Reasoning: Using distance formula:
$AB=$ $\sqrt{\left(3-\frac{3}{2}\right)^{2}+\left(\frac{3}{4}+\frac{7}{4}\right)^{2}}=\sqrt{8.5}$
$CD=$ $\sqrt{(3-1)^{2}+\left(- 2-\frac{1}{2}\right)^{2}}=\sqrt{10.25}$
Since $(8.5<$ $10.25)$, therefore $AB<$ $CD$.

Reasoning: The midpoint of two points on a number line is the average of their coordinates.

Reasoning: The opposite sides have equal lengths, so the diagonals bisect each other. Hence the shape is a parallelogram.

Reasoning: Using the distance formula, the side lengths squared are $AB^{2}=29$, $BC^{2}=29$, and $AC^{2}=58$. Since two sides are equal, it is isosceles, & since $29+29=58$, it satisfies the Pythagorean theorem, making it right.

Reasoning: Using the midpoint formula: $\left(\frac{x_{1} + x_{2}}{2},\frac{y_{1} + y_{2}}{2}\right)$
For $(c,- d)$ and $(- d,c)$:
$\left(\frac{c + (- d)}{2},\frac{- d + c}{2}\right)=\left(\frac{c - d}{2},\frac{- d + c}{2}\right)$

Reasoning: Using midpoint formula: $\left(\frac{x + (- 3)}{2},\frac{4 + y}{2}\right)=(2,3)$
Equating coordinates:
$\frac{x - 3}{2}=2$ $⟹$ $x=7$ ; $\frac{4 + y}{2}=3$ $⟹$ $y=2$
Therefore, the coordinates are $(7,2)$.

Reasoning: Equal ordered pairs have equal coordinates:
$\frac{a + 1}{2}=4$ $⟹$ $a+1=8$ $⟹$ $a=7$

Reasoning: $(-4,-(-6))=(-4,6)$
Here, $x$ is negative and $y$ is positive, so it lies in quadrant $II$.

Reasoning: On the line, $B$ lies between $A$ and $C$, so: $d(A,C)=d(A,B)+d(B,C)$
Therefore, $d(A,C)-d(B,C)=d(A,B)$
Since distance is symmetric, $d(A,B)=d(B,A)$. Option D is correct.

Reasoning: The abscissa means $x$-coordinate. Every point on the $y$-axis has $x$-coordinate $0$.

Reasoning: The minimum number of sides required to form a polygon is three (triangle).

Reasoning: Given $x0$. For $(- x,y)$, $x$ becomes positive & $y$ remains positive, so it lies in quadrant $I$.

Reasoning: Distance from $y$-axis $=x$-coordinate $=3$ units
Distance from $x$-axis $=y$-coordinate $=5$ units
Minimum distance is from $y$-axis.

Reasoning: The midpoint divides a line segment into two equal parts.
Therefore, $LN=2(MN)$
So, $LN : MN=2 :1$

Reasoning: Every perpendicular bisector of a line segment passes through its midpoint.

Reasoning: The diagonals are:
● $AC$: from $(2,3)$ to $(2,- 3)$ $⟹$ vertical line $x=2$
● $DB$: from $(0,0)$ to $(5,0)$ $⟹$ horizontal line $y=0$
Their intersection is where $x=2$ and $y=0$.

Reasoning: Since $S$ is the midpoint of $PQ$, using the distance formula for $SR$ gives: $SR=5$

Reasoning: Distance formula:
$20^{2}=c^{2}+(10+1)^{2}$ $⟹$ $400=c^{2}+121$
$c^{2}=279=9(31)$ $⟹$ $c=3 \sqrt{31}$

Reasoning: The hypotenuse is the longest side:
$d=\sqrt{(7+3)^{2}+(4-1)^{2}}=\sqrt{100+9}=\sqrt{109}$

Reasoning: The hypotenuse endpoints are $(4,0)$ and $(- 1,- 5)$.
Midpoint: $\left(\frac{4 - 1}{2},\frac{0 - 5}{2}\right)=\left(\frac{3}{2},-\frac{5}{2}\right)$
(xxi) If end points of a diameter of the circle are $(4,5)$ and $(6,9)$, then coordinates of its center will be:
A. $(7,5)$ B. $\left(1,2\right)$
C. $(5,7)$ D. $(0,0)$
Answer: C
The center is the midpoint of the diameter: $\left(\frac{4 + 6}{2},\frac{5 + 9}{2}\right)=(5,7)$
(xxii) If two points $(2,3)$ and $(4,5)$ are at equal distance from $(3,t)$ then, the value of $t$ is:
A. $5$ B. $4$
C. $3$ D. $7$
Answer: B
Equal distances: $(3-2)^{2}+(t-3)^{2}=(3-4)^{2}+(t-5)^{2}$
Solving gives: $t=4$
(xxiii) A point $P(x,y)$ is at equal distance from both the axes and lies in the third quadrant. If it is at a distance of $4$ units from $y$-axis, then its coordinates are:
A. $(- 4$, $4)$ B. $(4$, $- 4)$
C. $(4$, $4)$ D. $(- 4$, $- 4)$
Answer: D
Equal distance from both axes means: $\left|x\right|=\left|y\right|$
The third quadrant has both coordinates negative, so the point is:
$(- 4$, $- 4)$
(xxiv) A line $AB$intersects $x$-axis at $C$with a distance of $3$ units from origin and $y$-axis at $D$ with a distance of $6$ units from origin, then $d(C,D)$ is:
A. $3\sqrt{5}$ B. $3\sqrt{3}$
C. $3$ D. $45$
Answer: A
Take $C(3,0)$ and $D(0,6)$.
$CD=\sqrt{(3-0)^{2}+(0-6)^{2}}=\sqrt{9+36}=\sqrt{45}=3 \sqrt{5}$