Maths M.C.Q (1 Marks)

Let the given points are $A=\left(x_1, y_1\right)=(1,2), B=\left(x_2, y_2\right)=(0,0)$ and $C=\left(x_3, y_3\right)=(a, b)$.
$\because\text{Area of } \triangle\text{ABC }\triangle=\frac{1}{2}[\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)]$
$\therefore\triangle=\frac{1}{2}[1(0 - \text{b})+0(\text{b}-2)+\text{a}(2 -0)]$
$\triangle=\frac{1}{2}(\text{-b} + 0 +2\text{a})$
$\triangle=\frac{1}{2}(2\text{a}-\text{b})$
Since, the points $A(1, 2), B(0, 0)$ and $C(a, b)$ are collinear, then area of $\triangle\text{ABC}$ should be equal to zero.
$\text{i.e.,}\text{ area of }\triangle\text{ ABC}=0$
$\Rightarrow\frac{1}{2}(2\text{a} - \text{b})=0 $
$\Rightarrow2\text{a}-\text{b}=0$
$\Rightarrow2\text{a}=\text{b}$
Hence, the required relation is $2a = b.$

Given that, $\text{P}\Big(\frac{1}{2}, 4\Big)$ is the mid$-$point of the line segment the point $Q(-6, 5)$ and $R(-2, 3)$, which shows in the figure given below:

$\therefore\text{Mid-point of QR} = \text{P}\Big(\frac{-6-2}{2},\frac{5+3}{2}\Big) $
$\text{QR}=\text{P}(-4, 4)$
$\left[\right.$ Since, mid$-$point of line segment having points $\left(x_1, y_1\right)$ and $\left.\left(x_2, y_2\right)=\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)\right]$
But mid$-$point $\text{P}\Big(\frac{\text{a}}{3},4\Big)$ is given
$\therefore\Big(\frac{\text{a}}{3},4\Big)=(-4, 4)$
On comparing the coordinates, we get
$\frac{\text{a}}{3}=-4$
$\therefore\text{a}=-12$
Hence, the required value of a is $-12.$

Area of $\triangle\text{ABC}$ whose Vertices $\text{A}\equiv(\text{x}_1,\text{ y}_1),\text{B}\equiv(\text{x}_2,\text{ y}_2)$ and $\text{C}\equiv(\text{x}_3,\text{ y}_3)$ are given by
$\begin{vmatrix}\frac{1}{2}[\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)] \end{vmatrix}$
Here, $x_1=3, y_1=0, y_2=0, x_3=8$ and $y_3=4$
$=\begin{vmatrix}\frac{1}{2}[3(0-4)+7(4-0)+8(0-0)] \end{vmatrix}$
$=\begin{vmatrix} \frac{1}{2}(-12+28+0)\end{vmatrix}$
$=\begin{vmatrix}\frac{1}{2}(16) \end{vmatrix}$
$=8$
Hence, the required area of $\text{AABC}$ is $8.$

If $P ( x , y )$ divides the line segment joining $A \left( x _1, y _2\right)$ and $B \left( x _2, y_2\right)$ internally in the ratio m : n then
$\text{x}=\frac{1(3)+2(7)}{1+2},\text{y}=\frac{1(4)+2(-6)}{1+2} [$by section formula$]$
$\Rightarrow\text{x}=\frac{3+14}{3},\text{y}=\frac{-4+12}{3}$
$\Rightarrow\text{x}=\frac{17}{3},\text{y}=-\frac{8}{3}$
So, $\big(\text{x},\text{ y})=\Big(\frac{17}{3},-\frac{8}{3}\Big) $ lies in $IV$ quadrant.
$[$Since, in $IV$ quadrant, $x-$coordinate is positive and $y-$coordinate is negative$]$

Let the fourth vertex of parallelogram, $\text{D}\equiv(\text{x}_4,\text{ y}_4)$ and $L, M$ be the middle points of $AC$ and $BD$, respectively,
Then, $\text{L}\equiv\Big(\frac{-2+8}{2},\frac{3+3}{2}\Big)\equiv(3, 3)$
$\Big[$ Since, mid-point of any line segment which passes throught the points $(x_1, y_1)$ and $(x_2, y_2)$ $=\Big(\frac{\text{x}_1+\text{x}_2}{2},\frac{\text{y}_1+\text{y}_2}{2}\Big)\Big]$
and $\text{M}=\Big(\frac{6+\text{x}_4}{2},\frac{7+\text{y}_4}{2}\Big)$

Since, $\text{ABCD}$ is a parallelogram, therefore diagonals $AC$ and $BD$ will bisect each other.
Hence, $L$ and $M$ are the same points.
$\therefore3=\frac{6+\text{x}_4}{2}\text{ and }3=\frac{7+\text{y}_4}{2}$
$\Rightarrow6=6+\text{x}_4\text{ and }6=7+\text{y}_{4}$
$\Rightarrow\text{x}_{4}=0\text{ and }\text{y}_{4}$
$\therefore\text{x}_{4}\text{ and }\text{y}_{4}=-1$
Hence, the fourth vertex of parallelogram is $D (x_4, y_4) = D(0, -1).$

1. Distance between $(0, 0)$ and $\Big(\frac{-3}{4},1\Big)$

2. Distance between $(0, 0)$ and $2,\frac{7}{3}$

3. Distance between $(0, 0)$ and $\Big(5,\frac{-1}{2}\Big)$

4. Distance between $(0, 0)$ and $\Big(-6,\frac{5}{2}\Big)$

Let the coordinate of the point which is equidistant from the three vertices $0(0,0), A(0,2 y)$ and $B (2 x , 0)$ is $P ( h , k )$.
Then, $PO = PA = PB$
$\Rightarrow(P O)^2=(P A)^2=(P B)^2$
By distance formula, $\left[\sqrt{(h-0)^2+(k-0)^2}\right]^2=\left[\sqrt{(h-0)^2+(k-2 y)^2}\right]^2=\left[\sqrt{(h-2 x)^2+(k-0)^2}\right]^2$
$\Rightarrow h^2+k^2=h^2+(k-2 y)^2=(h-2 x)^2+k^2 \ldots \ldots . . .( (ii))$
Taking first two equations, we get
$h^2+k^2=h^2+(k-2 y)^2$
$k^2=k^2+4 y^2-4 y k$
$4 y(y-k)=0$
$y=k[\because y \neq 0]$
Taking first and third equations, we get
$h^2+h^2=(h-2 x)^2+k^2$
$h^2=h^2+4 x^2-4 x h$
$4 x(x-h)=0$
$x=h[\because x \neq 0]$
$\therefore$ Required points $=(h, k)=(x, y)$

Given that, the point $P(2, 1)$ lies on the line segment joining the points $A(4, 2)$ and $B(8, 4)$, which shows in the figure below:

Now, distance between $A(4, 2)$ and $(2, 1),$
$\text{AP}=\sqrt{(2-4)^2+(1-2)^2}$
$\left[\therefore\right.$ distance between two points $\left( x _1, y _1\right)$ and $\left.\left( x _2, y _2\right), d =\sqrt{\left( x _2- x _1\right)^2+\left( y _2- y _1\right)^2}\right]$
$\text{AP}=\sqrt{(-2)^2+(-1)^2}$
$\text{AP}=\sqrt{4+1}$
$\text{AP}=\sqrt{5}$
Distance between $A(4, 2)$ and $B(8, 4),$
$\text{AB}=\sqrt{(8-4)^2+(4-2)^2}$
$\text{AB}=\sqrt{(4)^2+(2)^2}$
$\text{AB}=\sqrt{16+4}$
$\text{AB}=\sqrt{20}$
$\text{AB}=2\sqrt{5}$
Distance between $B(8, 4)$ and $P(2, 1),$
$\text{BP}=\sqrt{(8-2)^2+(4-1)^2}$
$\text{BP}=\sqrt{(6)^2+(3)^2}$
$\text{BP}=\sqrt{36+9}$
$\text{BP}=\sqrt{45}$
$\text{BP}=3\sqrt{5}$
$\therefore\text{AB}=2\sqrt{5}=2\text{AP}$
$\text{AP}=\frac{\text{AB}}{2}$
Hence, required condition is $\text{AP}=\frac{\text{AB}}{2}.$

$\therefore$ Distance between the points $\left( x _1, y _2\right)$ and $\left( x _2, y _2\right)$
$d=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$
Here, $x_1=-6, y_1=8$ and $x_2=0, y_2=0$
$\therefore$ Distance between $P(-6, 8)$ and origin i.e., $O(0, 0)$
$\text{PO}=\sqrt{[0-(-6)]^2+(0-8)^2}$
$\text{PO}=\sqrt{(6)^2+(-8)^2}$
$\text{PO}=\sqrt{36+64}$
$\text{PO}=\sqrt{100}$
$\text{PO}=10$

Let $A(- 4, 0), B(4, 0), C(0, 3)$ are the given vertices.
Now, distance between $A (-4, 0)$ and $B (4, 0),$
$\text{AB}=\sqrt{[4-(-4)^2+(0-0)^2}$
$\Big[\ \therefore$ distance between two points $\left( x _1, y _1\right)$ and $\left( x _2, y _2\right)$, $\text{d}=\sqrt{(\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2}\Big]$
$\text{AB}=\sqrt{(4+4)^2}=\sqrt{8^2}=8$
Distance between B(4, 0) and C(0, 3),
$\text{BC}=\sqrt{(0-4)^2+(3-0)^2}=\sqrt{16+9}$
$\text{BC}=\sqrt{25}=5$
Distance between B(-4, 0) and C(0, 3),
$\text{AC}=\sqrt{[0-(-4)^2]+(3-0)^2}=\sqrt{16+9}$
$​​$$\text{AC}=\sqrt{25}=5$
$\because\text{BC}=\text{AC}$
Hence, $\triangle\text{ABC}$ is an isosceles triangle because an isosceles triangle has two sides equal.

Distance between the points $( x _1, y _1)$ and $( x _2, y _2)$
$\text{d}=\sqrt{(\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2}$
Here, $x_1=0, y_1=5$ and $x_2=-5, y_2=0$
Distance between the point $(0, 5)$ and $(-5, 0)$
$\text{d}=\sqrt{(-5-0)^2+(0-5)^2}$
$\text{d}=\sqrt{25+25}$
$\text{d}=\sqrt{50}$
$\text{d}=5\sqrt{2}$

We further, adding all the distance of a triangle to get the oerimeter of a triangle. We point the vertices of a triangle i. e., $(0, 4), (0, 0)$ and $(3, 0)$ on the paper shown as given below:

Now, perimeter of $\triangle\text{AOB} =$ Sum of the length of all its sides $= d(AO) + d(OB) + d(AB)$
$\therefore$ Distance between the points $\left( x _1, y _1\right)$ and $\left( x _2, y _2\right)$,
$\text{d}=\sqrt{(\text{x}_2-\text{x}_1)^2+(\text{y}_2-\text{y}_1)^2}$
$d =$ Distance between $A(0, 4)$ and $O(0, 0) +$ Distance between $O(0, 0)$ and $B(3, 0) +$ Distance between $A(0, 4)$ and $B(3, 0)$
$\text{d}\sqrt{(0-0)^2+(0-4)^2}+\sqrt{(3-0)^2+(0-0)^2}+\sqrt{(3-0)^2+(0-4)^2}$
$\text{d}=\sqrt{0+16}+\sqrt{9+0}+\sqrt{(3)^2+(4)^2}$
$\text{d}=4+3+\sqrt{9+16}$
$\text{d}=7+\sqrt{25}$
$\text{d}=7+5$
$\text{d}=12$
Hence, the required perimeter of triangle is $12.$

We know that, the perpendicular bisector of the any line segment divides the jjpe segment into two equal parts i.e., the perpendicular bisector of the line segment always passes through the mid$-$point of the line segment. Mid$-$point of the line segment joining the points $A (-2, -5)$ and $S(2, 5)$
$=\Big(\frac{-2+2}{2},\frac{-5+5}{2}\Big)=(0,0)$
$\Big[$ Since, mid-point of any line segment which passes throught the points $( x _1, y _1)$ and $( x _2, y _2)$ $=\Big(\frac{\text{x}_1+\text{x}_2}{2},\frac{\text{y}_1+\text{y}_2}{2}\Big)\Big]$
Hence, $(0, 0)$ is the required point lies on the perpendicular bisector of the lines segment.