sqrt{53} \)
\( CD = \sqrt{(7 - 5)^2 + (7 - 0)^2} = \sqrt{2^2 + 7^2} = \sqrt{4 + 49} = \sqrt{53} \)
\( DA = \sqrt{(0 - 7)^2 + (5 - 7)^2} = \sqrt{(-7)^2 + (-2)^2} = \sqrt{49 + 4} = \sqrt{53} \)

Since \( AB = BC = CD = DA = \sqrt{53} \), all four sides of the quadrilateral are equal.
Next, let us check the lengths of the diagonals \( AC \) and \( BD \):
\( AC = \sqrt{(5 - 0)^2 + (0 - 5)^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2} \)
\( BD = \sqrt{(7 - (-2))^2 + (7 - (-2))^2} = \sqrt{9^2 + 9^2} = \sqrt{162} = 9\sqrt{2} \)

Since the four sides are equal and the diagonals are unequal (\( AC \neq BD \)), the given points form the vertices of a rhombus.