Q: Point on the \$x\$-axis which is equidistant from \$(2, -5)\$ and \$(-2, 9)\$ is:

Since point on \$x-\$axis, then coordinate of the point is \$\left(x, 0\right )\$. According to the question this point \$\left(x, 0\right )\$ is equidistant from the points \$\left(2, -5\right)\$ and \$\left(-2, 9\right)\$.That is, distance from \$\left(x, 0\right )\$ and \$\left(2, -5\right)\$ \$=\$ distance from \$\left(x, 0\right )\$ and \$\left(-2, 9\right)\$.\$\sqrt{\left( 2-x \right)^2+\left( -5-0 \right)^2}=\sqrt{\left( -2-x \right)^2+\left( 9-0 \right)^2}\\ \Rightarrow \left( 2-x \right)^2+\left( -5 \right)^2=\left( -2-x \right)^2+\left( 9 \right)^2\\ \Rightarrow 4-4x+x^2 +25=4 +4x+4+81 \\ \Rightarrow 8x=25-81\\ \Rightarrow 8x=-56\\ \Rightarrow x =-7 \$So, point is \$\left(-7,0 \right)\$.

Q: If \$(1, 2), (4, y), (x, 6)\$ and \$(3, 5)\$ are the vertices of a parallelogram taken in order, find \$x\$ and \$y\$.

Let \$A(1,2)\$, \$B(4,y)\$,\$C(x,6)\$ and \$D(3,5)\$ are the vertices of a parallelogram \$ABCD\$\$.AC\$ and \$BD\$ are the diagonals .\$O\$ is the midpoint of \$AC\$ and \$BD.\$If \$O\$ is the mid-point of \$AC\$ ,then the coordinates of \$O\$ are \$=\left(\dfrac{1+x}{2},\dfrac{2+6}{2} \right) = \left(\dfrac{x+1}{2},4 \right)\$If \$O\$ is the mid-point of BD then coordinates of O are \$\left(\dfrac{4+3}{2},\dfrac{5+y}{2}\right) = \left(\dfrac{7}{2},\dfrac{5+y}{2} \right)\$Since both coordinates are of the same point \$O\$\$\therefore \dfrac{1+x}{2}=\dfrac{7}{2}\$\$\Rightarrow 1+x=7\$\$\Rightarrow x=7-1=6\$\$\therefore \dfrac{5+y}{2}=4\$\$\Rightarrow 5+y=8\$\$\Rightarrow y=8-5=3\$Hence \$ x=6\$ and \$y=3\$.

Q: Find the coordinates of the foot of perpendicular from the point \$(-1, 3)\$ to the line \$3x - 4y - 16\displaystyle = 0\$

Let \$(a,b)\$ be the coordinates of the foot of the perpendicular from the point \$(-1,3)\$ to the line \$3x -4y - 16\displaystyle = 0\$Slope of the line joining \$(-1,3)\$ and \$(a,b)\$, is \$\displaystyle m_{1}= \frac{b-3}{a+1}\$Slope of the line \$3x -4y - 16\displaystyle = 0\$ or \$\displaystyle y= \frac{3}{4}x-4,m_{2}= \frac{3}{4}\$Since these two lines are perpendicular,\$\displaystyle m_{1}m_{2}= -1\$\$\displaystyle \therefore \left \{ \frac{b-3}{a+1} \right \}\cdot \left \{ \frac{3}{4} \right \}= -1\$\$\displaystyle \Rightarrow \frac{3b-9}{4a+4}= -1\$\$\displaystyle \Rightarrow 3b-9= 4a-4\$\$\displaystyle \Rightarrow 4a+3b= 5\$...(1)Point(a,b) lies on line \$\displaystyle 3x-4y= 16\$\$\displaystyle \therefore 3a-4b= 16\$....(2)On solving equation (1) and (2) we obtain\$\displaystyle a= \frac{68}{25}\$ and \$b= -\cfrac{49}{25}\$Thus, the required coordinates of the foot of the perpendicular are \$\displaystyle \left \{ \frac{68}{25},\frac{49}{25} \right \}\$

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

Let the point \$P(x,y)\$ is equidistant from the points \$A(3,6)\$and \$B(-3,4)\$\$\therefore PA=PB\$By using the distance formula\$=\sqrt{ { \left( {x_2}-{ { x_1 } } \right) }^{ 2}+{ \left( {y_2 }-{ y_1 } \right) }^{ 2 } } \$ we have\$\Rightarrow \sqrt{(x-3)^2+(y-6)^2}=\sqrt{(x+3)^2+(y-4)^2}\$\$\Rightarrow (x-3)^2+(y-6)^2=(x+3)^2+(y-4)^2\$\$\Rightarrow x^2-6x+9+y^2-12y+36=x^2+6x+9+y^2-8y+16\$\$\Rightarrow -6x-6x-12y+8y+36-16=0\$\$\Rightarrow -12x-4y+20=0\$\$\Rightarrow 3x+y-5=0\$Hence this is a relation between \$x\$ and \$y\$.

Q: Find a relation between x and y such that the point (x,y) is equidistant from the points (7,1) and (3,5).

\$\textbf{Step 1: Find the relation between x and y}.\$ \$\text{Let P(x, y) be equidistant from the points A(7,1) and B(3,5).}\$ \$\text{So, AP = BP}\$ \$\text{Squaring on both sides, we get}\$ \$ \Rightarrow (AP)^2=(BP)^2\$ \$\text{Using, distance formula, }\$ \$\text{Distance between}\ (x_1,y_1)\ \text{and} \ (x_2,y_2)= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}.....(1)\$ \$\text{Now,}\$ \$ \Rightarrow (AP)^2=(BP)^2\$ \$\Rightarrow (x-7)^2+(y-1)^2=(x-3)^2+(y-5)^2\$ \$\textbf{[Using eq(1)]}\$ \$\Rightarrow x^2+49-14x+y^2+1-2y=x^2+9-6x+y^2+25-10y\$ \$\Rightarrow -14x+50-2y=-6x+34-10y\$ \$\Rightarrow -7x+25-y=-3x+17-5y\$ \$\Rightarrow -4x+8+4y=0\$ \$\Rightarrow 4x-4y=8\$ \$\Rightarrow 4(x-y)=8\$ \$\Rightarrow x-y=2\$\$\textbf{Hence, this is the required relation between x and y . }\$

Q: If \$A(-2, 1), B(a, 0), C(4, b)\$ and \$D(1, 2)\$ are the vertices of a parallelogram ABCD, find the values of a and b. Hence find the lengths of its sides.

Given sides of parallelogram \$A(-2 , 1) ,B(a,0) , C(4 ,b) , D(1 , 2)\$We know that diagonals of Parallelogram bisect each other. Mid-point let say \$O\$ of diagonal \$AC\$ is given by \$x=\left ( \dfrac{x_1+x_2}{2}\right ) and \ y=\left ( \dfrac{y_1+y_2}{2}\right )\$\$O\$ \$\left(\dfrac{-2+4}{2} , \dfrac{1+b}{2}\right)\$ ..................(1)Mid-point let say \$P\$ of diagonal \$BD\$ is given by \$P\$ \$\left(\dfrac{a+1}{2} , \dfrac{0+2}{2}\right)\$ ..................(2)Points \$O\$ and \$P\$ are same Equating the corresponding co-ordinates of both midpoints, we get\$\dfrac{-2+4}{2} = \dfrac{a+1}{2}\$ \$\Rightarrow a=1\$and\$\dfrac{1+b}{2} = \dfrac{0+2}{2}\$ \$\Rightarrow b=1\$Now the Given co-ordinates of the parallelogram are written as \$A(-2 , 1) ,B(1,0) , C(4 ,1) , D(1 , 2)\$By distance formula, \$\sqrt{ { \left( {x_2}-{ { x_1 } } \right) }^{ 2}+{ \left( {y_2 }-{ y_1 } \right) }^{ 2 } } \$ we can find the length of each side \$AB = \sqrt{(-2-1)^2+(1-0)^2}\$\$AB = \sqrt{(3)^2+(1)^2} = \sqrt{10}\$\$AB=CD\$ ...............(pair of opposite sides of the parallelogram are parallel and equal)\$BC = \sqrt{(4-1)^2+(1-0)^2}\$\$BC = \sqrt{(3)^2+(1)^2} = \sqrt{10}\$\$BC=AD\$ ...............(pair of opposite sides of the parallelogram are parallel and equal )\$\Rightarrow AB=BC=CD=AD = \sqrt{10}\$\$\Rightarrow ABCD\$ is a \$ Rhombus \$

Q: The perimeter of a triangle with vertices \$(0, 4) (0, 0)\$ and \$(3, 0)\$ is:

Let \$ABC\$ be a triangle having \$A(0,4); B(0,0); C(3,0)\$ Using distance formula, we get \$AB=\sqrt { { (0-0) }^{ 2 }+{ (0-4) }^{ 2 } } =\sqrt { 16 } =4\\ BC=\sqrt { { (3-0) }^{ 2 }+{ (0-0) }^{ 2 } } =\sqrt { 9 } =3\\ CA=\sqrt { { (0-3) }^{ 2 }+{ (4-0) }^{ 2 } } =\sqrt { 25 } =5\$Perimeter of \$\triangle ABC=AB+BC+CA=4+3+5=12\$

Q: Find the distance of a point \$P(x, y)\$ from the origin.

Let \$O\$ be the origin and \$Q\$ be the foot of the perpendicular dropped from \$P\$ onto the \$x\$ axis. So \$\Delta OPQ\$ is right-angled at \$Q\$.By definition of coordinates:\$OQ=x\$ coordinate of \$P=\$ distance of \$P\$ from \$y\$ axis \$= |x|\$Similarly, \$QP = |y|\$.Thus, by using Pythagoras theorem on \$\Delta OPQ\$, we get \$OP = \sqrt{OQ^2 + QP^2} = \sqrt{x^2 + y^2}\$ Thus the distance of the point \$P(x, y)\$ from the origin \$O(0, 0)\$ is \$\sqrt{x^2+y^2}\$

Q: If the distance between the points \$(4, p)\$ and \$(1, 0)\$ is \$5\$, then the value of \$p\$ is:

The given points are \$(1,0)\$ and \$ (4,p)\$ \$d=\sqrt { { \left( { x }_{ 2 }-{ x }_{ 1 } \right) }^{ 2 }+{ \left( { y }_{ 2 }-{ y }_{ 1 } \right) }^{ 2 } }\$\$\Rightarrow 5 =\sqrt{(4-1)^2+p^2}\$\$\Rightarrow 25 =(4-1)^2+p^2\$\$\Rightarrow 25-9=p^2\$\$\Rightarrow p^2=16\$\$\Rightarrow p=+4,-4\$

Q: Find the distance of the point \$P(-6,8)\$ from the origin.

The distance of the point \$P(-6,8)\$ from the origin is \$\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10\$ units.

Question 1

Point on the $x$-axis which is equidistant from $(2, -5)$ and $(-2, 9)$ is:

Accepted Answer

Since point on $x-$axis, then coordinate of the point is $\left(x, 0\right )$. According to the question this point $\left(x, 0\right )$ is equidistant from the points $\left(2, -5\right)$ and $\left(-2, 9\right)$.That is, distance from  $\left(x, 0\right )$ and $\left(2, -5\right)$ $=$ distance from $\left(x, 0\right )$ and $\left(-2, 9\right)$.$\sqrt{\left( 2-x \right)^2+\left( -5-0 \right)^2}=\sqrt{\left( -2-x \right)^2+\left( 9-0 \right)^2}\ \Rightarrow \left( 2-x \right)^2+\left( -5 \right)^2=\left( -2-x \right)^2+\left( 9 \right)^2\ \Rightarrow 4-4x+x^2 +25=4 +4x+4+81 \ \Rightarrow 8x=25-81\ \Rightarrow 8x=-56\ \Rightarrow x =-7 $So, point is $\left(-7,0 \right)$.

Question 2

If $(1, 2), (4, y), (x, 6)$ and $(3, 5)$ are the vertices of a parallelogram taken in order, find $x$ and $y$.

Accepted Answer

Let $A(1,2)$, $B(4,y)$,$C(x,6)$ and $D(3,5)$ are the vertices of a parallelogram $ABCD$$.AC$ and $BD$ are the diagonals .$O$ is the midpoint of $AC$ and $BD.$If $O$ is the mid-point of $AC$ ,then the coordinates of $O$ are $=\left(\dfrac{1+x}{2},\dfrac{2+6}{2} \right) = \left(\dfrac{x+1}{2},4 \right)$If $O$ is the mid-point of BD then coordinates of O are $\left(\dfrac{4+3}{2},\dfrac{5+y}{2}\right) = \left(\dfrac{7}{2},\dfrac{5+y}{2} \right)$Since both coordinates are of the same point $O$$\therefore \dfrac{1+x}{2}=\dfrac{7}{2}$$\Rightarrow 1+x=7$$\Rightarrow x=7-1=6$$\therefore \dfrac{5+y}{2}=4$$\Rightarrow 5+y=8$$\Rightarrow y=8-5=3$Hence $ x=6$ and $y=3$.

Question 3

Find the coordinates of the foot of perpendicular from the point $(-1, 3)$ to the line $3x -  4y -  16\displaystyle = 0$

Accepted Answer

Let $(a,b)$ be the coordinates of the foot of the perpendicular from &#10;the point $(-1,3)$ to the line $3x  -4y -  16\displaystyle = 0$Slope of the line joining $(-1,3)$ and $(a,b)$, is $\displaystyle m_{1}= \frac{b-3}{a+1}$Slope of the line $3x  -4y -  16\displaystyle = 0$ or $\displaystyle y= \frac{3}{4}x-4,m_{2}= \frac{3}{4}$Since these two lines are perpendicular,$\displaystyle m_{1}m_{2}= -1$$\displaystyle \therefore \left \{ \frac{b-3}{a+1} \right \}\cdot \left \{ \frac{3}{4} \right \}= -1$$\displaystyle \Rightarrow \frac{3b-9}{4a+4}= -1$$\displaystyle \Rightarrow 3b-9= 4a-4$$\displaystyle \Rightarrow 4a+3b= 5$...(1)Point(a,b) lies on line $\displaystyle 3x-4y= 16$$\displaystyle \therefore 3a-4b= 16$....(2)On solving equation (1) and (2) we obtain$\displaystyle a= \frac{68}{25}$ and $b= -\cfrac{49}{25}$Thus,&#10; the required coordinates of the foot of the perpendicular are &#10;$\displaystyle \left \{ \frac{68}{25},\frac{49}{25} \right \}$

Question 4

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

Accepted Answer

Let the point $P(x,y)$ is equidistant from the points $A(3,6)$and $B(-3,4)$$\therefore PA=PB$By using the distance formula$=\sqrt{ { \left( {x_2}-{ { x_1 } } \right)  }^{ 2}+{ \left( {y_2 }-{ y_1  } \right)  }^{ 2 } } $    we have$\Rightarrow \sqrt{(x-3)^2+(y-6)^2}=\sqrt{(x+3)^2+(y-4)^2}$$\Rightarrow (x-3)^2+(y-6)^2=(x+3)^2+(y-4)^2$$\Rightarrow x^2-6x+9+y^2-12y+36=x^2+6x+9+y^2-8y+16$$\Rightarrow -6x-6x-12y+8y+36-16=0$$\Rightarrow -12x-4y+20=0$$\Rightarrow 3x+y-5=0$Hence this is a relation between $x$ and $y$.

Question 5

Find a relation between x and y such that the point (x,y) is equidistant from the points (7,1) and (3,5).

Accepted Answer

$\textbf{Step 1: Find the relation between x and y}.$                $\text{Let P(x, y) be equidistant from the points A(7,1) and B(3,5).}$                $\text{So, AP = BP}$                $\text{Squaring on both sides, we get}$                $ \Rightarrow (AP)^2=(BP)^2$                $\text{Using, distance formula, }$                $\text{Distance between}\ (x_1,y_1)\ \text{and} \ (x_2,y_2)= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}.....(1)$                $\text{Now,}$                $ \Rightarrow (AP)^2=(BP)^2$                $\Rightarrow (x-7)^2+(y-1)^2=(x-3)^2+(y-5)^2$                $\textbf{[Using eq(1)]}$                $\Rightarrow x^2+49-14x+y^2+1-2y=x^2+9-6x+y^2+25-10y$                $\Rightarrow -14x+50-2y=-6x+34-10y$                $\Rightarrow -7x+25-y=-3x+17-5y$                $\Rightarrow -4x+8+4y=0$                $\Rightarrow 4x-4y=8$                $\Rightarrow 4(x-y)=8$                $\Rightarrow x-y=2$$\textbf{Hence, this is the required relation between x and y . }$

Question 6

If $A(-2, 1), B(a, 0), C(4, b)$ and $D(1, 2)$ are the vertices of a parallelogram ABCD, find the values of a and b. Hence find the lengths of its sides.

Accepted Answer

Given sides of parallelogram $A(-2 , 1) ,B(a,0) ,  C(4 ,b)  , D(1 , 2)$We know that diagonals of Parallelogram bisect each other. Mid-point let say  $O$ of diagonal $AC$ is given by $x=\left ( \dfrac{x_1+x_2}{2}\right ) and \ y=\left ( \dfrac{y_1+y_2}{2}\right )$$O$  $\left(\dfrac{-2+4}{2}  , \dfrac{1+b}{2}\right)$ ..................(1)Mid-point let say  $P$ of diagonal $BD$ is given by $P$  $\left(\dfrac{a+1}{2}  , \dfrac{0+2}{2}\right)$ ..................(2)Points $O$ and $P$ are same Equating the corresponding co-ordinates of both midpoints, we get$\dfrac{-2+4}{2}  = \dfrac{a+1}{2}$ $\Rightarrow a=1$and$\dfrac{1+b}{2}  = \dfrac{0+2}{2}$ $\Rightarrow b=1$Now the Given co-ordinates of the parallelogram are written as $A(-2 , 1) ,B(1,0) ,  C(4 ,1)  , D(1 , 2)$By distance formula, $\sqrt{ { \left( {x_2}-{ { x_1 } } \right)  }^{ 2}+{ \left( {y_2 }-{ y_1  } \right)  }^{ 2 } } $ we can find the length of each side $AB = \sqrt{(-2-1)^2+(1-0)^2}$$AB = \sqrt{(3)^2+(1)^2}  = \sqrt{10}$$AB=CD$   ...............(pair of opposite sides of the parallelogram are parallel and equal)$BC = \sqrt{(4-1)^2+(1-0)^2}$$BC = \sqrt{(3)^2+(1)^2}  = \sqrt{10}$$BC=AD$   ...............(pair of opposite sides of the parallelogram are parallel and equal   )$\Rightarrow AB=BC=CD=AD = \sqrt{10}$$\Rightarrow ABCD$  is a $ Rhombus $

Question 7

The perimeter of a triangle with vertices $(0, 4) (0, 0)$ and $(3, 0)$ is:

Accepted Answer

Let $ABC$ be a triangle having $A(0,4); B(0,0); C(3,0)$ Using distance formula, we get $AB=\sqrt { { (0-0) }^{ 2 }+{ (0-4) }^{ 2 } }  =\sqrt { 16 }  =4\ BC=\sqrt { { (3-0) }^{ 2 }+{ (0-0) }^{ 2 } }  =\sqrt { 9 }  =3\ CA=\sqrt { { (0-3) }^{ 2 }+{ (4-0) }^{ 2 } } =\sqrt { 25 }  =5$Perimeter of $\triangle ABC=AB+BC+CA=4+3+5=12$

Question 8

Find the distance of a point $P(x, y)$ from the origin.

Accepted Answer

Let $O$ be the origin and $Q$ be the foot of the perpendicular dropped from $P$ onto the $x$ axis. So $\Delta OPQ$ is right-angled at $Q$.By definition of coordinates:$OQ=x$ coordinate of $P=$ distance of $P$ from $y$ axis $= |x|$Similarly, $QP = |y|$.Thus, by using Pythagoras theorem on $\Delta OPQ$, we get $OP = \sqrt{OQ^2 + QP^2} = \sqrt{x^2 + y^2}$ Thus the distance of the point $P(x, y)$ from the origin $O(0, 0)$ is $\sqrt{x^2+y^2}$

Question 9

If the distance between the points $(4, p)$ and $(1, 0)$ is $5$, then the value of $p$ is:

Accepted Answer

The given points are $(1,0)$ and $ (4,p)$  $d=\sqrt { { \left( { x }_{ 2 }-{ x }_{ 1 } \right)  }^{ 2 }+{ \left( { y }_{ 2 }-{ y }_{ 1 } \right)  }^{ 2 } }$$\Rightarrow 5 =\sqrt{(4-1)^2+p^2}$$\Rightarrow 25 =(4-1)^2+p^2$$\Rightarrow 25-9=p^2$$\Rightarrow p^2=16$$\Rightarrow p=+4,-4$

Question 10

Find the distance of the point $P(-6,8)$ from the origin.

Accepted Answer

The distance of the point $P(-6,8)$ from the origin is $\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10$ units.