Note: If the points are $(x_1, y_1)$ and $(x_2, y_2)$, then the distance between them is equal to $\sqrt{(x_2-x_1)^2 + (y_2-x_1)^2 }$

Question 1: Find the distance between the following pair of points:

(i) $(-6,7)$ and $(-1, -5)$

(ii) $(a +b, b + c)$ and $(a -b, c -b)$

(iii) $(a \sin \alpha, - b \cos \alpha)$ and $(-a \cos \alpha, b \sin \alpha)$

(iv) $(a,0)$ and $(0,b)$

(i) Distance between the given points $(-6,7)$ and $(-1, -5)$

$= \sqrt{(-1+6)^2 + (-5-7)^2 }$

$\sqrt{5^2+12^2} = 13$

(i) Distance between the given points $(a +b, b + c)$ and $(a -b, c -b)$

$= \sqrt{(a-b-a-b)^2 + (c-b-b-c)^2 }$

$= \sqrt{(-2b)^2+(-2b)^2}$

$= \sqrt{4b^2 + 4b^2}$

$= 2\sqrt{2}b$

(i) Distance between the given points $(a \sin \alpha, - b \cos \alpha)$ and $(-a \cos \alpha, b \sin \alpha)$

$= \sqrt{(-a \cos \alpha - a \sin \alpha)^2 + (b \sin \alpha + b \cos \alpha}$

$= \sqrt{a^2(\cos \alpha + \sin \alpha)^2 + b^2 (\cos \alpha + \sin \alpha)^2}$

$= \sqrt{a^2 + b^2} (\cos \alpha + \sin \alpha)$

(i) Distance between the given points $(a,0)$ and $(0,b)$

$= \sqrt{(0-a)^2 + (b-0)^2 } = \sqrt{a^2+b^2} = 13$

$\\$

Question 2: Find the value of $a$ when the distance between the points $(3, a)$ and $(4,1)$ is $\sqrt{10}$.

Given: $\sqrt{10} = \sqrt{(4-3)^2 + (1-a)^2 }$

$\Rightarrow \sqrt{10} = \sqrt{1 + 1 + a^2-2a }$

Squaring both sides

$\Rightarrow 10 = 2 + a^2-2a$

$\Rightarrow a^2 - 2a -8 = 0$

$\Rightarrow (a-4)(a+2) = 0$

$\Rightarrow a = -2, 4$

$\\$

Question 3: If the points $(2, 1)$ and $(1,-2)$ are equidistant from the point $(x, y)$, show that $x+3y=0$.

Given: $\sqrt{(2-x)^2 + (1-y)^2 } = \sqrt{(1-x)^2 + (-2-y)^2 }$

$\Rightarrow 4 + x^2 - 4x + 1 + y^2 - 2y = 1 + x^2 - 2x + 4 + y^2+4y$

$\Rightarrow 5 - 4x - 2y = 5 - 2x + 4y$

$\Rightarrow -2x - 6y = 0$

$\Rightarrow x + 3y = 0$

$\\$

Question 4: Find the values of $x, y$ if the distances of the point $(x, y)$ from $(-3, 0)$ as well as from $(3,0)$ are $4$.

Given: $\sqrt{(-3-x)^2 + (0-y)^2 } = 4$

$9 + x^2 + 6x + y^2 = 16$

$x^2 + y^2 +6x = 7$ … … … … … i)

$\sqrt{(3-x)^2 + (0-y)^2 } = 4$

$9 + x^2 - 6x + y^2 = 1$

$x^2 + y^2 -6x = 7$ … … … … … ii)

From i) and ii) we get

$x^2 + y^2 +6x = x^2 + y^2 -6x$

$\Rightarrow 12 x = 0 \Rightarrow x = 0$

Therefore from i), $y^2 = 16-9 \Rightarrow y = \pm \sqrt{7}$

$\\$

Question 5: The length of a line segment is of $10$ units and the coordinates of one end-point are $(2,-3)$.If the abscissa of the other end is $10$, find the ordinate of the other end.

Given: $\sqrt{(10-2)^2 + (y+3)^2 } = 10$

$\Rightarrow 64 + y^2 + 9 + 6y = 100$

$\Rightarrow y^2 + 6y - 27 = 0$

$\Rightarrow (y+9)(y-3) = 0$

$\Rightarrow y = -9, 3$

$\\$

Question 6: Show that the points $(-4,-1), (-2,-4),(4,0)$ and $(2,3)$ are the vertices points of a rectangle.

Let the points be $A(-4,-1), B(-2,-4), C(4,0)$ and $D(2,3)$

Therefore sides of the figure

$AB = \sqrt{(-2+4)^2 + (-4+1)^2 }= \sqrt{4+9} = \sqrt{13}$

$BC = \sqrt{(4+2)^2 + (0+4)^2 }= \sqrt{36+16} = \sqrt{52}$

$CD = \sqrt{(4-2)^2 + (0-3)^2 }= \sqrt{4+9} = \sqrt{13}$

$DA = \sqrt{(2+4)^2 + (3+1)^2 }= \sqrt{36+16} = \sqrt{52}$

Diagonals of the figure

$AC= \sqrt{(4+4)^2 + (0+1)^2 }= \sqrt{64+1} = \sqrt{65}$

$BD = \sqrt{(2+2)^2 + (3+4)^2 }= \sqrt{16+49} = \sqrt{65}$

Hence, $AB = CD, BC = DA, AC=BD$.Therefore the figure is a rectangle since the opposite sides are equal and the diagonals are also equal.

$\\$

Question 7: Show that the points $A (1,- 2), B (3, 6), C (5, 10)$ and $D (3, 2)$ are the vertices of a parallelogram.

The points are $A (1,- 2), B (3, 6), C (5, 10)$ and $D (3, 2)$

Therefore sides of the figure

$AB = \sqrt{(3-1)^2 + (6+2)^2 }= \sqrt{4+64} = \sqrt{68}$

$BC = \sqrt{(5-3)^2 + (10-6)^2 }= \sqrt{4+16} = \sqrt{20}$

$CD = \sqrt{(5-3)^2 + (10-2)^2 }= \sqrt{4+64} = \sqrt{68}$

$DA = \sqrt{(3-1)^2 + (2+2)^2 }= \sqrt{4+16} = \sqrt{20}$

Diagonals of the figure

$AC= \sqrt{(5-1)^2 + (10+2)^2 }= \sqrt{16+144} = \sqrt{160}$

$BD = \sqrt{(3-3)^2 + (2-6)^2 }= \sqrt{0+16} = \sqrt{16}$

Hence, $AB = CD, BC = DA, AC \neq BD$.Therefore the figure is a parallelogram since the opposite sides are equal but the diagonals are unequal.

$\\$

Question 8: Prove that the points $A(1,7),B(4,2),C(-1,-1)$ and $D (4,4)$ are the vertices of a square.

The points are $A(1,7),B(4,2),C(-1,-1)$ and $D (4,4)$

Therefore sides of the figure

$AB = \sqrt{(4-1)^2 + (2-7)^2 }= \sqrt{9+25} = \sqrt{34}$

$BC = \sqrt{(-1-4)^2 + (-1-2)^2 }= \sqrt{25+9} = \sqrt{34}$

$CD = \sqrt{(-1+4)^2 + (-1-4)^2 }= \sqrt{9+25} = \sqrt{34}$

$DA = \sqrt{(-4-1)^2 + (4-7)^2 }= \sqrt{25+9} = \sqrt{34}$

Diagonals of the figure

$AC= \sqrt{(-1-1)^2 + (-1-7)^2 }= \sqrt{4+64} = \sqrt{68}$

$BD = \sqrt{(-4-4)^2 + (4-2)^2 }= \sqrt{64+4} = \sqrt{68}$

Hence, $AB = CD = BC = DA, AC = BD$.Therefore the figure is a square since the opposite sides are equal and also the diagonals are equal.

$\\$

Question 9: Prove that the points $(3, 0), (6,4)$ and $(- 1, 3)$ are vertices of a right-angled isosceles triangle.

Let the points be $A(3, 0), B(6,4)$ and $C(- 1, 3)$

Therefore sides of the figure

$AB = \sqrt{(6-3)^2 + (4-0)^2 }= \sqrt{9+16} = 5$

$BC = \sqrt{(-1-6)^2 + (3-4)^2 }= \sqrt{49+1} = 5\sqrt{2}$

$AC = \sqrt{(-1-3)^2 + (3-0)^2 }= \sqrt{16+9} = 5$

We see that $AC^2 + AB^2 = BC^2$

Therefore $\triangle ABC$ is a right angled triangle at $A$. Also $AB = AC$ therefore triangle is also an isosceles triangle.

$\\$

Question 10: Prove that $(2, -2), (-2, 1 )$ and $(5, 2)$ are the vertices of a right angled triangle. Find the area of the triangle and the length of the hypotenuse.

Let the points be $A(2, -2), B(-2, 1 )$ and $C(5, 2)$

Therefore sides of the figure

$AB = \sqrt{(-2-2)^2 + (1+2)^2 }= \sqrt{16+9} = 5$

$BC = \sqrt{(5+2)^2 + (2-1)^2 }= \sqrt{49+1} = \sqrt{50} = 5\sqrt{2}$

$AC = \sqrt{(5-2)^2 + (2+2)^2 }= \sqrt{9+16} = 5$

We see that $AB^2 + CA^2 = BC^2$

Therefore $\triangle ABC$ is a right angled triangle at $A$.

Area $=$ $\frac{1}{2}$ $\times AC \times AB =$ $\frac{1}{2}$ $\times 5 \times 5 = 12.5$ sq. units

Hypotenuse $= 5\sqrt{2}$

$\\$

Question 11: Prove that the points $(2a,4a),(2a,6a)$ and $(2a+\sqrt{3}a, 5a)$ are the vertices of an equilateral triangle.

Let the points be $A(2a,4a), B(2a,6a)$ and $C (2a+\sqrt{3}a, 5a)$

Therefore sides of the figure

$AB = \sqrt{(2a-2a)^2 + (6a-4a)^2 }= \sqrt{0+4a^2} = 2a$

$BC = \sqrt{(2a+\sqrt{3}a - 2a)^2 + (5a-6a)^2 }= \sqrt{3a^2+a^2} = 2a$

$AC = \sqrt{(2a-2a-\sqrt{3}a)^2 + (4a-5a)^2 }= \sqrt{3a^2+a^2} = 2a$

We see that $AB=BC=CA$

Therefore $\triangle ABC$ is an equilateral triangle.

$\\$

Question 12: Prove that the points $(2, 3), (-4, -6)$ and $(1, 3 / 2)$ do not form a triangle.

Let the points be $A(2, 3), B(-4, -6)$ and $C(1, 3 / 2)$

Therefore sides of the figure

$AB = \sqrt{(-4-2)^2 + (-6-3)^2 }= \sqrt{36+81} = \sqrt{117} = 10.817$

$BC = \sqrt{(1+4)^2 + (1.5+6)^2 }= \sqrt{25+56.25} =\sqrt{81.25} = 9.014$

$AC = \sqrt{(2-1)^2 + (3-1.5)^2 }= \sqrt{1+2.25} = \sqrt{3.25} = 1.803$

In a triangle, the sum of the length of to sides should be greater then the third side.

$AB + BC = 10.817 + 9.014 > 1.803 (CA) \rightarrow TRUE$

$BC + CA = 9.014 + 1.803 = 10.817 (AB) \rightarrow FALSE$

$CA + AB = 1.803 + 10.817 > 9.014 (BC) \rightarrow TRUE$

Because all three conditions are not TRUE, this is not a triangle.

$\\$

Question 13: An equilateral triangle has two vertices at the points $(3,4)$ and $(-2,3)$, find the coordinates of the third vertex.

Given vertices: $A(3, 4)$ and $B ( -2, 3)$

Let the third vertices be $O(x, y)$

$AB^2 = ( 3 + 2)^2 + ( 4-3)^2 = 25 +1 = 26$

$OA^2 = ( x+2)^2 + ( y - 3)^2$

$\Rightarrow OA^2 = x^2 + 4 + 4x + y^2 + 9 - 6y$

$\Rightarrow OA^2 = x^2 + y^2 + 4x - 6y + 13$

$OB^2 = ( x - 3)^2 + ( y - 4)^2$

$\Rightarrow OB^2 = x^2 + 9 - 6x + y^2 + 16 - 8 y$

$\Rightarrow OB^2 = x^2 + y^2 -6x -8y + 25$

$OA^2 = OB^2$

$\Rightarrow x^2 + y^2 + 4x -6y + 13 = x^2 + y^2 -6x -8y + 25$

$\Rightarrow 4x -6y + 13 = -6x-8y+25$

$\Rightarrow 10x + 2y = 12$

$\Rightarrow 5x + y = 16$

$\Rightarrow y = ( 6-5x)$

Also $OA^2 = AB^2$

$\Rightarrow x^2 + y^2 + 4x - 6y + 13 = 26$

$\Rightarrow x^2 + y^2+ 4x - 6y - 13 = 0$

Substituting

$x^2 + ( 6-5x)^2+ 4x - 6( 6-5x) - 13 = 0$

$\Rightarrow x^2 + 36 + 25x^2 - 60x + 4x - 36 + 30x - 13 = 0$

$\Rightarrow 26c^2 - 26x - 13 = 0$

$\Rightarrow 2x^2 - 2x - 1= 0$

$\Rightarrow x =$ $\frac{1 \pm \sqrt{3}}{2}$

Therefore $x =$ $\frac{1 + \sqrt{3}}{2}$ or $x =$ $\frac{1 - \sqrt{3}}{2}$

If $x =$ $\frac{1 + \sqrt{3}}{2}$ $\Rightarrow y = 6 - 5 ($ $\frac{1 + \sqrt{3}}{2}$ $) =$ $\frac{7 -5\sqrt{3}}{2}$

If $x =$ $\frac{1 - \sqrt{3}}{2}$ $\Rightarrow y = 6 - 5 ($ $\frac{1 - \sqrt{3}}{2}$ $) =$ $\frac{7 + 5\sqrt{3}}{2}$

Therefore the coordinate of the third vertices would be $\Big( \frac{1 + \sqrt{3}}{2}, \frac{7 -5\sqrt{3}}{2} \Big), \Big( \frac{1 + \sqrt{3}}{2}, \frac{7 +5\sqrt{3}}{2} \Big)$

$\\$

Question 14: Show that the quadrilateral whose vertices are $(2, -1), (3, 4), (-2,3)$ and $(-3, -2)$ is a rhombus.

Let the points be $A(2,-1), B(3,4), C(-2,-3)$ and $D(-3,-2)$

Therefore sides of the figure

$AB = \sqrt{(3-2)^2 + (4+1)^2 }= \sqrt{1+25} = \sqrt{26}$

$BC = \sqrt{(-2-3)^2 + (3-4)^2 }= \sqrt{25+1} = \sqrt{26}$

$CD = \sqrt{(-3+2)^2 + (-2+1)^2 }= \sqrt{1+25} = \sqrt{26}$

$DA = \sqrt{(-3-2)^2 + (-2+1)^2 }= \sqrt{25+1} = \sqrt{26}$

Diagonals of the figure

$AC= \sqrt{(-2-2)^2 + (3+1)^2 }= \sqrt{16+16} = \sqrt{32}$

$BD = \sqrt{(-3-3)^2 + (-2-4)^2 }= \sqrt{81+36} = \sqrt{117}$

Hence, $AB = BC = CD = CA$. Therefore we see that the sides are equal but the diagonals are not equal. Hence $ABCD$  is a Rhombus.

$\\$

Question 15: Two vertices of an isosceles triangle are $(2,0)$ and $(2,5)$. Find the third vertex if the length of the equal sides is $3$.

Given $A (2, 0), B ( 2, 5)$

Length of equal sides $= 3$

Let the third vertices be $( x, y)$

$AB = \sqrt{(2-2)^2 + (5-0)^2 }= \sqrt{25} = 5$

Therefore equal sides are $AC$ and $BC$

$AC^2 = (x-2)^2 + ( y-0)^2 = x^2 + 4 - 4x + y^2 = 9$

$\Rightarrow x^2 + y^2 - 4x = 5$   … … … … … i)

$BC^2 = (x-2)^2 + ( y - 5)^2 = x^2 + 4 - 4x + y^2 + 25 - 10y = 9$

$\Rightarrow x^2 + y^2 - 4x - 10 y = -20$   … … … … … ii)

Subtracting i) from ii)

$\Rightarrow -10y = -25$

$\Rightarrow y = 2.5$

Therefore from i)

$x^2 + ( 2.5)^2 - 4x = 5$

$\Rightarrow x^2 - 4x + 6.25 = 5$

$\Rightarrow x^2 - 4x + 1.25 = 0$

$\Rightarrow x =$ $\frac{4 \pm \sqrt{11}}{2}$

Therefore the coordinates are $\Big( \frac{4 + \sqrt{11}}{2}, \frac{5}{2} \Big) , \Big( \frac{4 - \sqrt{11}}{2}, \frac{5}{2} \Big)$

$\\$

Question 16: Which point on x-axis is equidistant from $(5, 9)$ and $(- 4,6)$?

Given $(5, 9)$ and $( -4, 6)$

Let $( x, 0)$ be the equidistant point

Therefore $(x-5)^2 + ( 0-9)^2 = ( x+4)^2 + ( 0-6)^2$

$\Rightarrow x^2 + 25 - 10 x + 81 = x^2 + 16 + 8x + 36$

$\Rightarrow 106 - 10x = 52 + 8x$

$\Rightarrow 54 = 18x$

$\Rightarrow x = 3$

Therefore the point is $(3, 0)$

$\\$

Question 17: Prove that the points $(- 2,5), (0, 1)$ and $(2,-3)$ are collinear.

Given $A ( -2, 5), B(0, 1), C ( 2, -3)$

$AB = \sqrt{(-2-0)^2 + (5-1)^2 }= \sqrt{4+16} = \sqrt{20} = 2\sqrt{5}$

$BC = \sqrt{(0-2)^2 + (1+3)^2 }= \sqrt{4+16} = \sqrt{20} = 2\sqrt{5}$

$AC = \sqrt{(2+2)^2 + (-3-5)^2 }= \sqrt{16+64} = \sqrt{80} = 4\sqrt{5}$

Therefore $AB + BC = AC$

Hence the points are collinear.

$\\$

Question 18: The coordinates of the point $P$ are $(-3,2)$. Find the coordinates of the point $Q$ which lies on the line joining $P$ and origin such that $OP = OQ$.

Given $P(-3, 2)$. Let $Q$ be $(x, y)$

$OP = OQ$

$\sqrt{(0+3)^2 + (0-2)^2 }= \sqrt{(0-x)^2 + (0-y)^2 }$

$9+4 = x^2 + y^2$

$\Rightarrow x^2 + y^2 = 13$

Now Slope of $OP =$ Slope of $OQ$

$\Rightarrow$ $\frac{0-2}{0+3}$ $=$ $\frac{0-y}{0-x}$

$\Rightarrow -$ $\frac{2}{3}$ $=$ $\frac{y}{x}$

$\Rightarrow y = -$ $\frac{2}{3}$ $x$

Substituting

$x^2 + (-$ $\frac{2}{3}$ $x)^2 = 13$

$\Rightarrow x^2 +$ $\frac{4}{9}$ $x^2 = 13$

$\Rightarrow x^2 = 9$

$\Rightarrow x = \pm 3$

If $x = 3, y = -2$

If $x = -3, y = 2$

Since $Q$ needs to be in the same line as $OP$, $Q$ is $(3, -2)$

$\\$

Question 19: Which point on y-axis is equidistant from $(2, 3)$ and $(4, 1)$?

Given points $A(2, 3), B(-4, 1)$

Let point $C$ be $(0, y)$

$AC = CB$

$\Rightarrow ( 0-2)^2 + (y-3)^2 = (0+4)^2 + ( y-1)^2$

$4 + y^2 + 9 - 6y = 16 + y^2 + 1 - 2y$

$\Rightarrow 13 - 6y = 17 - 2$

$\Rightarrow -4-4y = 0$

$\Rightarrow y = -1$

Hence the point is $(0, -1)$

$\\$

Question 20: The three vertices of a parallelogram are $(3, 4), (3,8)$ and $(9, 8)$. Find the fourth vertex.

Given points $A( 3, 4), B ( 3, 8), C (9, 8)$

Let $D$ be $(x, y)$

Since $ABCD$ is a parallelogram

$AB = CD$

$\Rightarrow ( 3-3)^2 + (8-4)^2 = ( 9-x)^2 + ( 8-y)^2$

$\Rightarrow 16 = 81 + x^2 -18x + 64 + y^2 - 16 y$

$\Rightarrow x^2 + y^2 - 18x - 16y + 129 = 0$  … … … … … i)

Also $AD = BC$

$\Rightarrow (3-x)^2 + ( 4 - y)^2 = ( 3-9)^2 + ( 8-8)^2$

$\Rightarrow 9 + x^2 - 6x + 16 + y^2 -8y = 36+0$

$\Rightarrow x^2 + y^2 -6x - 8y - 11 = 0$  … … … … … ii)

Slope of $AB =$ Slope of $CD$

$\frac{8-4}{3-3}$ $=$ $\frac{8-y}{9-x}$

$\Rightarrow (9-x) \times 4 = 0$

$\Rightarrow x = 9$

Hence $9^2 + y^2 -6(9) - 8y- 11=0$

$\Rightarrow y^2 - 8y + 16 = 0$

$\Rightarrow (y - 4)^2 = 0$

$\Rightarrow y = 4$

Hence $D(9, 4)$

$\\$

Question 21: Find the circumcenter of the triangle whose vertices are $(-2,-3),(-1,0), (7,-6)$.

Given $A(-2, -3), B(-1, 0), C(7, -6)$

Let $O(x, y)$ be the circumcenter. Therefore

$OA^2 = (x+2)^2 + ( y+3)^2$

$OB^2 = (x+1)^2 + ( y-0)^2$

$OC^2 = (x-7)^2 + (y+6)^2$

$OA^2 = OB^2$

$\Rightarrow (x+2)^2 + ( y+3)^2 = (x+1)^2 + ( y-0)^2$

$\Rightarrow x^2 + 4 + 4x + y^2 + 9 + 6y = x^2 + 1 + 2x + y^2$

$\Rightarrow 4x + 6y + 13 = 2x+1$

$\Rightarrow 2x+ 6y = -12$

$\Rightarrow x = -3y -6$  … … … … … i)

$OB^2 = OC^2$

$\Rightarrow (x+1)^2 + ( y-0)^2 = (x-7)^2 + (y+6)^2$

$\Rightarrow x^2 + 1 + 2x + y^2 = x^2 + 49 - 14x + y^2 + 36 + 12y$

$\Rightarrow 2x+1 = 85 - 14x + 12y$

$\Rightarrow 16x -12y = 84$

$\Rightarrow 4x-3y=21$   … … … … … ii)

Substituting i) in ii)

$4(-3y-6) - 3y = 21$

$\Rightarrow -12y -24 -3y = 21$

$\Rightarrow -15 y = 45$

$\Rightarrow y = -3$

Hence $x = -3(-3) - 6 =9-6=3$

Therefore circumcenter $O$ is $( 3 -3)$

$\\$

Question 22: Find the angle subtended at the origin by the line segment whose end points are $(0, 100)$ and $(10,0)$.

Given points $A(0, 100)$ and $B(10, 0)$

Point $A$ is on y axis and point $B$ is on x axis.

Therefore angle subtended by $A$ and $B$ on origin is $90^o$

$\\$

Question 23: Find the center of the circle passing through $(5, -8), (2,- 9)$ and $(2, 1)$.

Given $A(5, -8), B(2, -9), C(2, 1)$

Let center be $O(x, y)$. Therefore

$OA^2 = (x-5)^2 + ( y+8)^2$

$OB^2 = (x-2)^2 + ( y+9)^2$

$OC^2 = (x-2)^2 + (y-1)^2$

$OA^2 = OB^2$

$\Rightarrow (x-5)^2 + ( y+8)^2 = (x-2)^2 + ( y+9)^2$

$\Rightarrow x^2 + 25 - 10x +y^2 + 64 + 16y = x^2 + 4 - 4x + y^2 + 81 + 18y$

$\Rightarrow 89-10x+16y = 85-4x+18y$

$\Rightarrow 6x+2y = 4$

$\Rightarrow 3x+y= 2$   … … … … … i)

$OB^2 = OC^2$

$\Rightarrow (x-2)^2 + ( y+9)^2 = (x-2)^2 + (y-1)^2$

$\Rightarrow y^2 + 81 + 18y = y^2 + 1 - 2y$

$\Rightarrow 20y = -80$

$\Rightarrow y = -4$

Substituting in i) we get

$3x = 2 - (-4) = 6$

$\Rightarrow x = 2$

Hence the center of the circle is $O(2, 4)$

$\\$

Question 24: Find the value of $k$, if the point $P (0,2)$ is equidistant from $(3, k)$ and $(k,5)$.

Given $P(0, 2), A(3, k), B(k, 5)$

Since $PA = PB$

$\Rightarrow (3-0)^2 + (k-2)^2 = (k-0)^2 + (5-2)^2$

$\Rightarrow 9+k^2 + 4 - 4k = k^2 + 9$

$\Rightarrow 4-4k=0$

$\Rightarrow k = 1$

$\\$

Question 25: If two opposite vertices of a square arc $(5, 4)$ and $(1, -6)$, find the coordinates of its remaining two vertices.

Given $A(5,4 ), C(1, -6)$.

Let $B$ be $(x, y)$

Since $ABCD$ is a square, $AB = BC$

$(x-5)^2 + (y-4)^2 = (x-1)^2 + (y+6)^2$

$\Rightarrow x^2 + 25 -10x + y^2 + 16 -8y = x^2 + 1 - 2x + y^2 +36 + 12y$

$\Rightarrow 8x+ 20y = 4$

$\Rightarrow 2x+5y = 1$

$\Rightarrow y =$ $\frac{1-2x}{5}$   … … … … … i)

$AB^2 + BC^2 = AC^2$

$\Rightarrow x-5)^2 + ( y-4)^2 + ( x-1)^2 + (y+6)^2 = (1-5)^2 + ( -6 - 4)^2$

$\Rightarrow x^2 + 26 - 10x + y^2 + 16 - 8y + x^2 + 1 - 2x + y^2 + 36+ 12y = 16 + 100$

$\Rightarrow 2x^2 + 2y^2 + 78 - 12x + 4y = 116$

$\Rightarrow 2x^2 + 2y^2 - 12x + 4y = 38$

$\Rightarrow x^2 + y^2 -6x + 2y = 19$   … … … … … ii)

Substituting i) into ii) we get

$x^2 + \Big($ $\frac{1-2x}{5}$ $\Big)^2 -6x + 2 \Big($ $\frac{1-2x}{5}$ $\Big) = 19$

$\Rightarrow 25x^2 + 1 + 4x^2 - 4x - 150x + 10 - 20x = 475$

$\Rightarrow 29x^2 - 174x = 464$

$\Rightarrow x^2 - 6x = 16$

$\Rightarrow x^2 - 6x - 16 = 0$

$\Rightarrow (x-8)(x+2) = 0$

$\Rightarrow x = 8 \ or \ x = -2$

If $x = 8$  then $y = -3$

If $x = -2$, then $y = 1$

Hence the other two corners of the square are $(8, -3)$ and $( -2, 1)$

$\\$

Question 26: Show that the points $(-3, 2), (-5, -5), (2, -3)$ and $(4, 4)$ are the vertices of a rhombus. Find the area of this rhombus.

Let the points be $A(-3, 2), B(-5, -5), C(2,-3)$ and $D(4, 4)$

Therefore sides of the figure

$AB = \sqrt{(-5+3)^2 + (-5-2)^2 }= \sqrt{4+49} = \sqrt{53}$

$BC = \sqrt{(2+5)^2 + (-3+5)^2 }= \sqrt{49+4} = \sqrt{53}$

$CD = \sqrt{(4-2)^2 + (4+3)^2 }= \sqrt{4+49} = \sqrt{53}$

$DA = \sqrt{(4+3)^2 + (4-2)^2 }= \sqrt{49+4} = \sqrt{53}$

Diagonals of the figure

$AC= \sqrt{(2+3)^2 + (-3-2)^2 }= \sqrt{25+25} = \sqrt{50} = 5 \sqrt{2}$

$BD = \sqrt{(4+5)^2 + (4+5)^2 }= \sqrt{81+81} = \sqrt{162} = 9 \sqrt{9}$

Hence, $AB = BC = CD = CA$. Therefore we see that the sides are equal but the diagonals are not equal. Hence $ABCD$  is a Rhombus.

Area $=$ $\frac{1}{2}$ ( Product of lengths of diagonal)

$=$ $\frac{1}{2}$ $5 \sqrt{2} \times \sqrt{9} = 45$ sq. units

$\\$

Question 27: Find the coordinates of the circumcenter of the triangle whose vertices are $(3, 0), (-1,-6)$ and $(4,-1)$. Also, find its circumradius.

Given: $A ( 3, 0), B ( -1, -6) , C ( 4, -1)$

Let $O(x, y)$ be the center

$OA^2 = ( x-3)^2 + (y-0)^2$

$OB^2 = (x+1)^2 +( y + 6)^2$

$OC^2 = ( x-4)^2 + ( y+1)^2$

$OA^2 = OB^2$

$\Rightarrow ( x-3)^2 + (y-0)^2 = (x+1)^2 +( y + 6)^2$

$\Rightarrow x^2 + 9 - 6x + y^2 = x^2 + 1 + 2x + y^2 + 36 + 12 y$

$\Rightarrow 9-6x = 1 + 2x + 36 + 12y$

$\Rightarrow 8x + 12y + 28 = 0$

$\Rightarrow 2x + 3y + 7 = 0$   … … … … … i)

$OB^2 = OC^2$

$\Rightarrow (x+1)^2 +( y + 6)^2 = ( x-4)^2 + ( y+1)^2$

$\Rightarrow x^2 + 1 + 2x + y^2 + 36 12 y = x^2 + 16 -8x + y + 1 + 2y$

$\Rightarrow 37 + 2x + 12y = 17 - 8x + 2y$

$\Rightarrow 10x + 10 y + 20 = 0$

$\Rightarrow x + y + 2 = 0$  … … … … … ii)

Solving i) and ii) we get $y = -3$

Therefore $x = 1$

Hence the center is $(1, -3)$

Therefore Circumradius $= \sqrt{ (1-3)^2 + ( -3-0)^2} = \sqrt{13}$ units

$\\$

Question 28: Find a point on the x-axis which is equidistant from the points $(7 6)$ and $(-3,4)$.

Given $A ( 7, 6) , B ( -3, 4)$

Let the point be $O (x, 0)$

Given $OA = OB \Rightarrow OA^2 = OB^2$

$(x-7)^2 + ( 0-6)^2 = ( x+3)^2 + ( 0-4)^2$

$x^2 + 49 - 14x + 36 = x^2 + 9 + 6x + 16$

$85 - 14x = 6x + 25$

$20x = 60$

$x = 3$

Therefore point is $(3, 0)$

$\\$

Question 29: (i) show that the points $A(5,6), 8(1, 5), C(2,1)$ and $D(6,2)$ are the vertices of a square.

(ii) Prove that the points $A (2,3), B(-2,2),C(-1,-2)$, and $D(3,-1)$ are the vertices of a square $ABCD$.

(iii) Name the type of $\triangle PQR$  formed by the points $P(\sqrt{2}, \sqrt{2}), Q( -\sqrt{2}, -\sqrt{2})$, and $R(-\sqrt{6}, \sqrt{6})$

i) Let the points be $A(5,6), B(1, 5), C(2, 1)$ and $D(6, 2)$

Therefore sides of the figure

$AB = \sqrt{(5-1)^2 + (6-5)^2 }= \sqrt{16+1} = \sqrt{17}$

$BC = \sqrt{(1-2)^2 + (5-1)^2 }= \sqrt{1+16} = \sqrt{17}$

$CD = \sqrt{(2-6)^2 + (1-2)^2 }= \sqrt{16+1} = \sqrt{17}$

$DA = \sqrt{(6-5)^2 + (2-6)^2 }= \sqrt{1+16} = \sqrt{17}$

Diagonals of the figure

$AC= \sqrt{(5-2)^2 + (6-1)^2 }= \sqrt{9+25} = \sqrt{34}$

$BD = \sqrt{(1-6)^2 + (5-2)^2 }= \sqrt{25+9} = \sqrt{34}$

Hence, $AB = BC=CD=DA$. Also diagonals are equal. Therefore the figure is a square.

ii) Let the points be $A(2,3), B(-2,2), C(-1,-2)$ and $D(3, -1)$

Therefore sides of the figure

$AB = \sqrt{(2+2)^2 + (3-2)^2 }= \sqrt{16+1} = \sqrt{17}$

$BC = \sqrt{(-2+1)^2 + (2+2)^2 }= \sqrt{1+16} = \sqrt{17}$

$CD = \sqrt{(-1-3)^2 + (-2+1)^2 }= \sqrt{16+1} = \sqrt{17}$

$DA = \sqrt{(3-2)^2 + (-1-3)^2 }= \sqrt{1+16} = \sqrt{17}$

Diagonals of the figure

$AC= \sqrt{(2+1)^2 + (3+2)^2 }= \sqrt{9+25} = \sqrt{34}$

$BD = \sqrt{(-2-3)^2 + (2+1)^2 }= \sqrt{25+9} = \sqrt{34}$

Hence, $AB = BC=CD=DA$. Also diagonals are equal. Therefore the figure is a square.

iii) Let the points be $P(\sqrt{2}, \sqrt{2}), Q( -\sqrt{2}, -\sqrt{2}), R(-\sqrt{6}, \sqrt{6})$

Therefore sides of the figure

$PQ = \sqrt{(\sqrt{2} + \sqrt{2})^2 + (\sqrt{2} + \sqrt{2})^2 }= \sqrt{8+8} = \sqrt{16}$

$QR = \sqrt{(-\sqrt{2} + \sqrt{6})^2 + (-\sqrt{2} - \sqrt{6})^2 }= \sqrt{2+ 6 - 2 \sqrt{12}+ 2 + 6 + 2\sqrt{12}} = \sqrt{16}$

$RP = \sqrt{(-\sqrt{6} - \sqrt{2})^2 + (\sqrt{6} - \sqrt{2})^2 }= \sqrt{6+ 2 + 2 \sqrt{12}+ 6 + 2 - 2\sqrt{12}} = \sqrt{16}$

Hence, $PQ = QR = RP$. Therefore the figure is an equilateral triangle.

$\\$

Question 30: Find the point on x-axis which is equidistant from the points $(-2, 5)$ and $(2,-3)$.

Given $A ( -2, 5), B ( 2, -3)$

Let the equidistant point be $O(x, 0)$

$OA = OB \Rightarrow OA^2 = OB^2$

$\Rightarrow (x+2)^2 + ( 0-5)^2 = (x-2)^2 + (0+3)^2$

$\Rightarrow x^2 + 4 + 4x + 25 = x^2 + 4 - 4x + 9$

$\Rightarrow 8x = -29 + 13$

$\Rightarrow 8x = -16$

$\Rightarrow x = -2$

Therefore point is $( -2, 0)$

$\\$

Question 31: Find the value of $x$ such that $PQ = QR$ where the coordinates of $P, Q$ and $R$ are $(6,-1), (1,3)$ and $(x,8)$ respectively.

Given: $P(6, -1), Q(1, 3), R(x, -8)$

$PQ = QR \Rightarrow PQ^2 = QR^2$

$\Rightarrow (1-6)^2 + ( 3+1)^2 = ( x-1)^2 + ( 8-3)^2$

$\Rightarrow 25 + 16 = x^2 + 1 - 2x + 25$

$\Rightarrow x^2 - 2x - 15 = 0$

$\Rightarrow (x-5)(x+3) = 0$

$\Rightarrow x = 5 \ or \ -3$

$\\$

Question 32: Prove, that the points $(0, 0), (5,5)$ and $(-5, 5)$ are the vertices of a right isosceles triangle.

Let the points be $A(0,0), B(5,5), C(-5, 5)$

Therefore sides of the figure

$AB = \sqrt{(5-0)^2 + (5-0)^2 }= \sqrt{25+25} = \sqrt{50}$

$BC = \sqrt{(-5-5)^2 + (5-5)^2 }= \sqrt{100}$

$CA = \sqrt{(-5-0)^2 + (5-0)^2 }= \sqrt{25+25} = \sqrt{50}$

Since $AB = CA, \triangle ABC$ is an Isosceles triangle.

$\\$

Question 33: If the point $P(x, y)$ is equidistant from the points $A(5,1)$ and $B(1,5)$, prove that $x=y$.

Given: $A(5, 1), B(1, 5) , P(x, y)$

$AP = BP \Rightarrow AP^2 = BP^2$

$\Rightarrow ( x-5)^2 + ( y - 1)^2 = (x-1)^2 + ( y - 5)^2$

$\Rightarrow x^2 + 25 - 10x + y^2 + 1 - 2y = x^2 + 1 - 2x + y^2 + 25 - 10y$

$\Rightarrow 26 - 10x - 2y = 26 - 2x - 10y$

$\Rightarrow -8x = - 8y$

$\Rightarrow x = y$

$\\$

Question 34: 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 $PR$.

Given: $P(5, -3), R(x, 6) , Q (0, 1)$

$QP = QR \Rightarrow QP^2 = QR^2$

$\Rightarrow (0-5)^2 + ( 1+3)^2 = (0-x)^2 + ( 1-6)^2$

$\Rightarrow 25 + 16 = x^2 + 25$

$\Rightarrow x = \pm 4$

When $x = 4, QR = \sqrt{4^2 + 25} = \sqrt{16 + 25} = \sqrt{41}$ units

When $x = -4, QR = \sqrt{(-4)^2 + 25} = \sqrt{16 + 25} = \sqrt{41}$ units

$\\$

Question 35: Find the values of $y$ for which the distance between the points $P (2, -3)$ and $Q (10, y)$ is $10$ units.

Given: $P(2, -3), Q(10, y) , PQ =10$

$(10-2)^2 + (y+3)^2 = 10^2$

$\Rightarrow 64 + ( y + 3)^2 = 100$

$\Rightarrow ( y + 3) ^2 = 36$

$\Rightarrow y + 3 = \pm 6$

$\Rightarrow y = 3, -9$

$\\$

Question 36: Find the center of the circle passing through $(6, -6), (3,-7)$ and $(3, 3)$.

Given: $A ( 6, -6), B(3, -7), C ( 3,3)$

Let center be $O(x, y)$

$OA^2 = ( x-6)^2 + ( y+6)^2$

$OB^2 = ( x-3)^2 + ( y+7)^2$

$OC^2 = ( x-3)^2 + ( y-3)^2$

$OA^2 = OB^2$

$\Rightarrow ( x-6)^2 + ( y+6)^2 = ( x-3)^2 + ( y+7)^2$

$\Rightarrow x^2 + 36 - 12 x + y^2 + 36 + 12 y = x^2 + 9 - 6x + y^2 + 49 + 14 y$

$\Rightarrow 72-12x+12y = 58 - 6x +14y$

$\Rightarrow 14 = 6x + 2y$

$\Rightarrow 3x+y = 7$   … … … … … i)

$OB^2 = OC^2$

$\Rightarrow ( x-3)^2 + ( y+7)^2 = ( x-3)^2 + ( y-3)^2$

$\Rightarrow y^2 + 49 + 14y = y^2 +9 - 16y$

$\Rightarrow 20y = - 40$

$\Rightarrow y = -2$   … … … … … ii)

Substituting in i)

$3x + ( -2) = 7$

$\Rightarrow 3x = 9$

$\Rightarrow x = 3$

Therefore center is $( 3, -2)$

$\\$

Question 37: Two opposite vertices of a square are $(-1,2)$ and $(3,2)$. Find the coordinates of other two vertices.

Given $A ( -1, 2), C(3, 2) , B(x, y)$

$AB^2 = BC^2$

$\Rightarrow (x+1)^2 + ( y - 2)^2 = ( x-3)^2 + ( y - 2)^2$

$\Rightarrow x^2 + 1 + 2x + y^2 + 4 - 4x = x^2 + 9 -6x + y^2 + 4 - 4y$

$\Rightarrow 1+2x = 9-6x$

$\Rightarrow 8x = 8$

$\Rightarrow x= 1$   … … … … … i)

Also

$AB^2 + BC^2 = AC^2$

$\Rightarrow (x+1)^2 + ( y - 2)^2 + ( x-3)^2 + ( y - 2)^2 = ( 3+1)^2 + ( 2- 2)^2$

$\Rightarrow x^2 + 1 + 2x + y^2 + 4 - 4y + x^2 + 9 - 6x + y^2 + 4 - 4y = 16$

$\Rightarrow 2x^2 + 2y^2 - 4x - 8y + 2 = 0$

From i), $x = 1$

$\Rightarrow 2 + 2y^2 - 4 - 8y + 2 = 0$

$\Rightarrow 2y^2 - 8y= 0$

$\Rightarrow y^2 - 4y=0$

$\Rightarrow y( y - 4) = 0$

$\Rightarrow y = 0 \ or \ y = 4$

Hence the coordinates are $B( 1, 4), D(1, 0)$

$\\$

Question 38: Name the quadrilateral formed, if any, by the following points, and give reasons for your answers:

(i) $A (-1, - 2), B(1, 0), C (-1, 2),D (-3, 0)$

(ii) $A (-3, 5), B (3, 1),C (0, 3), D (-1, -4)$

(iii) $A (4,5), B (7 ,6), C (4,3), D (1,2)$

i)    Let the points be $A(-1, -2), B(1,0), C(-1,2)$ and $D(-3,0)$

Therefore sides of the figure

$AB = \sqrt{(1+1)^2 + (0+2)^2 }= \sqrt{4+4} = \sqrt{8}$

$BC = \sqrt{(-1-1)^2 + (2-0)^2 }= \sqrt{4+4} = \sqrt{8}$

$CD = \sqrt{(-3+1)^2 + (0-2)^2 }= \sqrt{4+4} = \sqrt{8}$

$DA = \sqrt{(-3+1)^2 + (0+2)^2 }= \sqrt{4+4} = \sqrt{8}$

Diagonals of the figure

$AC= \sqrt{(-1+1)^2 + (2+2)^2 }= \sqrt{16} = 4$

$BD = \sqrt{(-3-1)^2 + (0-0)^2 }= \sqrt{16} = 4$

All four sides are equal and also the diagonals are equal. Hence this is a square.

ii)   Let the points be $A(-3, 5), B(3, 1), C(0, 3)$ and $D(-1,-4)$

Therefore sides of the figure

$AB = \sqrt{(3+3)^2 + (1-5)^2 }= \sqrt{81+16} = \sqrt{97}$

$BC = \sqrt{(0-3)^2 + (3-1)^2 }= \sqrt{9+4} = \sqrt{13}$

$CD = \sqrt{(-1-0)^2 + (-4-3)^2 }= \sqrt{1+49} = \sqrt{50}$

$DA = \sqrt{(-1+3)^2 + (-4-5)^2 }= \sqrt{4+81} = \sqrt{85}$

All four sides are unequal . Hence this is a quadrilateral.

iii)  Let the points be $A(4,5), B(7,6), C(4,3)$ and $D(1,2)$

Therefore sides of the figure

$AB = \sqrt{(7-4)^2 + (6-5)^2 }= \sqrt{9+1} = \sqrt{10}$

$BC = \sqrt{(4-7)^2 + (3-6)^2 }= \sqrt{9+9} = \sqrt{18}$

$CD = \sqrt{(1-4)^2 + (2-3)^2 }= \sqrt{9+1} = \sqrt{10}$

$DA = \sqrt{(1-4)^2 + (2-5)^2 }= \sqrt{9+9} = \sqrt{18}$

Diagonals of the figure

$AC= \sqrt{(4-4)^2 + (3-5)^2 }= \sqrt{4} = 2$

$BD = \sqrt{(1-7)^2 + (2-6)^2 }= \sqrt{36+16} = \sqrt{52}$

All four sides are equal and but the diagonals are unequal, this is a Rhombus.

$\\$

Question 39: Find the equation of the perpendicular bisector of the line segment joining points $(7,1)$ and $(3,5)$.

Given: $A ( 7, 1), B ( 3, 5)$

Mid point of $AB = \Big($ $\frac{7+3}{2}$ $,$ $\frac{1+5}{2}$ $\Big) = (5, 3)$

Slope of $AB =$ $\frac{5-1}{3-7}$ $=$ $\frac{4}{-4}$ $= -1$

Therefore slope of perpendicular bisector $= 1$

Therefore equation of perpendicular bisector  is

$y - 3 = 1 ( x - 5)$

$\Rightarrow y = x - 2$

$\\$

Question 40: Prove that the points $(3, 0), (4, 5), (-1, 4)$ and $(-2, -1)$, taken in order, form a rhombus. Also, find its area.

Let the points be $A(3,0), B(4,5), C(-1,4)$ and $D(-2,-1)$

Therefore sides of the figure

$AB = \sqrt{(4-3)^2 + (5-0)^2 }= \sqrt{1+25} = \sqrt{26}$

$BC = \sqrt{(-1-4)^2 + (4-5)^2 }= \sqrt{25+1} = \sqrt{26}$

$CD = \sqrt{(-2+1)^2 + (-1-4)^2 }= \sqrt{1+25} = \sqrt{26}$

$DA = \sqrt{(-2-3)^2 + (-1-0)^2 }= \sqrt{25+1} = \sqrt{26}$

Diagonals of the figure

$AC= \sqrt{(-1-3)^2 + (4-0)^2 }= \sqrt{16+16} = 4\sqrt{2}$

$BD = \sqrt{(-2-4)^2 + (-1-5)^2 }= \sqrt{36+36} = \sqrt{72} = 6\sqrt{2}$

All four sides are equal and but the diagonals are unequal, this is a Rhombus.

Area of the Rhombus $=$ $\frac{1}{2}$ (product of diagonals)

$=$ $\frac{1}{2}$ $\times 4\sqrt{2} \times 6\sqrt{2} = 24$ sq. units.

$\\$

Question 41: In the seating arrangement of desks in a classroom three students Rohini, Sandhya and Bina are seated at $A (3, 1), B (6,4)$ and $C (8, 6)$. Do you think they are seated in a line?

Given: $A ( 3, 1), B ( 6, 4) , C(8, 6)$

$AB = \sqrt{(6-3)^2 + (4-1)^2 }= \sqrt{9+9} = 3\sqrt{2}$

$BC = \sqrt{(8-6)^2 + (6-4)^2 }= \sqrt{4+4} = 2\sqrt{2}$

$CA = \sqrt{(3-8)^2 + (1-6)^2 }= \sqrt{25+25} = 5\sqrt{2}$

$\Rightarrow AB + BC = CA$

Hence they are all in a straight line.

$\\$

Question 42: Find a point on y-axis which is equidistant from the points $(5 , - 2)$ and $(- 3, 2)$ .

Given: $A ( 5, -2), B ( -3, 2)$

Let the equidistant point be $O(0, y)$

Given $OA = OB \Rightarrow OA^2 = OB^2$

Therefore $(0-5)^2 + ( y + 2)^2 = ( 0+3)^2 + ( y - 2)^2$

$\Rightarrow 25 + y^2 + 4 + 4y = 9 + y^2 + 4 - 4y$

$\Rightarrow 29 + 4y = 13 - 4y$

$\Rightarrow 8y = - 16$

$\Rightarrow y = -2$

Therefore the point is $(0, -2)$

$\\$

Question 43: Find a relation between $x$ and $y$ such that the point $(x, y)$ is equidistant from the points $(3,6)$ and $(-3,4)$.

Given $A ( 3, 6), B ( -3, 4)$

Let equidistant point $O(x, y)$

Given $OA = OB \Rightarrow OA^2 = OB^2$

$\Rightarrow (x-3)^2 + ( y - 6)^2 = (x+3)^2 + ( y -4)^2$

$\Rightarrow x^2 + 9 - 6x + y^2 + 36 - 12y = x^2 + 9 + 6x + y^2 + 16 - 4y$

$\Rightarrow 45 - 6x - 12y = 25 + 6x - 8y$

$\Rightarrow 12x + 4y = 20$

$\Rightarrow 3x + y = 5$

$\\$

Question 44: If a point $A(0,2)$ is equidistant from the points $B (3,p)$ and $C(p,5)$, then find the value of $p$.

Given: $A(0, 2), B(3, p), C(p, 5)$

Since $AB = AC \Rightarrow AB^2 = AC^2$

$\Rightarrow (3-0)^2 + ( p-2)^2 = ( p - 0)^2 + ( 5-2)^2$

$\Rightarrow 9+p^2 +4 - 4p = p^2 + 9$

$\Rightarrow 13-4p = 9$

$\Rightarrow 4p = 4$

$\Rightarrow p = 1$

$\\$

Question 45: Prove that the points $(7, 10), (-2,5)$ and $(3, -4)$ are the vertices of an isosceles right triangle.

Given: $A ( 7, 10), B ( -2, 5) , C(3, -4)$

$AB = \sqrt{(-2-7)^2 + (5-10)^2 }= \sqrt{81+25} = \sqrt{106}$

$BC = \sqrt{(3+2)^2 + (-4-5)^2 }= \sqrt{25+81} = \sqrt{106}$

$CA = \sqrt{(7-3)^2 + (10+4)^2 }= \sqrt{16+196} = \sqrt{212}$

Therefore $AB = AC$ ( i.e. two sides of the triangle are equal)

Also $AB^2 + BC^2 = AC^2$

$\Rightarrow 106 + 106 = 212$

Therefore $\triangle ABC$ is a right angled isosceles triangle.

$\\$

Question 46: If the point $P(x, 3)$ is equidistant from the points $A(7,-1)$ and $B(6,8)$, find the value of $x$ and find the distance $AP$.

Give: $P(x, 3), A ( 7, -1), B(6, 8)$

Since $PA = PB \Rightarrow PA^2 = PB^2$

$\Rightarrow (7-x)^2 + ( -1 -3)^2 = (6-x)^2 + (8-3)^2$

$\Rightarrow 49 + x^2 - 14x + 16 = 36 + x^2 - 12x + 25$

$65 - 14x = 61 - 12x$

$4 = 2x \ or \ x = 2$

Hence the point is $(2, 3)$

$AP = \sqrt{(7-2)^2 + ( -1-3)^2} = \sqrt{25+16} = \sqrt{41}$

$\\$

Question 47: It $A(3,y)$ is equidistant from points $P (8,-3)$ and $Q(7,6)$, find the value of $y$ and find the distance $AP$.

Given: $A ( 3, y), P(8, -3) , Q(7,6)$

Since $AP = AQ \Rightarrow AP^2 = AQ^2$

$\Rightarrow (8-3)^2 + ( -3-y)^2 = ( 7-3)^2 + ( 6-y)^2$

$\Rightarrow 25 + 9 + y^2 + 6y = 16 + 36 + y^2 - 12y$

$\Rightarrow 34 + 6y = 52 - 12y$

$\Rightarrow 18 y = 18$

$\Rightarrow y = 1$

Therefore $A ( 3, 1)$

$AP = \sqrt{(8-3)^2 + ( -3-1)^2} = \sqrt{25+16} = \sqrt{41}$

$\\$

Question 48: If $(0, - 3)$ and $(0, 3)$ are the two vertices of an equilateral triangle, find the coordinates of its third vertex.

Given: $A (0, -3), B(0, 3)$

Let third vertices be $C(x, y)$

$AC^2 = ( x-0)^2 + (y+3)^2$

$AB^2 = ( 0-0)^2 + ( 3+3)^2 = 36$

$BC^2 = (x-0)^2 + (y-3)^2$

Since $AC^2 = AB^2$

$\Rightarrow x^2 + (y+3)^2 = 36$

$\Rightarrow x^2 + y^2 + 9 + 6y = 36$

or $\Rightarrow x^2 + y^2 + 6y = 27$  … … … … … i)

$AC^2 = BC^2$

$\Rightarrow x^2 + (y+3)^2 = x^2 + ( y-3)^2$

$\Rightarrow x^2 + y^2 +9 + 6y = x^2 + y^2 + 9 - 6y$

$\Rightarrow 9 + 6y = 9 - 6y$

$\Rightarrow 12y = 0$

$\Rightarrow y = 0$

Substituting in i)

$x^2 = 27 \Rightarrow x = \pm 3\sqrt{3}$

Therefore $C$ could be $( 3\sqrt{3}, 0)$ or $(- 3\sqrt{3}, 0)$

$\\$

Question 49: If the point $P (2, 2)$ is equidistant from the point s $A (-2, k)$ and $B (-2k, 3)$, find $k$. Also, find the length of $AP$.

Given: $P(2, 2), A ( -2, k), B ( -2k, -3)$

$AP^2 = ( -2-2)^2 + ( k-2)^2 = 16 + ( k-2)^2$

$BP^2 = (-2k-2)^2 + ( -3-2)^2 = ( -2k-2)^2 + 25$

$AP^2 = BP^2$

$\Rightarrow 16 + ( k-2)^2 = ( -2k-2)^2 + 25$

$\Rightarrow k^2 + 4 - 4k = 9 + 4k^2 + 4 + 8k$

$\Rightarrow 3k^2 + 12k + 9 = 0$

$\Rightarrow k^2 + 4k + 3 = 0$

$\Rightarrow (k+3)(k+1)=0$

$\Rightarrow k = -3 \ or \ k = -1$

If $k = -3 \Rightarrow AP = \sqrt{(-2-2)^2+(-3-2)^2} = \sqrt{16+25} = \sqrt{41}$

If $k = -1 \Rightarrow AP = \sqrt{(-2-2)^2+(-1-2)^2} = \sqrt{16+9} = \sqrt{25} = 5$

$\\$

Question 50: If the point $A(0,2)$ is equidistant from the points $B(3,p)$ and $C(p,5)$, find $p$. Also, find the length of $AB$.

Given: $A(0, 2), B(3, p), C(p, 5)$

$AB^2 = ( 3-0)^2 + ( p-2)^2$

$AC^2 = (p-0)^2 + ( 5-2)^2$

$AB^2 = AC^2$

$(3-0)^2 + ( p - 2)^2 = ( p-0)^2 + (5-2)^2$

$9+p^2+4 - 4p = p^2 + 9$

$4-4p = 0$

$p=1$

Therefore $AB = \sqrt{9 + ( 1-2)^2 } = \sqrt{10}$

$\\$

Question 51: If the point $P (k -1,2)$ is equidistant from the points $A(3,k)$ and $B(k, 5)$, find the values of $k$.

Given: $P(k-1, 2), A(3, k), B(k, 5)$

$AP^2 = ( 3-k+1)^2 + ( k-2)^2 = (4-k)^2 + ( k-2)^2$

$BP^2 = (k-k+1)^2 + ( 5-2)^2 = 1+9 = 10$

$AP^2 = BP^2$

$\Rightarrow (4-k)^2 + ( k-2)^2 =10$

$\Rightarrow 16 +k^2 -8k + k^2 + 4 - 4k = 10$

$\Rightarrow 2k^2 - 12k + 20 = 0$

$\Rightarrow k^2 -6k + 5 = 0$

$\Rightarrow (k-1)(k-5) = 0$

$\Rightarrow k = 1 \ or \ k = 5$

$\\$

Question 52: If $(-4,3)$ and $(4,3)$ are two vertices of an equilateral triangle, find the coordinates of the third vertex, given that the origin lies in the (i) interior, (ii) exterior of the triangle.

Given: $A (-4, 3), B(4, 3)$

Let third vertices be $C(x, y)$

$AB^2 = (4+4)^2 + ( 3- 3)^2 = 64$

$BC^2 = ( x-4)^2 + ( y - 3)^2$

$AC^2 = ( x+ 4)^2 + ( y - 3)^2$

$BC^2 = AC^2$

$\Rightarrow ( x-4)^2 + ( y - 3)^2 = ( x+ 4)^2 + ( y - 3)^2$

$\Rightarrow x^2 + 16 - 8x = x^2 + 16 + 8x$

$\Rightarrow 16x = 0$

$\Rightarrow x = 0$

$BC^2 = AB^2$

$\Rightarrow ( x-4)^2 + ( y - 3)^2 = 64$

$\Rightarrow 16 + (y-3)^2 = 64$

$\Rightarrow (y-3)^2 = 48$

$\Rightarrow y - 3 = \pm 4\sqrt{3}$

$\Rightarrow y = 3 \pm 4\sqrt{3}$

Hence $C$ is i) $(0, 3 - 4\sqrt{3})$ and ii) $( 0, 3+4\sqrt{3})$

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Question 53: Ayush starts walking from his house to office. Instead of going to the office directly, he goes to a bank first, from there to his daughter’s school and then reaches the office. What is the extra distance traveled by Ayush in reaching the office? (Assume that all distances covered are in straight hires). If the house is situated at $(2,4)$, bank at $(5,8),$ school at $(13,14)$ and office at $(13, 26)$ and coordinates are in kilometers.

Given: House $( 2, 4)$, Bank $( 5, 8)$, School $( 13, 14)$, Office $(13, 26)$

Distance from House to Office $= \sqrt{(13-2)^2 + (26-4)^2} = \sqrt{121+ 484} = \sqrt{605} = 24.6$ km

Distance of House to Bank $= \sqrt{(5-2)^2 + ( 8-4)^2} = \sqrt{9+16} = 5$ km

Distance of Bank to School $= \sqrt{(13-5)^2+ (14-8)^2} = \sqrt{64+36} = 10$ km

Distance from school to office $= \sqrt{(13-13)^2 + (26-14)^2} = \sqrt{144} = 12$ km

Therefore total distance traveled $= 5 + 10 + 12 = 27$ km

Extra distance $= 27 - 24.6 = 2.4$ km

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Question 54: The center of a circle is $(2a, 1)$. Find the values of $a$ if the circle passes through the point $(11,-a)$ and has diameter $10\sqrt{2}$ units.

Given: $O(2a, 1), A(11, -a)$

Radius $= 5\sqrt{2}$

Therefore $(5\sqrt{2})^2 = (11-2a)^2 + ( -a -1)^2$

$50 = 121 + 4a^2 - 44 a + a^2 +1 + 2a$

$5a^2 - 42a + 72 = 0$

$a = 6 \ or \ 2.4$

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Question 55: Find a point which is equidistant from the points $A(-5,4)$ and $B (-1,6)$. How many such points are there?

Given: $A(-5, 4), B(-1, 6)$

Let $C(x, y)$ be equidistant

$AC^2 = BC^2$

$\Rightarrow (x+5)^2 + ( y-4)^2 = (x+1)^2 + ( y - 6)^2$

$\Rightarrow x^2 + 25 + 10x + y^2 + 16 - 8y = x^2 + 1 + 2x + y^2 + 36 - 12y$

$\Rightarrow 41+ 10x - 8y = 37 + 2x - 12y$

$\Rightarrow 8x + 4y = -4$

$\Rightarrow 2x + y + 1 = 0$   … … … … … i)

Therefore all points satisfying the equation i) will be equidistant from the two given points. Hence we have infinite such points.

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Question 56: The points $A (2,9), B (a,5)$ and $C (5, 5)$ are the vertices of a $\triangle ABC$ right angled at $B$. Find the values of $a$ and hence the area of $\triangle ABC$.

Given: $A(2, 9), B(a, 5), C(5, 5)$

$AB^2 = ( a-2)^2 + (5-9)^2 = (a-2)^2 + 16$

$BC^2 = ( 5-a)^2 + (5-5)^2 = (5-a)^2$

$AC^2 = ( 2-5)^2 + (9-5)^2 = 9 + 16 = 25$

$AB^2 + BC^2 = AC^2$

$\Rightarrow (a-2)^2 + 16 + (5-a)^2 = 25$

$\Rightarrow a^2 + 4 - 4a + 16 + 25 + a^2 - 10a = 25$

$\Rightarrow 2a^2 - 14a + 20 = 0$

$\Rightarrow a^2 - 7a + 10 = 0$

$\Rightarrow ( a-5)(a-2) = 0$

$\Rightarrow a = 5 \ or \ 2$

When $a = 5, AB^2 = 25$ and $BC^2 = 0. BC \neq 0$ hence $a \neq 5$

When $a = 2, AB^2 = 16$ and $BC^2 = 9$. Hence $a = 2$

Therefore Area $=$ $\frac{1}{2}$ $\times 4 \times 3 = 6$ sq units.