CBSE Class 10 Mathematics Coordinate Geometry Assignment Set 19

Question. Prove that the points \((-2,5)\), \((0,1)\) and \((2,-3)\) are collinear.
Answer: Let the given points be \(A(-2,5)\), \(B(0,1)\), and \(C(2,-3)\).
Using the distance formula, \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\):
\(AB = \sqrt{(0 - (-2))^2 + (1 - 5)^2} = \sqrt{2^2 + (-4)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}\)
\(BC = \sqrt{(2 - 0)^2 + (-3 - 1)^2} = \sqrt{2^2 + (-4)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5}\)
\(AC = \sqrt{(2 - (-2))^2 + (-3 - 5)^2} = \sqrt{4^2 + (-8)^2} = \sqrt{16 + 64} = \sqrt{80} = 4\sqrt{5}\)
Since \(AB + BC = 2\sqrt{5} + 2\sqrt{5} = 4\sqrt{5} = AC\), the points are collinear.

Question. Show that the points \(A(0,0)\), \(B(0,5)\), \(C(6,5)\), \(D(6,0)\) form a rectangle.
Answer: Let us find the lengths of the sides using the distance formula:
\(AB = \sqrt{(0 - 0)^2 + (5 - 0)^2} = \sqrt{25} = 5\)
\(BC = \sqrt{(6 - 0)^2 + (5 - 5)^2} = \sqrt{36} = 6\)
\(CD = \sqrt{(6 - 6)^2 + (0 - 5)^2} = \sqrt{25} = 5\)
\(DA = \sqrt{(0 - 6)^2 + (0 - 0)^2} = \sqrt{36} = 6\)
Since opposite sides are equal, i.e., \(AB = CD = 5\) and \(BC = DA = 6\), \(ABCD\) is a parallelogram.
Now, let us find the lengths of the diagonals:
\(AC = \sqrt{(6 - 0)^2 + (5 - 0)^2} = \sqrt{36 + 25} = \sqrt{61}\)
\(BD = \sqrt{(6 - 0)^2 + (0 - 5)^2} = \sqrt{36 + 25} = \sqrt{61}\)
Since the diagonals are also equal (\(AC = BD = \sqrt{61}\)), \(ABCD\) is a rectangle.

Question. Show that the points \(A(3,2)\), \(B(0,5)\), \(C(-3,2)\) and \(D(0,-1)\) are the vertices of a square.
Answer: Let us find the lengths of the sides using the distance formula:
\(AB = \sqrt{(0 - 3)^2 + (5 - 2)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}\)
\(BC = \sqrt{(-3 - 0)^2 + (2 - 5)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}\)
\(CD = \sqrt{(0 - (-3))^2 + (-1 - 2)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}\)
\(DA = \sqrt{(3 - 0)^2 + (2 - (-1))^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}\)
Since all four sides are equal (\(AB = BC = CD = DA = 3\sqrt{2}\)), \(ABCD\) is a rhombus.
Now, let us find the lengths of the diagonals:
\(AC = \sqrt{(-3 - 3)^2 + (2 - 2)^2} = \sqrt{(-6)^2 + 0} = \sqrt{36} = 6\)
\(BD = \sqrt{(0 - 0)^2 + (-1 - 5)^2} = \sqrt{0 + (-6)^2} = \sqrt{36} = 6\)
Since the diagonals are also equal (\(AC = BD = 6\)), \(ABCD\) is a square.

Question. Show that the points \(A(2,1)\), \(B(5,4)\), \(C(4,7)\) and \(D(1,4)\) are the vertices of a parallelogram.
Answer: Let us find the lengths of the sides using the distance formula:
\(AB = \sqrt{(5 - 2)^2 + (4 - 1)^2} = \sqrt{3^2 + 3^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}\)
\(BC = \sqrt{(4 - 5)^2 + (7 - 4)^2} = \sqrt{(-1)^2 + 3^2} = \sqrt{1 + 9} = \sqrt{10}\)
\(CD = \sqrt{(1 - 4)^2 + (4 - 7)^2} = \sqrt{(-3)^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}\)
\(DA = \sqrt{(2 - 1)^2 + (1 - 4)^2} = \sqrt{1^2 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10}\)
Since the opposite sides are equal (\(AB = CD = 3\sqrt{2}\) and \(BC = DA = \sqrt{10}\)), the points \(A\), \(B\), \(C\), and \(D\) are the vertices of a parallelogram.

Question. Prove that the points \(A(3,0)\); \(B(6,4)\) and \(C(-1,3)\) are the vertices of a right angled triangle. Are these vertices of an isosceles triangle also?
Answer: Let us find the squares of the lengths of the sides using the distance formula:
\(AB^2 = (6 - 3)^2 + (4 - 0)^2 = 3^2 + 4^2 = 9 + 16 = 25\)
\(BC^2 = (-1 - 6)^2 + (3 - 4)^2 = (-7)^2 + (-1)^2 = 49 + 1 = 50\)
\(CA^2 = (3 - (-1))^2 + (0 - 3)^2 = 4^2 + (-3)^2 = 16 + 9 = 25\)
Here, \(AB^2 + CA^2 = 25 + 25 = 50 = BC^2\).
Since the sum of the squares of two sides is equal to the square of the third side, by the converse of Pythagoras' theorem, \(\triangle ABC\) is a right-angled triangle (right-angled at \(A\)).
Also, since \(AB^2 = CA^2 \implies AB = CA = 5\), two sides of the triangle are equal.
Thus, these are also the vertices of an isosceles triangle.

Question. Show that the following points do not form any triangle.
\(A(4,2)\); \(B(7,5)\); \(C(9,7)\)
Answer: Let us find the area of \(\triangle ABC\) using the area formula:
\(\text{Area} = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|\)
\(\text{Area} = \frac{1}{2} |4(5 - 7) + 7(7 - 2) + 9(2 - 5)|\)
\(\text{Area} = \frac{1}{2} |4(-2) + 7(5) + 9(-3)|\)
\(\text{Area} = \frac{1}{2} |-8 + 35 - 27| = \frac{1}{2} |35 - 35| = 0\)
Since the area of the triangle is \(0\), the points \(A\), \(B\), and \(C\) are collinear and do not form any triangle.

Question. Find the Co-ordinates of the point on the axis of \(y\) which is equidistant from the points \((-1,2)\) and \((3,4)\)
Answer: Let the required point on the y-axis be \(P(0, y)\).
According to the question, \(P\) is equidistant from \(A(-1, 2)\) and \(B(3, 4)\).
\(PA = PB \implies PA^2 = PB^2\)
\((0 - (-1))^2 + (y - 2)^2 = (0 - 3)^2 + (y - 4)^2\)
\(1^2 + y^2 - 4y + 4 = (-3)^2 + y^2 - 8y + 16\)
\(y^2 - 4y + 5 = y^2 - 8y + 25\)
\(-4y + 8y = 25 - 5\)
\(4y = 20 \implies y = 5\)
Thus, the coordinates of the point are \((0,5)\).

Question. The distance \(PQ\) between the points \(P(5,-9)\) and \(Q(11,y)\) is 10 units. Find the value of \(y\).
Answer: Given that the distance \(PQ = 10\) units.
By the distance formula:
\(PQ^2 = 10^2\)
\((11 - 5)^2 + (y - (-9))^2 = 100\)
\(6^2 + (y + 9)^2 = 100\)
\(36 + (y + 9)^2 = 100\)
\((y + 9)^2 = 64\)
\(y + 9 = \pm 8\)
If \(y + 9 = 8 \implies y = -1\)
If \(y + 9 = -8 \implies y = -17\)
Thus, the values of \(y\) are \(-17\) or \(-1\).

Question. If two vertices of an equilateral triangle are \(A(0,0)\); \(B(3,0)\), find the third vertex.
Answer: Let the third vertex be \(C(x, y)\).
Since \(\triangle ABC\) is equilateral, \(AC = BC = AB\).
Using the distance formula, \(AB = \sqrt{(3 - 0)^2 + (0 - 0)^2} = 3\).
Since \(AC = 3 \implies AC^2 = 9\):
\(x^2 + y^2 = 9\) — (1)
Since \(BC = 3 \implies BC^2 = 9\):
\((x - 3)^2 + (y - 0)^2 = 9\)
\(x^2 - 6x + 9 + y^2 = 9\)
\(x^2 - 6x + y^2 = 0\) — (2)
Subtracting (2) from (1):
\(6x = 9 \implies x = \frac{3}{2}\)
Substituting \(x = \frac{3}{2}\) into (1):
\((\frac{3}{2})^2 + y^2 = 9\)
\(\frac{9}{4} + y^2 = 9 \implies y^2 = 9 - \frac{9}{4} = \frac{27}{4}\)
\(y = \pm \frac{3\sqrt{3}}{2}\)
Thus, the third vertex is \((\frac{3}{2}, \frac{3\sqrt{3}}{2})\) or \((\frac{3}{2}, -\frac{3\sqrt{3}}{2})\).

Question. Find the point which is equidistant from the points \(A(6,6)\); \(B(-1,7)\); \(C(5,7)\)
Answer: Let the required point be \(P(x, y)\).
Since \(P\) is equidistant from \(A\), \(B\), and \(C\), we have \(PA^2 = PB^2 = PC^2\).
First, equate \(PB^2 = PC^2\):
\((x - (-1))^2 + (y - 7)^2 = (x - 5)^2 + (y - 7)^2\)
\((x + 1)^2 = (x - 5)^2\)
\(x^2 + 2x + 1 = x^2 - 10x + 25\)
\(12x = 24 \implies x = 2\)
Next, equate \(PA^2 = PC^2\):
\((x - 6)^2 + (y - 6)^2 = (x - 5)^2 + (y - 7)^2\)
Substituting \(x = 2\):
\((2 - 6)^2 + (y - 6)^2 = (2 - 5)^2 + (y - 7)^2\)
\((-4)^2 + y^2 - 12y + 36 = (-3)^2 + y^2 - 14y + 49\)
\(16 + y^2 - 12y + 36 = 9 + y^2 - 14y + 49\)
\(52 - 12y = 58 - 14y\)
\(2y = 6 \implies y = 3\)
Thus, the coordinates of the required point are \((2,3)\).