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 $(3, 0)$

Link to the rest of Exercise 19.4