Question 15 Marks
If $(-3, 2), (1, -2)$ and $(5, 6)$ are the midpoints of the sides of a triangle, find the coordinates of the vertices of the triangle.
Answer
let $A\left(x_1, y_1\right), B\left(x_2, y_2\right)$ and $O\left(x_3, y_3\right)$ be the coordinates of the vertices of $\triangle A B C$.
$D$ is the midpoint of $A B<$
$ D (-3,2)= D \left(\frac{ x _1+ x _2}{2}, \frac{ y _1+ y _2}{2}\right)$
$\frac{ x _1+ x _2}{2}=-3, \frac{ y _1+ y _2}{2}$
$x _1+ x _2=-6\ldots(1)$
$Y_1+Y_2=4\ldots(2)$
Similarly
$x_2+x_3=2\ldots(3) $
$Y 2+Y 3=-4\ldots(4)$
$x_1+x_3=10\ldots(5)$
$Y_1+Y_3=12\ldots(6)$
Adding $(1), (3)$ and $(5)$
$2\left(x_1+x_2+x_t\right)=6 $
$x_1+x_2+x_3=3$
$-6+x_3=3 $
$x_3=9$
From $(3)$
$x_2+9=2 $
$x_2=9$
From $(3)$
$x_2+9=2$
$x_2=-7$
From $(5)$
$x_1+9=10$
$x_1=1$
Adding $(2), ( 4)$ and $(6)$
$2\left(y_1+Y_2+y_3\right)=12$
$y_1+Y_2+Y_3=6 $
$4+y_3=6$
$Y_3=2$
from $(4)$
$y _2+2=-4 $
$Y _2=-6$
from $(6)$
$Y_1+2=12$
$Y_1=10$
The coordinates of the vertices of $\triangle ABC$ are $(9,2),(1,10)$ and $(-7,-6)$. View full question & answer→Question 25 Marks
If the midpoints of the sides ofa triangle are $(-2, 3), (4, -3), (4, 5),$ find its vertices.
Answer
Let $P\left(x_1, y_1\right), Q\left(x_2, y_2\right)$ and $R\left(x_3, y_3\right)$ be the coordinates of the vertices of . $\triangle P Q R$
Midpoint of PQ is D
$D (-2,3)= D \left(\frac{ x _1+ x _2}{2}, \frac{ y _1+ y _2}{2}\right)$
$\frac{ x _1+ x _2}{2}=-2, \frac{ y _1+ y _2}{2}=3 $
$X _1+ X _2=-4 \quad \ldots . .(1), \quad Y _1+ y _2=6\ldots(2)$
similarly,
$x_2+x_3=8 \ldots \ldots(3), \quad y_2+y_3=-6\ldots(4)$
$x_1+x_3=8 \quad \ldots .(5), \quad y_1+y_3=10\ldots(6)$
Adding $(1), (3)$ and $(5)$
$2\left(x_1+x_2+x_3\right)=12$
$x_1+x_2+x_3=6$
$-4+x_3=6$
$x_3=10$
Adding $(2), (4)$ and $(6)$
$ 2\left(Y_1+Y_2+Y_3\right)=10$
$1+Y_2+Y_3=5$
$6+Y_3=5$
$Y_3=-1$ View full question & answer→Question 35 Marks
The points $(2, -1), (-1, 4)$ and $(-2, 2)$ are midpoints of the sides ofa triangle. Find its vertices.
Answer
Let $A\left(x_1, y_1\right), B\left(x_2, y_2\right)$ and $C\left(x_3, y_3\right)$ be the coordinates of the vertices of $\triangle A B C$.
Midpoint of $A B$, i.e. $D$
$D(2,1)=D\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$
$2=\frac{ x _1+ x _2}{2}, \frac{ y _1+ y _2}{2}=-1$
$x_1+x_2=4\ldots(1)$
$Y_1+Y_2=-2\ldots(2)$
Similarly,
$x_1+x_3=-2\ldots(3)$
$y_1+y_3=8\ldots(4)$
$x_2+x_3=-4\ldots(5)$
$Y_2+Y_3=4\ldots(6)$
Adding (1), (3) and (5)
$2\left(x_1+x_2+x_3\right)=-2 $
$x_1+x_2+x_3=-1 $
$4+x_3=-1 \ldots(\text { from }(1)) $
$x_3=-5$
From (3)
$x_1-5=-2 $
$x_1=3$
From (5)
$x_2=-5=-4 $
$x_2=1$
Adding (2),(4) and (6)
$2\left(Y_1+Y_2+Y_3\right)=10 $
$Y_1+Y_2+Y_3=5$
$-2+Y_3=5 \quad \text { [from(2)] }$
$y_3=7$
From (4)
$y_1+7=8 $
$y_1=1$
From (6)
$y_2+7=4$
$y_2=-3$
The coordinates of the vertices of A.ABC are $(3,1),(1,-3)$ and $(-5,7)$ View full question & answer→Question 45 Marks
Let $A(-a, 0), B(0, a)$ and $C(\alpha, \beta)$ be the vertices of the $L 1 A B C$ and $G$ be its centroid. Prove that $G A^2+G B^2+G C^2=\frac{1}{3}\left(A B^2+B C^2+C A^2\right)$
Answer
Coordinates of $G$ are,
$G(x, y)=G\left(\frac{-a+0+a}{3}, \frac{0+a+b}{3}\right)=G\left(0, \frac{a+b}{3}\right) $
$G A^2=(0+a)^2+\left(\frac{a+b}{3}-0\right)^2 $
$G A^2=\frac{9 a^2+a^2+b^2+2 a b}{9}=\frac{10 a^2+b^2+2 a b}{9} $
$G B^2=(0-0)^2+\left(\frac{a+b}{3}-a\right)^2 $
$G B^2=\left(\frac{b-2 a}{3}\right)^2=\frac{b^2+4 a^2-4 a b}{9} $
$G C^2=(0-a)^2+\left(\frac{a+b}{3}-b\right)^2 $
$G C^2=a^2+\left(\frac{a-2 b}{3}\right)^2=\frac{9 a^2+a^2+4 b^2-4 a b}{9} $
$G A^2+G B^2+G C^2=\frac{10 a^2+b^2+2 a b+b^2+4 a^2-4 a b+10 a^2+4 b^2-4 a b}{9} $
$=\frac{24 a^2+6 b^2-6 a b}{9}$
$G A^2+G B^2+G C^2=\frac{1}{3}\left(8 a^2+2 b^2-2 a b\right)\ldots(1) $
$A B^2=(-a-0)^2+(0-a)^2=2 a^2 $
$B C^2=(0-a)^2+(a-b)^2=a^2+a^2+b^2-2 a b=2 a^2+b^2-2 a b $
$A C^2=(-a-a)^2+(0-b)^2=4 a^2+b^2$
from $(1)$ and $(2)$
$G A^2+G B^2+G C^2=\frac{1}{3}\left(A B^2+B^2+C A^2\right)$ View full question & answer→Question 55 Marks
Prove that the points $A(-5, 4), B(-1, -2)$ and $C(S, 2)$ are the vertices of an isosceles right-angled triangle. Find the coordinates of D so that ABCD is a square.
Answer
$A B=\sqrt{(-1+5)^2(-2-4)^2}=\sqrt{16+36}=\sqrt{52} \text { units } $
$B C=\sqrt{(-1-5)^2+(-2-2)^2}=\sqrt{36+36}=\sqrt{52} \text { units } $
$A C=\sqrt{(5+5)^2+(2-4)^2}=\sqrt{100+4}=\sqrt{104} \text { units } $
$A B^2+B C^2=52+52=104 $
$A C^2=104 $
$\because A B=A C \text { and } A B^2+B C^2=A C^2$
$\therefore ABC$ is an isosceles right angled triangle.
Let the coordinates of $D$ be $( x , y )$
If $A B C D$ is a square,
Midpoint of $AC =$ mid point of $BD$
$O\left(\frac{-5+5}{2}, \frac{4+2}{2}\right)=O\left(\frac{x-1}{2}, \frac{y-2}{2}\right) $
$O=\frac{x-1}{2}, 3=\frac{y-2}{2} $
$x=1, y=8$
Coordinates of $D$ are $(1,8)$ View full question & answer→Question 65 Marks
ABC is a triangle whose vertices are A(-4, 2), B(O, 2) and C(-2, -4). D. E and Fare the midpoint of the sides BC, CA and AB respectively. Prove that the centroid of the Δ ABC coincides with the centroid of the Δ DEF.
Answer
Let $D, E$ and $F$ be the midpoints of the sides $A B, A C$ and $B C$ of $\triangle A B C$ respectively.
$
\therefore AD : DB = BF : FC = AE : EC =1: 1
$
Coordinates of D are,
$
D ( x , y )= D \left(\frac{0-4}{2}, \frac{2+2}{3}\right)= D (-2,2)
$
Similarly ,
$
E(a, b)=E\left(\frac{-4-2}{2}, \frac{2-4}{2}\right)=E(-3,-1)
$
and,
$
F(p, q)=F\left(\frac{0-2}{2}, \frac{2-4}{2}\right)=F(-1,-1)
$
Coordinates of centroid of $\triangle ABC$ are,
$
=\left(\frac{-4-2+0}{3}, \frac{2-4+2}{3}\right)=(-2,0)
$
Coordinates of centroid of $\triangle D E F$ are ,
$
=\left(\frac{-2-3-1}{3}, \frac{2-1-1}{3}\right)=(-2,0)
$
Thud the centroid of $\triangle D E F$ coincides with centroid of $\triangle D E F$. View full question & answer→Question 75 Marks
$(4, 2)$ and $(-1, 5)$ are the adjacent vertices ofa parallelogram. $(-3, 2)$ are the coordinates of the points of intersection of its diagonals. Find the coordinates of the other two vertices.
Answer
Let the coordinates of $C$ and $D$ be $( x , y )$ and $( a , b )$ respectively.
Midpoint of $AC$ is $O$ coordinates of $O$ are ,
$O(-3,2)=O\left(\frac{4+ x }{2}, \frac{2+ y }{2}\right) $
$-3=\frac{4+ x }{2}, 2=\frac{2+ y }{2} $
$-6=4+ x , 4=2+ y $
$x =-10, y =2 $
$C (-10,2)$
Similarly, coordinates of midpoint of DB, i.e. O are,
$O(-3,2)=O\left(\frac{a-1}{2}, \frac{b+5}{2}\right) $
$-3=\frac{a-1}{2}, 2=\frac{b+5}{2} $
$-6=a-1,4=b+5 $
$a=-5, b=-1 $
$D(-5,-1)$
Thus, the coordinates of each other two vertices are $(-10,2)$ and $(-5-1)$ View full question & answer→Question 85 Marks
Find the ratio in which the point $P (2, 4)$ divides the line joining points $(-3, 1)$ and $(7, 6).$
Answer
Let the point $P$ divides $A B$ in the ratio $k: 1$.
Coordinates of $P$ are
$x=\frac{7 k-3}{k+1} $
$y=\frac{6 k+1}{k+1}$
But given, $P(x, y)=P(2,4)$
$\therefore 2=\frac{7 k -3}{ k +1}$
$\Rightarrow 2 k +2=7 k -3$
$\Rightarrow 5=5 k$
$\Rightarrow k =1$
$k : 1=1: 1$
or $4=\frac{6 k +1}{ k +1}$
$4 k +4=6 k +1$
$\Rightarrow 3=2 k$
$\Rightarrow k =\frac{3}{2}$
$k: 1=3: 2$ View full question & answer→Question 95 Marks
Show that the lines x = O and y = O trisect the line segment formed by joining the points (-10, -4) and (5, 8). Find the points of trisection.
Answer
Let $P(x, 0)$ lies on the line $y=0$ i.e. $x$-axis and divides the line segment $A B$ in the ratio
Coordinates of $P$ are,
$P ( x , 0)= P \left(\frac{5 k -10}{ k +1}, \frac{8 k -4}{ k +1}\right)$
$\Rightarrow 0=\frac{8 k -4}{ k +1}, \frac{5 k -10}{ k +1}= x$
$\Rightarrow 8 k =4, \frac{5\left(\frac{1}{2}\right)-10}{\frac{1}{2}+ x }= x \ldots \ldots . . . \text { from (1) }$
$\Rightarrow k =\frac{1}{2} \ldots \ldots . .(1), x =-5$
Hence $P(-5,0)$ divides $A B$ in the ratio 1: 2 .
Let $Q(0, y)$ lies on the line $x=0$ i.e. $y$ - axis and
divides the line segment $A B$ in the ratio $k_1: 1$.
Coordinates of $Q$ are
$Q(0, y)=Q\left(\frac{5 k_1-10}{k_1+1}, \frac{8 k_1-4}{k_1+1}\right)$
$0=\frac{5 k_1-10}{k_1+1}, y=\frac{8 k_1-4}{k_1+1}$
$\Rightarrow 5 k_1=10, y=\frac{8(2)-4}{2+1} \quad \ldots . . . \text { from (2) }$
$\Rightarrow k_1=2 \quad \ldots . \text { (2) } \quad y=4$
Hence, $Q(O, 4)$ divides in the ratio 2 : 1.
Hence proved Paid Qare the points of trisection of AB. View full question & answer→Question 105 Marks
Show that the line segment joining the points $(-3, 10)$ and $(6, -5)$ is trisected by the coordinates axis.
Answer
Let the coordinates of two points $x$-axis and $y$-axis be $P(x, O)$ and $G(0, y)$ respectively.
Let $P$ divides $A B$ in the ratio $k: 1$.
Coordinates of $P$ are
$ P(x, 0)=P\left(\frac{6 k-3}{k+1}, \frac{-5 k+10}{k+1}\right)$
$\Rightarrow 0=\frac{-5 k+10}{k+1}$
$\Rightarrow 5 k=10$
$\Rightarrow k=2 $
Hence $P$ divides $A B$ in the ratio $2: 1 .$
Let $Q$ divides $A B$ in the ratio $k_1: 1$.
Coordinates of $Q$ are,
$ Q (0, y )= Q \left(\frac{6 k _1-3}{ k +1}, \frac{-5 k +10}{ k +1}\right)$
$\Rightarrow 0=\frac{6 k _1-3}{ k +1}$
$\Rightarrow 6 k _1=3$
$\Rightarrow k _1=\frac{1}{2} $
Hence $Q$ divides $A B$ in the ratio $1:2$
Hence proved, $P$ and $Q $are the points of trisection. View full question & answer→Question 115 Marks
A (2, 5), B (-1, 2) and C (5, 8) are the vertices of triangle ABC. Point P and Q lie on AB and AC respectively, such that AP: PB = AQ: QC = 1: 2. Calculate the coordinates of P and Q. Also, show that 3PQ = BC.
Answer
$AP : PB =1: 2$
Coordinates of $P$ are,
$
P(x, y)=P\left(\frac{-1+4}{2+1}, \frac{10+2}{2+1}\right)=P(1,4)
$
$AQ : QC =1: 2$
Coordinates of $Q$ are,
$
Q ( a , b )= Q \left(\frac{4+5}{2+1}, \frac{10+8}{2+1}\right)= Q (3,6)
$
Coordinates of P and Q are $(1,4)$ and $(3,6)$
$
P Q=\sqrt{(3-1)^2+(6-4)^2}=\sqrt{4+4}=2 \sqrt{2} \text { units }
$
$
BC =\sqrt{(5+1)^2+(8-2)^2}=\sqrt{36+36}=6 \sqrt{2} \text { units }
$
Hence proved, $3 PQ = BC$ View full question & answer→Question 125 Marks
Find the points of trisection of the segment joining $A ( -3, 7)$ and $B (3, -2).$
Answer
Let $P ( x , y )$ and $Q ( a , b )$ be the pcint of trisection of the line segment $AB$.
$AP : PB =1: 2$
Coordinates of $P$ are
$x=\frac{1 \times 3+2 \times-3}{1+2}=-1 $
$y=\frac{1 \times-2+2 \times 7}{1+2}=4$
$P(-1,4)$
$AQ : QB =2: 1$
coordinates of $Q$ are,
$a=\frac{2 \times 3=1 \times-3}{2+1}=1$
$b=\frac{2 \times-2+1 \times 7}{2+1}=1$
$Q(1,1)$
$\therefore$ The points of trisection ae $(-1,4)$ and $(1,1)$. View full question & answer→Question 135 Marks
The origin $o (0, O), P (-6, 9)$ and $Q (12, -3)$ are vertices of triangle $OPQ.$ Point M divides $OP$ in the ratio $1: 2$ and point $N$ divides $OQ$ in the ratio $1: 2.$ Find the coordinates of points M and N. Also, show that $3MN = PQ.$
Answer
It is given that $M$ divides $O P$ in the ratio $1: 2$ and point $N$ divides $O Q$ in the ratio $1: 2$.
Using section formula, the coordinates of $M$ are
$\left(\frac{-6+0}{3}, \frac{9+0}{3}\right)=(-2,3)$
Using section formula, the coordinates of $N$ are
$\left(\frac{12+0}{3}, \frac{-3+0}{3}\right)=(4,-1)$
Thus, the ooordinates of $M$ and $N$ are $(-2,3)$ and $(4,-1)$ respectively.
Now, using distance formula, we have:
$ P Q=\sqrt{(-6-12)^2+(9+3)^2}=\sqrt{324+144}=\sqrt{468}$
$M N=\sqrt{(4+2)^2+(-1-3)^2}=\sqrt{36+36}=\sqrt{52} $
It can be observed that :
$PQ =\sqrt{468}=\sqrt{9 \times 52}=3 \sqrt{52}=3 MN$
Hence proved. View full question & answer→Question 145 Marks
Find the coordinate of $O$, the centre of a circle passing through $P(3,0), Q(2, \sqrt{5})$ and $R(-2 \sqrt{2}$ $,-1)$. Also find its radius.
Answer
Let $O(x, y)$ be the centre of the circle
$ O P=O Q \text { (radii of same circle) }$
$\Rightarrow O P^2=O Q^2$
$\left(\sqrt{(x-3)^2+(y-0)^2}\right)^2=\left(\sqrt{(x-2)^2+(y-\sqrt{5})^2}\right)^2$
$\Rightarrow x^2+9-6 x+y^2=x^2+4-4 x+y^2+5-2 \sqrt{5 y}$
$\Rightarrow-2 x+2 \sqrt{5} y=0$
$\Rightarrow-x+\sqrt{5} y=0\ldots(1) $
Similarly,$O Q=O R$
$ \Rightarrow O Q^2=O R^2$
$\Rightarrow(x-2)^2+(y-\sqrt{5})^2=(x+2 \sqrt{2})^2+(y+1)^2$
$\Rightarrow x^2+4-4 x+y^2+5-2 \sqrt{5} y=x^2+8+4 \sqrt{2}+y^2+1+2 y$
$\Rightarrow-4 x-4 \sqrt{2} x-2 \sqrt{5} y=0$
$\Rightarrow-2 x-2 \sqrt{2} x-\sqrt{5} y-y=0 \ldots . . .(2) $
Putting $x=\sqrt{5} y$ from (1) and (2)
$ -2 \sqrt{5} y-2 \sqrt{10} y-\sqrt{5} y-y=0$
$(-3 \sqrt{5}-2 \sqrt{10}-1) y=0$
$y=0 $
from (1)
$x=\sqrt{5}(0)=0$
$\Rightarrow x =0$
Thus, coordinates of $O$ are $(0,0)$.
Radius $=\sqrt{(0-3)^2+(0-0)^2}=\sqrt{9}=3$ units View full question & answer→Question 155 Marks
Find the coordinate of $O,$ the centre of a circle passing through $A (8 , 12) , B (11 , 3),$ and $C (0 , 14).$ Also , find its radius.
Answer
Let $O(x, y)$ be the centre of the circle.
$ O A=O B \text { (radii of the same circle) }$
$\Rightarrow O A^2=O B^2$
$(x-8)^2+(y-12)^2=(x-11)^2+(y-3)^2$
$\Rightarrow x^2+64-16 x+y^2+144-24 y=x^2+121-22 x+y^2+9-6 y$
$\Rightarrow 6 x-18 y+78=0$
$\Rightarrow x-3 y+13=0$
$\text { similarly, } O B=O C$
$\therefore O B^2=O C^2 $
$(x-11)^2+(y-3)^2=(x-0)^2+(y-14)^2$
$\Rightarrow x^2+121-22 x+y^2+9-6 y=x^2+y^2+196-28 y$
$\Rightarrow-22 x+22 y-66=0$
$\Rightarrow-x+y-3=0\ldots(2)$
$x-3 y+13=0\ldots(1)$
solving (1) \& (2) we get,
$ -2 y+10=0$
$\Rightarrow y=5 $
from (1)
$x-15+13=0$
$\Rightarrow x =2$
Thus, coordinates of $O$ are $(2,5)$
Radius $=\sqrt{(2-8)^2+(5-12)^2}=\sqrt{36+49}=\sqrt{85}$ units View full question & answer→Question 165 Marks
Find the point on the x-axis equidistant from the points $(5,4)$ and $(-2,3).$
AnswerLet the point on x-axis be $P (x,0)$
Given ,
$PA = PB$
$PA^2 = PB^2$
$(x - 5)^2 + (0 - 4)^2 = (x + 2)^2 + (0 - 3)^2$
$x^2 + 25 - 10x + 16 = x^2 + 4 + 4x + 9$
$\Rightarrow - 14 x + 28 = 0$
$\Rightarrow 14 x = 28$
$\Rightarrow x = 2$
$\therefore $ The point on x-axis is $(2 , 0)$ View full question & answer→Question 175 Marks
ABC is an equilateral triangle . If the coordinates of A and B are $(1 , 1)$ and $(- 1 , -1),$ find the coordinates of C.
Answer
$A B C$ is an equilateral triangle.
$\therefore A C=B C \text { and } A B=B C $
$\Rightarrow A C^2=B C^2 \text { and } A B^2=B C^2 $
$(x-1)^2+(y-1)^2=(x+1)^2+(y+1)^2 $
$\Rightarrow x^2+1-2 x+y^2+1-2 y=x^2+1+2 x+y^2+1+2 y $
$\Rightarrow-4 x-4 y=0 $
$\Rightarrow-4 x=4 y $
$\Rightarrow x=-y\ldots(1) $
$(1+1)^2+(1+1)^2=(x+1)^2+(y+1)^2 $
$\Rightarrow 8=x^2+1+2 x+y^2+1+2 y $
$\Rightarrow 8=y^2+1+2 x+y^2+1+2 y $
$\Rightarrow 2 y^2-6=0 $
$\Rightarrow y^2=3 $
$\Rightarrow y= \pm \sqrt{3}$
From $(1)$
$\therefore x= \pm \sqrt{3}$ View full question & answer→Question 185 Marks
$PQR$ is an isosceles triangle . If two of its vertices are $P (2 , 0)$ and $Q (2 , 5),$ find the coordinates of $R$ if the length of each of the two equal sides is $3.$
Answer
$ P Q=c$
$\therefore P R=Q R=3 \text { units } $
Let the coordinates of $R$ be on,
$ PR =\sqrt{( x -2)^2+( y -0)^2}$
$\Rightarrow 3=\sqrt{ x ^2+4-4 x + y ^2} $
squaring both sides ,
$ \Rightarrow 9= x ^2-4 x + y ^2+4$
$\Rightarrow x ^2-4 x + y ^2-5=0$
$\Rightarrow x ^2+ y ^2-4 x =5\ldots(1) $
$ \text { QR }=\sqrt{(x-2)^2+(y-5)^2}$
$\Rightarrow 3=\sqrt{x^2+4-4 x+y^2+25-10 y}$
$\Rightarrow 9=x^2+y^2-4 x-10 y+29$
$\Rightarrow 0=x^2+y^2-4 x-10 y+29 $
$\text { From (1) } 0=5-10 y+20$
$10 y=25$
$y=\frac{5}{2}$
$\Rightarrow x^2+\frac{25}{4}-4 x-5=0$
$\Rightarrow 4 x^2+25-16 x-20=0$
$\Rightarrow 4 x^2-16 x+5=0$
$D=(-16)^2-4(4)(5)$
$=256-80$
$=176$
$ \sqrt{ d }=\sqrt{176}=4 \sqrt{11}$
$x =\frac{16 \pm 4 \sqrt{11}}{2 \times 4}$
$=\frac{4 \pm 4 \sqrt{11}}{2}$
$=2+\frac{\sqrt{11}}{2}, 2-\frac{\sqrt{11}}{2} $
The coordinates of $R$ are $\left(2-\frac{\sqrt{11}}{2}, \frac{5}{2}\right)$ or $\left(2+\frac{\sqrt{11}}{2}, \frac{5}{2}\right)$ View full question & answer→Question 195 Marks
$ABCD$ is a square . If the coordinates of $A$ and $C$ are $(5 , 4)$ and $(-1 , 6) ;$ find the coordinates of $B$ and $D.$
Answer
Given $A B C D$ is a square.
$\therefore A B=B C$ (all sides of a square are equal)
$\sqrt{( x -5)^2+( y -4)^2}=\sqrt{( x +1)^2+( y -6)^2}$
squaring both sides,
$ x^2+25-10 x+y^2+16-8 y=x^2+1+2 x+y^2+36-12 y$
$\Rightarrow-12 x+4 y+4=0$
$\Rightarrow-3 x+y+1=0$
$y=3 x-1\ldots(1) $
Also, each angle in a square measures $90^{\circ}$
By pythogoras theorem,
$A B^2+B C^2=A C^2$
$ \Rightarrow(5-x)^2+(4-y)^2+(x+1)^2+(y-6)^2=36+4$
$\Rightarrow 25+x^2-10 x+16+y^2-8 y+x^2+1+2 x+y^2+36-12 y$
$\Rightarrow 2 x^2+2 y^2-8 x-20 y+38=0$
$\Rightarrow x^2+y^2-4 x-10 y+19=0$
$\Rightarrow x^2+(3 x-1)^2-4 x-10(3 x-1)+19=0$
$\Rightarrow x^2+9 x^2+1-6 x-4 x-30 x+10 \quad 10=0$
$\Rightarrow 10 x^2-40 x+30=0$
$\Rightarrow x^2-4 x+3=0$
$\Rightarrow x^2-3 x-x+3=0$
$\Rightarrow x(x-3)-1(x-3)=0$
$\Rightarrow(x-1)(x-3)=0$
$x=1,3 $
When,
$ x=1, y=3(1)-1=2 \quad(1,2)$
$x=3, y=3(3)-1=8 \quad(3,8) $
Thus, coordinates of $B$ and $D$ are $(1,2)$ and $(3,8)$ View full question & answer→Question 205 Marks
Prove that the points (0 , 2) , (1 , 1) , (4 , 4) and (3 , 5) are the vertices of a rectangle.
Answer
$A B=\sqrt{(0-1)^2+(2-1)^2}=\sqrt{2}$ units
$BC =\sqrt{(1-4)^2+(1-4)^2}=3 \sqrt{2}$ units
$C D=\sqrt{(4-3)^2+(4-5)^2}=\sqrt{2}$ units
$DA =\sqrt{(3-0)^2+(5-2)^2}=3 \sqrt{2}$ units
$A C=\sqrt{(4-0)^2+(4-2)^2}=\sqrt{20}=2 \sqrt{5}$ units
$BC =\sqrt{(3-1)^2+(5-1)^2}=\sqrt{20}=2 \sqrt{5}$ units
$\because A B=C D$ and $B C=D A$
Also, $AC = BD$
$\therefore ABCD$ is a rectangle. View full question & answer→Question 215 Marks
Prove that the points $(4 , 6) , (- 1 , 5) , (- 2, 0)$ and $(3 , 1)$ are the vertices of a rhombus.
Answer
$AB =\sqrt{(4+1)^2+(6-5)^2}=\sqrt{25+1}=\sqrt{26} \text { units } $
$BC =\sqrt{(-1+2)^2+(5-0)^2}=\sqrt{1+25}=\sqrt{26} \text { units } $
$CD =\sqrt{(-2-3)^2+(0-1)^2}=\sqrt{25+1}=\sqrt{26} \text { units } $
$DA =\sqrt{(3-4)^2+(1-6)^2}=\sqrt{1+25}=\sqrt{26} \text { units } $
$AC =\sqrt{(4+2)^2+(6-0)^2}=\sqrt{36+36}=36 \sqrt{2} \text { units } $
$BD =\sqrt{(-1-3)^2+(5-1)^2}=\sqrt{36+36}=16 \sqrt{2} \text { units } $
$\because AB = BC = CD = DA \text { and } AC \neq BD$
$\therefore ABCD$ is a rhombus. View full question & answer→Question 225 Marks
Prove that the points $(5 , 3) , (1 , 2), (2 , -2)$ and $(6 ,-1)$ are the vertices of a square.
Answer
$A B=\sqrt{(5-1)^2+(3-2)^2}=\sqrt{16+1}=\sqrt{17} \text { units } $
$B C=\sqrt{(1-2)^2+(2+2)^2}=\sqrt{1+16}=\sqrt{17} \text { units } $
$C D=\sqrt{(6-2)^2+(-1+2)^2}=\sqrt{16+1}=\sqrt{17} \text { units } $
$D A=\sqrt{(6-5)^2+(-1-3)^2}=\sqrt{1+16}=\sqrt{17} \text { units } $
$A C=\sqrt{(5-2)^2+(3+2)^2}=\sqrt{9+25}=\sqrt{34} \text { units } $
$B D=\sqrt{(6-1)^2+(-1-2)^2}=\sqrt{25+9}=\sqrt{34} \text { units } $
$\because A B=B C=C D=D A \text { and } A C=B D$
$\therefore ABCD$ is a square. View full question & answer→Question 235 Marks
Find the relation between x and y if the point M (x,y) is equidistant from R (0,9) and T (14 , 11).
Answer
$
\begin{aligned}
& \text { Given : } M R=M T \\
& \therefore M R^2=M T^2 \\
& (x-0)^2+(y-9)^2=(x-14)^2+(y-11)^2 \\
& x^2+y^2+81-18 y=x^2+196-28 x+y^2+121-22 y \\
& 81-18 y=196-28 x+121-22 y \\
& 28 x-18 y+22 y=196+121-81 \\
& 28 x+4 y=236 \\
& 7 x+y-58=0
\end{aligned}
$ View full question & answer→Question 245 Marks
Prove that the points $(7 , 10) , (-2 , 5)$ and $(3 , -4)$ are vertices of an isosceles right angled triangle.
Answer
$A B=\sqrt{(7+2)^2+(10-5)^2}=\sqrt{81+25}=\sqrt{106} \text { units } $
$B C=\sqrt{(-2-3)^2+(5+4)^2}=\sqrt{25+81}=\sqrt{106} \text { units } $
$A C=\sqrt{(7-3)^2+(10+4)^2}=\sqrt{16+196}=\sqrt{212} \text { units } $
$\because A B=B C$
$\therefore ABC$ is an isosceleles triangle
$A B^2+B C^2=100+106=212 $
$A C^2=212 $
$\because A B^2+B C^2=A C^2$
$\therefore ABC$ is also a right angled triangle. View full question & answer→Question 255 Marks
$x (1,2),Y (3, -4)$ and $z (5,-6)$ are the vertices of a triangle . Find the circumcentre and the circumradius of the triangle.
Answer
Circumcentre of $\triangle X Y Z$ will pass through the vertices $X, Y$ and $Z$
$ O X=O Y \quad \text { (radi of same circle), }$
$\Rightarrow O X^2=O Y^2$
$(a-1)^2+(b-2)^2=(a-3)^2+(b+4)^2$
$\Rightarrow 1-2 a+4-4 b=9-6 a+16+8 b$
$\Rightarrow 4 a-12 b=20$
$\Rightarrow a-3 b=5\ldots(1) $
$O Y=O Z$ (radii of same circle)
$O Y^2=O Z^2$
$ (a-3)^2+(b+4)^2=(a-5)^2+(b+6)^2$
$\Rightarrow 9-6 a+16+8 b=25-10 a+36+12 b$
$\Rightarrow 4 a-4 b=36$
$\Rightarrow a-b=9 \quad \ldots \ldots . .(2)$
$a-3 b=5\ldots(1)$
$\underline{a-b=9}$
$-2 b=-4$
$b=2$
$a=11 $
Circumcentre of $\triangle XYZ$ is $O (11,2)$
Circumradius $=\sqrt{(11-1)^2+(2-2)^2}=\sqrt{100}=10$ units View full question & answer→Question 265 Marks
$P(5 , -8) , Q (2 , -9)$ and $R(2 , 1)$ are the vertices of a triangle. Find tyhe circumcentre and the circumradius of the triangle.
Answer
Circumcircle of $\triangle PQR$ will pass through its vertices $P , Q$ and $R$.
$ O P=O Q$
$\Rightarrow O P^2=O Q^2$
$(x-5)^2+(y+8)^2=(x-2)^2+(y+9)^2$
$\Rightarrow 25-10 x+64+16 y=4-4 x+81+18 y$
$C-6 x-2 y+4=0$
$O Q=O R \quad \ldots \text { (radii of square circle) }$
$O Q^2=O R^2 $
$ (x-2)^2+(y+9)^2=(x-2)^2+(y-1)^2$
$\Rightarrow 81+18 y=1-2 y$
$\Rightarrow 20 y=-80$
$y=-4 \quad \ldots \ldots .(2)$
$-6 x+8+4=0 \quad \ldots . .[\text { from }(2)]$
$\Rightarrow-6 x=-12$
$\Rightarrow x=2 $
Circumcentre of $\triangle PQR$ is $O (2,-4)$
$ \text { Circumcentre }=\sqrt{(2-5)^2+(-4+8)^2}$
$=\sqrt{9+16}=\sqrt{25}=5 \text { units } $ View full question & answer→Question 275 Marks
$A(-2, -3), B(-1, 0)$ and $C(7, -6)$ are the vertices of a triangle. Find the circumcentre and the circumradius of the triangle.
Answer
Circumcircle of MBCwill pass through the vertices $A, B$ and $C . A B=O B$ (radii of same circle)
$\Rightarrow O A^2=O B^2 $
$(x+2)^2+(y+3)^2=(x+1)^2+Y^2 $
$\Rightarrow 4 x+4+9+6 y=2 x+1 $
$\Rightarrow 2 x+6 y=-12 \ldots . .(1)$
$O B=O C$ (radii of same circle)
$\Rightarrow OB ^2= OC ^2$
$(x+1)^2+y^2=(x-7)^2+(Y+6)^2$
$\Rightarrow 2 x+1=49-144 x+36+12$
$\Rightarrow 16 x-12=8 y$
$\Rightarrow 4 x-3 y=21\ldots(2)$
On solving $(1)$ and $(2)$
$4 x+12 y=-24 $
$4 x-3 y=21 $
$\underline{-\quad +\quad -} $
$15 y=-45 $
$y=-3 $
$\text { from (1) } $
$2 x+6(-3)=-12 $
$\Rightarrow 2 x=6 $
$\Rightarrow x=3$
Circumcentre of $\triangle ABC$ is $(3,-3)$ and
Circumradius $=\sqrt{(3+2)^2+(-3+3)^2}=\sqrt{25}=5$ units View full question & answer→Question 285 Marks
The centre of a circle passing through $P(8, 5)$ is $(x+l , x-4).$ Find the coordinates of the centre if the diameter of the circle is $20$ units.
Answer
Given diameter of the circle $=20$ units.
$\therefore$ radius $=10$ units
$OP =10$
$\sqrt{(x+1-8)^2+(x-4-5)^2}=10$
squaring both sides,
$x^2+49-14 x+x^2=81-18 x=100 $
$\Rightarrow 2 x^2-32 x+30=0 $
$\Rightarrow x^2-16 x+15=0 $
$\Rightarrow x^2-15 x-x+15=0 $
$\Rightarrow(x-15)(x-1)=0 $
$\Rightarrow x=15 \text { or } 1$
Coordinates of 0 when $x=15$ are $(16,11)$
Coordinates of 0 when $x=1$ are $(2,-3)$ View full question & answer→Question 295 Marks
Find the coordinates of $O,$ the centre passing through $A( -2, -3), B(-1, 0)$ and $C(7, 6).$ Also, find its radius.
Answer
let o $(x, y)$ be the centre of the circle.
$ O A=O B \quad \text { (radii of square circle) }$
$\Rightarrow O A^2=O B^2$
$(x+2)^2+(y+3)^2=(x+1)^2+(y-0)^2$
$\Rightarrow x^2+4+4 x+y^2+9+6 y=x^2+1+2 x+y^2$
$\Rightarrow 2 x+6 y+12=0$
$\Rightarrow x+3 y+6=0\ldots(1) $
Similarly, $O B=O C$
$\Rightarrow OB ^2= OC ^2$
$ (x+1)^2+(y+0)^2=(x+7)^2+(y-6)^2$
$\Rightarrow x^2+1+2 x+y^2=x^2+49-14 x+y^2+36+12 y$
$\Rightarrow 16 x-12 y-84=0$
$\Rightarrow 4 x-3 y-21=0 \ldots .(2)$
$\Rightarrow 4 x+12 y+24=0 \ldots .(1) $
on solving (1)\& (2) we get,
$ -15 y-45=0$
$\Rightarrow y=-3 $
from $(1)$
$ x-9+6=0$
$\Rightarrow x=3 $
Thus coordinate of $O$ are $(3,-3)$
$ \text { Radius }=\sqrt{(3+2)^2+(-3+3)^2}$
$=\sqrt{25+0}$
$=\sqrt{25} \text { units }$
$=5 \text { units } $
View full question & answer→