The correct option is option (c) $p=\pm 4$ The given points are $A(4, p)$ and $B(1,0)$ and $A B=5$. => $\left(x_{1}=4, y_{1}=p\right)$ and $\left(x_{2}=1, y_{2}=0\right)$ $\mathrm{AB}=5$...
The correct option is option (c) 4 units $\mathrm{A}(0,3), 0(0,0)$ and $\mathrm{B}(5,0)$ are the three vertices of a rectangle; let $\mathrm{C}$ be the fourth vertex Then, the length of the...
The correct option is option (c) 8 sq units The given points are $A(3,0), B(7,0)$ and $C(8,4)$. $\left(x_{1}=3, y_{1}=0\right),\left(x_{2}=7, y_{2}=0\right)$ and $\left(x_{3}=8, y_{3}=4\right)$ Area...
The correct option is option (c) $2 \mathrm{a}=\mathrm{b}$ The given points are $A(1,2), 0(0,0)$ and $C(a, b)$ $\left(x_{1}=1, y_{1}=2\right),\left(x_{2}=0, y_{2}=0\right)$ and $\left(x_{3}=a,...
The correct option is option $(b) k=6$ The given points are $A(2,3), B(5, k)$ and $C(6,7)$ $\left(x_{1}=2, y_{1}=3\right),\left(x_{2}=5, y_{2}=k\right)$ and $\left(x_{3}=6, y_{3}=7\right)$. Points...
The correct option is option (d) right - angled Let $\mathrm{P}(0,6), \mathrm{Q}(-5,3)$ and $\mathrm{R}(3,1)$ be the given points. Then, $\mathrm{PQ}=\sqrt{(-5-0)^{2}+(3-6)^{2}}$...
The correct option is option (a) isosceles Let $\mathrm{A}(-4,0), 8(4,0)$ and $\mathrm{C}(0,3)$ be the given points. Then, $\mathrm{AB}=\sqrt{(4+4)^{2}+(0-0)^{2}}$ $=\sqrt{(8)^{2}+(0)^{2}}$...
The correct option is option (c) $(-4,-15)$ Two vertices of $\triangle \mathrm{ABC}$ are $\mathrm{A}(-1,4)$ and $\mathrm{B}(5,2)$. Let the third vertex be $C(a, b)$. => the coordinates of its...
The correct option is option (d) $(4,0)$ The given point are $A(-1,0), B(5,-2)$ and $C(8,2)$. $\left(x_{1}=-1, y=0\right),\left(x_{2}=5, y=-2\right)$ and $\left(x_{3}=8, y_{3}=2\right)$ Let $G(x,...
The correct option is option (c) $\left(\frac{7}{2}, \frac{9}{2}\right)$ $\mathrm{D}$ is the midpoint of BC => the coordinates of $\mathrm{D}$ are $ \begin{aligned}...
The correct option is option (b) $2: 9$ Let the line $2 \mathrm{x}+\mathrm{y}-4=0$ divide the line segment in the ratio $\mathrm{k}: 1$ at the point $\mathrm{P}$. Usingsection formula the...
The correct option is option (b) -1 The given ports are $A(-3, b)$ and $B(1, b+4)$. => $\left(x_{1}=-3, y_{1}=b\right)$ and $\left(x_{2}=1, y_{2}=b+4\right)$ $\mathrm{x}=\frac{[(-3)+1]}{2}$...
The correct option is option(d) $1: 2$ Let $\mathrm{A} \mathrm{B}$ be divided by the $\mathrm{y}$-axis in the ratio $\mathrm{k}: 1$ at the point $\mathrm{P}$. using section formula, the coordinates...
The correct option is option (c) $1: 2$ Let $A B$ be divided by the $x$-axis in the ratio $k: 1$ at the point $P$. using section formula, the coordinates of $P$ are $\mathrm{P}\left(\frac{5...
The correct option is option (b) 4 units The $y$-coordinate the distance of the point from the $x$-axis => the y-coordinate is 4 .
The correct option is option (c) $(3,0)$ Let $p(x, 0)$ be the point on $x$-axis. Then as per the question $\mathrm{AP}=\mathrm{BP} \Rightarrow \mathrm{AP}^{2}=\mathrm{BP}^{2}$...
The correct option is option (b) 26 The given points are $A(-6,7)$ and $B(-1,-5)$. So $\mathrm{AB}=\sqrt{(-6+1)^{2}+(7+5)^{2}}$ $=\sqrt{(-5)^{2}+(12)^{2}}$ $=\sqrt{25+144}$ $=\sqrt{169}$ $=13$ =>...
The correct option is option (d) IV Let $(x, y)$ be the coordinates of $P$. Then, $x=\frac{2 \times 5+3 \times 2}{2+3}=\frac{10+6}{5}=\frac{16}{5}$ $y=\frac{2 \times 2+3...
The correct option is option (a) $(2,5)$ Let $(x, y)$ be the coordinates of $A$. then, $0=\frac{-2+x}{2} \Rightarrow x=2$ $4=\frac{3+y}{2} \Rightarrow y=8-3=5$ Therefore, the coordinates of A are...
The correct option is option (c) $-4$ $\mathrm{AQ}: \mathrm{BQ}=2: 1$. $\mathrm{y}=\frac{2 \times(-5)+1 \times(-2)}{2+1}$ $\begin{aligned} &=\frac{-10-2}{3} \\ &=-4 \end{aligned}...
The correct option is option (a) $(-6,7)$ Let $(\mathrm{x}, \mathrm{y})$ be the coordinates of the other end of the diameter. $-2=\frac{2+x}{2} \Rightarrow x=-6$ $5=\frac{3+y}{2} \Rightarrow...
The correct option is option (b) $(3,5)$ Since, the point P divides the me segment pining the points $A(1,3)$ and $B(4,6)$ in the ratio $2: 1$. Then, Coordinates of $P=\left(\frac{2 \times 4+1...
The correct option is option(a) 5 $A C$ and $B D$ are two diagonals of the rectangle $A B C D$. So $\mathrm{BD}=\sqrt{(4-0)^{2}+(0-3)^{2}}$ $=\sqrt{(4)^{2}+(-3)^{2}}$ $=\sqrt{16+9}$ $=\sqrt{25}$...
The correct option is option (a) $-8$ The point $P\left(\frac{a}{2}, 4\right)$ is the midpoint of the line segment joining the points $A(-6,5)$ and $\mathrm{B}(-2,3)$ $\frac{a}{2}=\frac{-6-2}{2}$...
The correct option is option (b) $\frac{1}{2} a b$ Let $A\left(x_{1}=a, y_{1}=0\right), 0\left(x_{2}=0, y_{2}=0\right)$ and $B\left(x_{3}=0, y_{3}=b\right)$ be the given vertices. So $...
The correct option is option (c) 6 Let $A\left(x_{1}=5, y_{1}=0\right), B\left(x_{2}=8, y_{2}=0\right)$ and $C\left(x_{3}=0, y_{3}=4\right)$ be the vertices of the triangle. Then, $\text { Area...
The correct option is option (a) -63 Let $A\left(x_{1}=x, y_{1}=2\right), B\left(x_{2}=-3, y_{2}=-4\right)$ and $C\left(x_{3}=7, y_{3}=-5\right)$ be collinear points. Then...
The correct option is option (b) 4 The diagonals of a parallelogram bisect each other. The vertices of the $11 \mathrm{gm} \mathrm{ABCD}$ are $\mathrm{A}(1,3), \mathrm{B}(-1,2) \text { and } C(2,5)...
The correct option is option (d) 12 Let $\mathrm{A}(0,4), \mathrm{B}(0,0)$ and $\mathrm{C}(3,0)$ be the given vertices. $\mathrm{AB}=\sqrt{(0-0)^{2}+(4-0)^{2}}=\sqrt{16}=4$...
The correct option is option (c) $\frac{16}{5}$ The point $C(k, 4)$ dives the join of the points $A(2,6)$ and $B(5,1)$ in the ratio $2: 3$. So $\mathrm{k}=\frac{2 \times 5+3 \times...
The correct option is option b) 7 Since $\mathrm{R}(5,6)$ is the midpoint of the line segment AB joining the points $\mathrm{A}(6,5)$ and $\mathrm{B}(4, \mathrm{y})$, therefore $\frac{5+y}{2}=6$...
The correct option is option (b) $(2,0)$ Let $\mathrm{P}(\mathrm{x}, 0)$ the point on $\mathrm{x}$-axis, then $\mathrm{AP}=\mathrm{BP} \Rightarrow \mathrm{AP}^{2}=\mathrm{BP}^{2}$...
The correct option is option(c) 4 The distance of a point $(x . y)$ from $x$-axis is $|y|$. Since, the point is $(-3,4)$. => its distance from $\mathrm{x}$-axis is $|4|=4$
The correct option is option (d) 10 The distance of a point $(x, y)$ from the origin $0(0,0)$ is $\sqrt{x^{2}+y^{2}}$ Let $\mathrm{P}(\mathrm{x}=-6, \mathrm{y}=8)$ be the gen point. Then $0...
The points are $A(2,3), B(4, k)$ and $C(6,-3)$ $\left(x_{1}=2, y_{1}=3\right),\left(x_{2}=4, y_{2}=k\right)$ and $\left(x_{3}=6, y_{3}=-3\right)$ Since, the points $\mathrm{A}, \mathrm{B}$ and...
Let the required ratio be $\mathrm{k}: 1$ Using section formula, the coordinates of $\mathrm{C}$ are $C\left(\frac{7 k+2}{k+1}, \frac{8 k+3}{k+1}\right)$ $ \begin{aligned} &\frac{7...
The points are $A(2,2), B(-4,-4)$ and $C(5,-8)$. $\left(x_{1}=2, y_{1}=2\right),\left(x_{2}=-4, y_{2}=-4\right)$ and $\left(x_{3}=5, y_{3}=-8\right)$ $G(x, y)$ be the centroid of $\Delta A B C$...
The given points are $A(a, b), B(b, c)$ and $C(c, a)$ $\left(\mathrm{x}_{1}=\mathrm{a}, \mathrm{y}_{1}=\mathrm{b}\right),\left(\mathrm{x}_{2}=\mathrm{b}, \mathrm{y}_{2}=\mathrm{c}\right)$ and...
Let the point $P(x, y)$ be equidistant from the points $A(7,1)$ and $B(3,5)$ $\mathrm{PA}=\mathrm{PB}$ $\Rightarrow \mathrm{PA}^{2}=\mathrm{PB}^{2}$...
The points $A(4,3)$ and $B(x, 5)$ lie on the circle with center $O(2,3)$. => $O A=0 B$ $ \begin{aligned} &\Rightarrow \sqrt{(x-2)^{2}+(5-3)^{2}}=\sqrt{(4-2)^{2}+(3-3)^{2}} \\...
The points $(3, a)$ lies on the line $2 x-3 y=5$. If $(3, a)$ lies on the line $2 x-3 y=5$ => $2 x-3 y=5$ $\Rightarrow(2 \times 3)-(3 \times a)=5$ $\Rightarrow 6-3 a=5$ $\Rightarrow 3...
points are $\mathrm{A}\left(\frac{-8}{5}, 2\right)$ and $\mathrm{B}\left(\frac{2}{5}, 2\right)$ => $\left(x_{1}=\frac{-8}{5}, y_{1}=2\right)$ and $\left(x_{2}=\frac{2}{5}, y_{2}=2\right)$ $...
Let $P(x, 0)$ be the point on $x$-axis. Then $\mathrm{AP}=\mathrm{BP} \Rightarrow \mathrm{AP}^{2}=\mathrm{BP}^{2}$ $\Rightarrow(x+1)^{2}+(0-0)^{2}=(x-5)^{2}+(0-0)^{2}$ $\Rightarrow x^{2}+2...
Since, the point $C(k, 4)$ divides the join of $A(2,6)$ and $B(5,1)$ in ratio $2: 3$. So $ \begin{aligned} &\mathrm{k}=\frac{2 \times 5+3 \times 2}{2+3} \\ &=\frac{10+6}{5} \\...
The given vertices are $A(7,-3), B(5,3)$ and $C(3,-1)$. $\mathrm{D}$ and $\mathrm{E}$ are the midpoints of $\mathrm{BC}$ and $\mathrm{AC}$ respectively. => Coordinates of $D=\left(\frac{5+3}{2},...
The vertices of the rectangle $\mathrm{ABCD}$ are $\mathrm{A}(2,-1), \mathrm{B}(5,-1), \mathrm{C}(5,6)$ and $\mathrm{D}(2,6)$. Coordinates of midpoint of $\mathrm{AC}=\left(\frac{2+5}{2},...
Let $k$ be the ratio in which the point $P(x, 2)$ divides the line joining the points $\mathrm{A}\left(\mathrm{x}_{1}=12, \mathrm{y}_{1}=5\right) \text { and } \mathrm{B}\left(\mathrm{x}_{2}=4,...
The given points are $P(k-1,2), A(3, k)$ and $B(k, 5)$. $ \begin{aligned} &\because \mathrm{AP}=\mathrm{BP} \\ &\therefore \mathrm{AP}^{2}=\mathrm{BP}^{2} \\...
The given vertices are $B(4,0), C(4,3)$ and $D(0,3)$ Since, BD one of the diagonals So $ \begin{aligned} &B D=\sqrt{(4-0)^{2}+(0-3)^{2}} \\ &=\sqrt{(4)^{2}+(-3)^{2}} \\ &=\sqrt{16+9} \\...
The given ports are $A(0,2), B(3, p)$ and $C(p, 5)$. $\begin{aligned} &A B=A C \Rightarrow A B^{2}=A C^{2} \\ &\Rightarrow(3-0)^{2}+(p-2)^{2}=(p-0)^{2}+(5-2)^{2} \\ &\Rightarrow...
The given points are $A(-1, y), 8(5,7)$ and $0(2,-3 y)$. Since, AO and $B 0$ are the radii of the circle. So $\mathrm{AO}=\mathrm{BO} \Rightarrow \mathrm{AO}^{2}=\mathrm{BO}^{2}$...
Let $A\left(x_{1}=0, y_{1}=-1\right), B\left(x_{2}=2, y_{2}=1\right)$ and $C\left(x_{3}=0, y_{3}=3\right)$ be the given points. => $\text { Area }(\Delta...
Let $\mathrm{A}\left(\mathrm{x}_{1}=3, \mathrm{y}_{1}=9\right), \mathrm{B}\left(\mathrm{x}_{2}=\mathrm{a}, \mathrm{y}_{3}=\mathrm{b}\right)$ and $\mathrm{C}\left(\mathrm{x}_{3}=4,...
$A(a, 0), B(0, b)$ and $C(1,1)$ $\left(x_{1}=a, y_{1}=0\right) \cdot\left(x_{2}=0, y_{2}=b\right)$ and $\left(x_{3}=1, y_{3}=1\right)$. Since, the points are collinear. So,...
Let $A\left(x_{1}=x, y_{1}=y\right), B\left(x_{2}=-5, y_{2}=7\right)$ and $C\left(x_{3}=-4, y_{3}=5\right)$ be the given points The given points are collinear if...
Let $A\left(x_{1}=2, y_{1}=1\right), B\left(x_{2}=x, y_{2}=y\right)$ and $C\left(x_{3}=7, y_{3}=5\right)$ be the given points The given points are collinear if...
Let $A\left(x_{1}=8, y_{1}=1\right), B\left(x_{2}=3, y_{2}=-2 k\right)$ and $C\left(x_{3}=k, y_{3}=-5\right)$ be the given points They are collinear iff...
Let $\mathrm{A}\left(\mathrm{x}_{1}=-3, \mathrm{y}_{1}=9\right), \mathrm{B}\left(\mathrm{x}_{2}=2, \mathrm{y}_{2}=\mathrm{y}\right)$ and $\mathrm{C}\left(\mathrm{x}_{3}=4, \mathrm{y}_{3}=-5\right)$...
$\mathrm{P}(1,4), \mathrm{Q}(3, \mathrm{y})$ and $\mathrm{R}(-3,16)$ are the given points. Then: $\left(x_{1}=1, y_{1}=4\right),\left(x_{2}=3, y_{2}=y\right)$ and $\left(x_{3}=-3, y_{3}=16\right)$...
$\mathrm{A}(-3,12), \mathrm{B}(7,6)$ and $C(\mathrm{x}, 9)$ are the given points. Then: $\left(x_{1}=-3, y_{1}=12\right),\left(x_{2}=7, y_{2}=6\right) \text { and }\left(x_{3}=x, y_{3}=9\right)$...
Let $A\left(x_{1}, y_{1}\right)=A(x, 2), B\left(x_{2}, y_{2}\right)=B(-3,-4)$ and $C\left(x_{3}, y_{3}\right)=C(7,-5)$. => the condition for three collinear points is...
(iii) Let $\mathrm{A}\left(\mathrm{x}_{1}=5, \mathrm{y}_{1}=1\right), \mathrm{B}\left(\mathrm{x}_{2}=1, \mathrm{y}_{2}=-1\right)$ and $\mathrm{C}\left(\mathrm{x}_{3}=11, \mathrm{y}_{3}=4\right)$ be...
(i) Let $A\left(x_{1}=2, y_{1}=-2\right), B\left(x_{2}=-3, y_{2}=8\right)$ and $C\left(x_{3}=-1, y_{3}=4\right)$ be the given points. $\text { Now }...
Let $A\left(x_{1}=-2, y_{1}=5\right), B\left(x_{2}=k, y_{2}=-4\right)$ and $C\left(x_{3}=2 k+1, y_{3}=10\right)$ be the vertices of the triangle, So $$ \begin{aligned} &\text { Area }(\Delta...
Let $A\left(x_{1}, y_{1}\right)=A(k+1,1), B\left(x_{2}, y_{2}\right)=B(4,-3)$ and $C\left(x_{3}, y_{3}\right)=C(7,-k)$ now Area $(\Delta...
Let $A\left(x_{1}, y_{1}\right)=A(1,-3), B\left(x_{2}, y_{2}\right)=B(4, p)$ and $C\left(x_{3}, y_{3}\right)=C(-9,7)$ Now Area $(\Delta...
Let $(x, y)$ be the coordinates of $D$ and $\left(x^{\prime}, y^{\prime}\right)$ be thee coordinates of $E$. since, the diagonals of a parallelogram bisect each other at the same point, therefore $$...
Let $\left(x_{2}, y_{2}\right)$ and $\left(x_{3}, y_{3}\right)$ be the coordinates of $B$ and $C$ respectively. Since, the coordinates of $A$ are $(1,-4)$, therefore $\frac{1+\mathrm{x}_{2}}{2}=2...
The vertices of the triangle are $A(7,-3), B(5,3), C(3,-1)$ Coordinates of $D=\left(\frac{5+3}{2}, \frac{3-1}{2}\right)=(4,1)$ For the area of the triangle ADC, let $\mathrm{A}\left(\mathrm{x}_{1},...
The verticals of the triangle are $A(2,1), B(4,3)$ and $C(2,5)$. Coordinates of midpoint of $\mathrm{AB}=\mathrm{P}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)=\left(\frac{2+4}{2},...
joining $A$ and $C$, we get two triangles $A B C$ and $A C D$. Let $A\left(x_{1}, y_{1}\right)=A(-5,7), B\left(x_{2}, y_{2}\right)=B(-4,-5), C\left(x_{3}, y_{3}\right)=C(-1,-6)$ and....
joining $A$ and $C$, we get two triangles $A B C$ and $A C D$. Let $A\left(x_{1}, y_{1}\right)=A(-3,-1), B\left(x_{2}, y_{2}\right)=B(-2,-4), C\left(x_{3}, y_{3}\right)=C(4,-1)$ and. Then...
joining P and R, we get two triangles PQR and PRS. Let $\mathrm{P}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)=\mathrm{P}(-5,-3), \mathrm{Q}\left(\mathrm{x}_{2},...
By joining $A$ and $C$, we get two triangles $A B C$ and $A C D$. $\mathrm{A}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)=\mathrm{A}(3,-1), \mathrm{B}\left(\mathrm{x}_{2},...
(iii) $\mathrm{A}(3,8), \mathrm{B}(-4,2)$ and $\mathrm{C}(5,-1)$ are verticals of $\triangle \mathrm{ABC}$. Then, $\left(x_{1}=3, y_{1}=8\right),\left(x_{2}=-4, y_{2}=2\right) \text { and...
(i) $\mathrm{A}(1,2), \mathrm{B}(-2,3)$ and $\mathrm{C}(-3,-4)$ are the vertices of $\triangle \mathrm{ABC}$. Then, $\left(x_{1}=1, y_{1}=2\right),\left(x_{2}=-2, y_{2}=3\right) \text { and...
The midpoint of $\mathrm{AB}$ is $\left(\frac{-10-2}{2}, \frac{4+10}{2}\right)=\mathrm{P}(-6,2)$ Let $k$ be the ratio in which P divides $C D$. So...
Since, the points $\mathrm{P}, \mathrm{Q}, \mathrm{R}$ and $\mathrm{S}$ are the midpoint of $\mathrm{AB}, \mathrm{BC}, \mathrm{CD}$ and $\mathrm{DA}$ respectively. Then Coordinates of...
Let $k$ be the ratio in which $P(-1, y)$ divides the line segment joining the points $\mathrm{A}(-3,10) \text { and } \mathrm{B}(6,-8)$ $$ \begin{aligned} &(-1, y)=\left(\frac{k(6)-3}{k+1},...
Let $(0, y)$ be the coordinates of $B$. Then $0=\frac{-3+y}{2} \Rightarrow y=3$ => the coordinates of $\mathrm{B}$ are $(0,3)$ Since, $A B=B C=A C$ and by symmetry the coordinates of $A$ lies on...
Let, $(x, 0)$ be the coordinates of $R$. $0=\frac{-4+x}{2} \Rightarrow x=4$ Thus, the coordinates of $\mathrm{R}$ are $(4,0)$. Since, $P Q=Q R=P R$ and the coordinates of $P$ lies on $y$-axis. Let...
The line segment joining the points $\mathrm{A}(3,-3)$ and $\mathrm{B}(-2,7)$ is divided by $\mathrm{x}$-axis. Let the required ratio be $\mathrm{k}: 1$. So, $0=\frac{\mathrm{k}(7)-3}{\mathrm{k}+1}...
Let the point $P\left(\frac{1}{2}, y\right)$ divides the line segment joining the points $A(3,-5)$ and $B(-7,9)$ in the ratio $\mathrm{k}: 1$. Then $\left(\frac{1}{2},...
$y$-axis divides the e segment pining the points $(-4,7)$ and $(3,-7)$ in the ratio $k: 1$. Then $0=\frac{3 \mathrm{k}-4}{\mathrm{k}+1}$ $\Rightarrow 3 \mathrm{k}=4$ $\Rightarrow...
Let $A(1,-2), B(3,6)$ and $C(5,10)$ be the three vertices of a parallelogram $A B C D$ and the fourth vertex be $D(a, b)$ Join AC and BD intersecting at 0 . Since, the diagonals of a parallelogram...
The points are $P(a,-11), Q(5, b), R(2,15)$ and $S(1,1)$ Join PR and QS, intersecting at 0 . Since,the diagonals of a parallelogram bisect each other Therefore, 0 is the midpoint of PR as well as $Q...
The points are $A(3,1), B(0,-2), C(1,1)$ and $D(4,4)$ Join $\mathrm{AC}$ and $\mathrm{BD}$, intersecting at 0 . Since, the diagonals of a parallelogram bisect each other. Midpoint of...
Two vertices of $\triangle \mathrm{ABC}$ are $\mathrm{B}(-3,1)$ and $\mathrm{C}(0,-2)$. Let the third vertex be $\mathrm{A}(\mathrm{a}, \mathrm{b})$. => the coordinates of its centroid are...
Two vertices of $\triangle \mathrm{ABC}$ are $\mathrm{A}(1,-6)$ and $\mathrm{B}(-5,2)$. Let the third vertex be $\mathrm{C}(\mathrm{a}, \mathrm{b})$. => the coordinates of its centroid are...
$\left(x_{1}=-1, y_{1}=0\right),\left(x_{2}=5, y_{2}=-2\right)$ and $\left(x_{3}=8, y_{3}=2\right)$ Let $G(x, y)$ be the centroid of the $\triangle \mathrm{ABC}$. Then,...
The vertices of $\triangle \mathrm{ABC}$ are $\mathrm{A}(0,-1), \mathrm{B}(2,1)$ and $\mathrm{C}(0,3)$. Let $A D, B E$ and $C F$ be the medians of $\triangle A B C$. Let $\mathrm{D}$ be the midpoint...
Let the line $x-y-2=0$ divide the line segment joining the points $A(3,-1)$ and $B(8,9)$ in the ratio $\mathrm{k}: 1$ at $\mathrm{P}$. => the coordinates of $\mathrm{P}$ are...
Let $A B$ be divided by the $x$-axis in the ratio $k: 1$ at the point $P$. using section formula the coordination of $P$ are $P=\left(\frac{5 k+2}{k+1}, \frac{6 k-3}{k+1}\right)$ But P lies on the...
Let the point $P(-3, k)$ divide the line $A B$ in the ratio $s: 1$ using the section formula: $\mathrm{x}=\frac{\mathrm{mx}_{1}+\mathrm{nx}_{1}}{\mathrm{~m}+\mathrm{n}},...
Let the point $\mathrm{P}(\mathrm{m}, 6)$ divide the line $\mathrm{AB}$ in the ratio $\mathrm{k}: 1$. using the section formula: $\mathrm{x}=\frac{\mathrm{mx}_{2}+\mathrm{n}...
Let $k: 1$ be the ratio in which the point $P\left(\frac{3}{4}, \frac{5}{12}\right)$ divides the line segment joining the points $A\left(\frac{1}{2}, \frac{3}{2}\right)$ and $(2,-5)$. Then...
Let the point $P(2,5)$ divide $A B$ in the ratio $k: 1$. using section formula, the coordinates of $\mathrm{P}$ are $x=\frac{-6 k+8}{k+1}, y=\frac{9 k+2}{k+1}$ It is given that the coordinates of...
$C(2,-3)$ is the center of the given circle. Let $A(a, b)$ and $B(1,4)$ be the two end-points of the given diameter $\mathrm{AB}$. Then, the coordinates of $\mathrm{C}$ are $x=\frac{a+1}{2},...
The given points are $A(-2,9)$ and $B(6,3)$ => $C(x, y)$ is the midpoint of $A B$. $$ \begin{aligned} &x=\frac{x_{1}+x_{2}}{2}, y=\frac{y_{1}+y_{2}}{2} \\ &\Rightarrow x=\frac{-2+6}{2},...
points are $A(2 a, 4)$ and $B(-2,3 b)$ Let $C(1,2 a+1)$ be the mid-point of $A B$. $\mathrm{x}=\frac{\mathrm{x}_{1}+\mathrm{x}_{2}}{2}, \mathrm{y}=\frac{\mathrm{y}_{1}+\mathrm{y}_{2}}{2}$...
points are $A(6,-5)$ and $B(-2,11)$. Let $(x, y)$ be the midpoint of $A B$. Then, $\mathrm{x}=\frac{\mathrm{x}_{1}+\mathrm{x}_{2}}{2}, \mathrm{y}=\frac{\mathrm{y}_{1}+\mathrm{y}_{2}}{2}$...
(i) points are $\mathrm{A}(3,0)$ and $\mathrm{B}(-5,4)$. Let $(x, y)$ be the midpoint of $A B$. Then: $x=\frac{x_{1}+x_{2}}{2}, y=\frac{y_{1}+y_{2}}{2}$ $\Rightarrow \mathrm{x}=\frac{3+(-5)}{2},...
Let $P$ and $Q$ be the points of trisection of $A B$. => P divides $A B$ in the radio $1: 2$ So, the coordinates of $P$ are $$ \begin{aligned} &x=\frac{\left(m x_{2}+n x_{1}\right)}{(m+n)},...
points are $A(1,6)$ and $B(5,-2)$. Then, $P(x, y)$ is a point that devices the line $A B$ in the ratio $1: 3$ Using the section formula: $$ \begin{aligned} &x=\frac{\left(m x_{2}+n...
Since, the points $P, Q, R$ and $S$ divide the line segment joining the points $\mathrm{A}(1,2)$ and $\mathrm{B}(6,7)$ in five equal parts, so...
Let the coordinates of $A$ be $(x, y)$. Here $\frac{P A}{P Q}=\frac{2}{5} .$ So, $\mathrm{PA}+\mathrm{AQ}=\mathrm{PQ}$ $\Rightarrow \mathrm{PA}+\mathrm{AQ}=\frac{5 \mathrm{PA}}{2}...
The coordinates of the points $\mathrm{A}$ and Bare $(-2,-2)$ and $(2,-4)$ respectively, where $\mathrm{AP}=\frac{3}{7} \mathrm{AB}$ and $\mathrm{P}$ lies on the line segment $\mathrm{AB}$. So...
The end points of $A B$ are $A(-5,11)$ and $B(4,-7)$. => $\left(x_{1}=-5, y_{1}=11\right)$ and $\left(x_{2}=4, y_{2}=-7\right)$ $m=7$ and $n=2$ Let the required point be $P(x, y)$. using section...
The end points of $\mathrm{AB}$ are $\mathrm{A}(-1,7)$ and $\mathrm{B}(4,-3)$. => $\left(x_{1}=-1, y_{1}=7\right)$ and $\left(x_{2}=4, y_{2}=-3\right)$ $m=2$ and $n=3$ Let the required point be...
(iii) The points are $\mathrm{A}(0,-4), \mathrm{B}(6,2) \mathrm{C}(3,5)$ and $\mathrm{D}(-3,-1)$. $\mathrm{AB}=\sqrt{(6-0)^{2}+\{2-(-4)\}^{2}}=\sqrt{(6)^{2}+(6)^{2}}=\sqrt{36+36}=\sqrt{72}=6...
(i) The points are $A(-4,-1), B(-2,-4) C(4,0)$ and $D(2,3)$ $$ \begin{aligned} &\mathrm{AB}=\sqrt{\{-2-(-4)\}^{2}+\{-4-(-1)\}^{2}}=\sqrt{(2)^{2}+(-3)^{2}}=\sqrt{4+9}=\sqrt{13} \text { units } \\...
vertices are $A(1,2), B(4,3), C(6,6)$ and $D(3,5)$. $$ \begin{aligned} &\text { AB }=\sqrt{(1-4)^{2}+(2-3)^{2}}=\sqrt{(-3)^{2}+(-1)^{2}} \\ &=\sqrt{9+1}=\sqrt{10} \\ &B...
points are $A(2,1), B(5,2), C(6,4)$ and $D(3,3)$ $$ \begin{aligned} &\mathrm{AB}=\sqrt{(5-2)^{2}+(2-1)^{2}}=\sqrt{(3)^{2}+(1)^{2}}=\sqrt{9+1}=\sqrt{10} \text { units } \\...
points are $A(6,1), B(8,2), C(9,4)$ and $D(7,3)$. $\mathrm{AB}=\sqrt{(6-8)^{2}+(1-2)^{2}}=\sqrt{(-2)^{2}+(-1)^{2}}$ $=\sqrt{4+1}=\sqrt{5}$...
points are $\mathrm{A}(3,0), \mathrm{B}(4,5), \mathrm{C}(-1,4)$ and $\mathrm{D}(-2,-1)$ $\mathrm{AB}=\sqrt{(3-4)^{2}+(0-5)^{2}}=\sqrt{(-1)^{2}+(-5)^{2}}$ $=\sqrt{1+25}=\sqrt{26}$...
points are $A(-3,2), B(-5,-5), C(2,-3)$ and $D(4,4)$. $\mathrm{AB}=\sqrt{(-5+3)^{2}+(-5-2)^{2}}=\sqrt{(-2)^{2}+(-7)^{2}}=\sqrt{4+49}=\sqrt{53}$ units...
(iii) points are $\mathrm{P}(0,-2), \mathrm{Q}(3,1), \mathrm{R}(0,4)$ and $\mathrm{S}(-3,1)$ $$ \begin{aligned} &\mathrm{PQ}=\sqrt{(3-0)^{2}+(1+2)^{2}}=\sqrt{(3)^{2}+(3)^{2}}=\sqrt{9+9}=\sqrt{18}=3...
(i) The given points are $\mathrm{A}(3,2), \mathrm{B}(0,5), \mathrm{C}(-3,2)$ and $\mathrm{D}(0,-1)$. $\mathrm{AB}=\sqrt{(0-3)^{2}+(5-2)^{2}}=\sqrt{(-3)^{2}+(3)^{2}}=\sqrt{9+9}=\sqrt{18}=3 \sqrt{2}$...
The given points are $0(0,0) \mathrm{A}(3, \sqrt{3})$ and $\mathrm{B}(3,-\sqrt{3})$. $0 A=\sqrt{(3-0)^{2}+\{(\sqrt{3})-0\}^{2}}=\sqrt{(3)^{2}+(\sqrt{3})^{2}}=\sqrt{9+3}=\sqrt{12}=2 \sqrt{3}$ units...
Let points be $\mathrm{A}(-5,6) \mathrm{B}(3,0)$ and $\mathrm{C}(9,8)$ $\mathrm{AB}=\sqrt{(3-(-5))^{2}+(0-6)^{2}}=\sqrt{(8)^{2}+(-6)^{2}}=\sqrt{64+36}=\sqrt{100}=10$ units...
Let points be $\mathrm{A}(-3,-3), \mathrm{B}(3,3)$ and $C(-3 \sqrt{3}, 3 \sqrt{3})$. Now $$ \begin{aligned} \mathrm{AB} &=\sqrt{(-3-3)^{2}+(-3-3)^{2}}=\sqrt{(-6)^{2}+(-6)^{2}} \\...
Points are $\mathrm{A}(2,4), \mathrm{B}(2,6)$ and $\mathrm{C}(2+\sqrt{3}, 5)$. Now $A B=\sqrt{(2-2)^{2}+(4-6)^{2}}=\sqrt{(0)^{2}+(-2)^{2}}$ $=\sqrt{0+4}=2$ $$ \begin{aligned} \mathrm{BC}...
$\because \angle \mathrm{B}=90^{\circ}$ $\therefore \mathrm{AC}^{2}=\mathrm{AB}^{2}+\mathrm{BC}^{2}$ $\Rightarrow(5+2)^{2}+(2-t)^{2}=(5-2)^{2}+(2+2)^{2}+(2+2)^{2}+(-2-t)^{2}$...
points are $A(3,0), B(6,4)$ and $C(-1,3)$. Now, $\mathrm{AB}=\sqrt{(3-6)^{2}+(0-4)^{2}}=\sqrt{(-3)^{2}+(-4)^{2}}$ $=\sqrt{9+16}=\sqrt{25}=5$...
The given points are $A(7,10), B(-2,5)$ and $C(3,-4)$. $$ \begin{aligned} &\mathrm{AB}=\sqrt{(-2-7)^{2}+(5-10)^{2}}=\sqrt{(-9)^{2}+(-5)^{2}}=\sqrt{81+25}=\sqrt{106} \\...
(iii) Let $\mathrm{A}(-1,-1), \mathrm{B}(2,3)$ and $\mathrm{C}(8,11)$ be the give points. Then $$ \begin{aligned} &\mathrm{AB}=\sqrt{(2+1)^{2}+(3+1)^{2}}=\sqrt{(3)^{2}+(4)^{2}}=\sqrt{25}=5 \text {...
(i) Let $\mathrm{A}(1,-1), \mathrm{B}(5,2)$ and $\mathrm{C}(9,5)$ be the give points. Then $\mathrm{AB}=\sqrt{(5-1)^{2}+(2+1)^{2}}=\sqrt{4^{2}+3^{2}}=\sqrt{25}=5 \text { units }$...
$\sqrt{(x-a-b)^{2}+(y-b+a)^{2}}=\sqrt{(x-a+b)^{2}+(y-a-b)^{2}}$...
$\mathrm{AP}=\mathrm{BP}$ $\Rightarrow \sqrt{(2+2)^{2}+(2+k)^{2}}=\sqrt{(2+2 k)^{2}+(2+3)^{2}}$ $\Rightarrow \sqrt{(4)^{2}+(2-\mathrm{k})^{2}}=\sqrt{(2+2 \mathrm{k})^{2}+(5)^{2}}$ $\Rightarrow...
$\mathrm{AC}=\mathrm{BC}$ $\Rightarrow \sqrt{(-2-3)^{2}+(3+1)^{2}}=\sqrt{(-2-x)^{2}+(3-8)^{2}}$ $\Rightarrow \sqrt{(5)^{2}+(4)^{2}}=\sqrt{(x+2)^{2}+(-5)^{2}}$ $\Rightarrow 25+16=(x+2)^{2}+25$...
the points $A(4,3)$ and $B(x, 5)$ lie on a circle with centerO $(2,3)$ Then $O A=0 B$ $(O A)^{2}=(O B)^{2}$ $\Rightarrow(4-2)^{2}+(3-3)^{2}=(x-2)^{2}+(5-3)^{2}$...
Let the required point be $P(x, y)$. Then $A P=B P=C P$ i.e, $(\mathrm{AP})^{2}=(\mathrm{BP})^{2}=(\mathrm{CP})^{2}$ => $(\mathrm{AP})^{2}=(\mathrm{BP})^{2}$ $$ \begin{aligned}...
points are $A(6,-1)$ and $B(2,3)$. The point $P(x, y)$ equidistant from the points A and B So, PA = PB $$ \begin{aligned} &\text { Also, }(P A)^{2}=(P B)^{2} \\...
$\mathrm{AP}=\mathrm{BP}$ $\Rightarrow \sqrt{(x-5)^{2}+(y-1)^{2}}=\sqrt{(x+1)^{2}+(y-5)^{2}}$ $\Rightarrow(x-5)^{2}+(y-1)^{2}=(x+1)^{2}+(y-5)^{2}$ $\text { (Squaring both sides) }$ $\Rightarrow...
Let $P(0, y)$ be a point on the $y$-axis. Then we have, $\mathrm{AP}=\mathrm{BP}$ $\Rightarrow \sqrt{(0-6)^{2}+(y-5)^{2}}=\sqrt{(0+4)^{2}+(y-3)^{2}}$ $\Rightarrow...
Let $\mathrm{P}(\mathrm{x}, 0)$ be the point on the $\mathrm{x}$-axis. Then as per the question we have $\mathrm{AP}=10$ $$ \begin{aligned} &\Rightarrow \sqrt{(x-11)^{2}+(0+8)^{2}}=10 \\...
Let $(\mathrm{x}, 0)$ be the point on the $\mathrm{x}$ axis. Then as per the question, we have $\sqrt{(x-2)^{2}+(0+5)^{2}}=\sqrt{(x+2)^{2}+(0-9)^{2}}$ $\Rightarrow...
$\mathrm{AB}=\mathrm{AC}$ $\Rightarrow \sqrt{(0-3)^{2}+(2-p)^{2}}=\sqrt{(0-p)^{2}+(2-5)^{2}}$ $\Rightarrow \sqrt{(-3)^{2}+(2-p)^{2}}=\sqrt{(-p)^{2}+(-3)^{2}}$ Squaring both sides, we get...
$A B=A C$ $\Rightarrow \sqrt{(x-8)^{2}+(2+2)^{2}}=\sqrt{(x-2)^{2}+(2+2)^{2}}$ Squaring both sides, we get $(x-8)^{2}+4^{2}=(x-2)^{2}+4^{2}$ $\Rightarrow x^{2}-16 x+64+16=x^{2}+4-4 x+16$ $\Rightarrow...
The given points are $\mathrm{P}(\mathrm{x}, 4)$ and $\mathrm{Q}(9,10)$. $ \therefore P Q=\sqrt{(x-9)^{2}+(4-10)^{2}} $ $$ \begin{aligned} &=\sqrt{(x-9)^{2}+(-6)^{2}} \\ &=\sqrt{x^{2}-18 x+81+36} \\...
The given points are $A(2,-3)$ and $B(10, y)$ $$ \begin{aligned} &\therefore \mathrm{AB}=\sqrt{(2-10)^{2}+(-3-\mathrm{y})^{2}} \\ &=\sqrt{(-8)^{2}+(-3-\mathrm{y})^{2}} \\...
$A B=5$ units Therefore, $(\mathrm{AB})^{2}=25$ units $\Rightarrow(5-a)^{2}+\{3-(-1)\}^{2}=25$ $\Rightarrow(5-a)^{2}+(3+1)^{2}=25$ $\Rightarrow(5-a)^{2}+(4)^{2}=25$ $$ \begin{aligned}...
(iii) $\quad C(-4,-6)$ Let $O(0,0)$ be the origin $$ \begin{aligned} & C=\sqrt{(-4-0)^{2}+(-6-0)^{2}} \\ &=\sqrt{(-4)^{2}+(-6)^{2}} \\ &=\sqrt{16+36} \\ &=\sqrt{52} \\ &=\sqrt{4 \times 13} \\ &=2...
(v) $\quad \mathrm{P}(\mathrm{a}+\mathrm{b}, \mathrm{a}-\mathrm{b})$ and $\mathrm{Q}(\mathrm{a}-\mathrm{b}, \mathrm{a}+\mathrm{b})$ points are $P(a+b, a-b)$ and $Q(a-b, a+b)$ => $\left(x_{1}=a+b,...
(i) (2, 3), (4, 1) (ii) (-5, 7), (-1, 3)(iii) (a, b), (- a, – b) We know that formula to find the distance (d) between two points $\left( {{x}_{1}},{{y}_{1}} \right)$and $\left( {{x}_{2}},{{y}_{2}}...