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Solution

Midpoint = ((16+36)/2, (4+6)/2) = (26, 5).

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Solution

Slope of $y = x$ is 1 → perpendicular slope = −1. Equation: $y - 2 = -1(x - 3) \Rightarrow x + y = 5$.

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Solution

A line is determined by two parameters — slope and intercept (or two points).

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Solution

Solving equal distance condition gives point (1,1).

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Solution

Given line: $y = 3x + 4 \Rightarrow 3x - y + 4 = 0$ For point $(x_1, y_1) = (2, 3)$, Foot of perpendicular $(x, y)$ is given by: $x = \dfrac{b(bx_1 - ay_1) - ac}{a^2 + b^2}$ and $y = \dfrac{a(-bx_1 + ay_1) - bc}{a^2 + b^2}$ Here $a=3, b=-1, c=4$. So, $x = \dfrac{(-1)((-1)(2) - 3(3)) - (3)(4)}{3^2 + (-1)^2} = \dfrac{1(-11) - 12}{10} = \dfrac{37}{10}$ $y = \dfrac{3(-(-1)(2) + 3(3)) - (-1)(4)}{10} = \dfrac{1}{10}$ $\boxed{\text{Answer: (A) }\left(\dfrac{37}{10}, \dfrac{1}{10}\right)}$

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Solution

Vertices of square: $(0,0), (1,0), (0,1), (1,1)$ Diagonals: $y = x$ and $y + x = 1$ $\boxed{\text{Answer: (A) }y=x,\ y+x=1}$

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Solution

Let $P(0, y)$. $\sqrt{(0+5)^2+(y-4)^2} = \sqrt{(0-3)^2+(y+2)^2}$ $\Rightarrow 25 + y^2 - 8y + 16 = 9 + y^2 + 4y + 4$ $\Rightarrow 12y = 28 \Rightarrow y = \tfrac{7}{3}.$

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Solution

For collinearity, slopes must be equal: $\dfrac{1-(-1)}{2-x} = \dfrac{5-1}{4-2} \Rightarrow \dfrac{2}{2-x} = 2 \Rightarrow 1 = 2 - x \Rightarrow x = 1.$

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Solution

A line parallel to the $x$-axis has no $x$-term ⇒ coefficient of $x$ must be $0$: $k-3=0 \Rightarrow k=3$. Also need coefficient of $y \ne 0$: $-(4-k^2)\ne 0 \Rightarrow k\ne \pm 2$ (satisfied).

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Solution

Set distances equal and square: $(x-a-b)^{2}+(y-b+a)^{2}=(x-a+b)^{2}+(y-a-b)^{2}$. Simplify ⇒ $-4b(x-a)+4a(y-b)=0 \Rightarrow bx=ay$.

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Solution

Parallel lines: $2x+6y+c=0$. Intercept points: $( -\tfrac{c}{2},0)$ and $(0,-\tfrac{c}{6})$. Distance between them $=\sqrt{\left(\tfrac{c}{2}\right)^{2}+\left(\tfrac{c}{6}\right)^{2}} =|c|\,\dfrac{\sqrt{10}}{6}$. Set equal to $10$ ⇒ $|c|=6\sqrt{10}$ ⇒ two values $c=\pm6\sqrt{10}$.

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Solution

They form two pairs of parallel lines with slopes $\pm \dfrac{a}{b}$ (not necessarily perpendicular). Under unequal scaling they form a rectangle (a square only if $a=b$).

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Solution

Vertices are $(0,1)$, $(1,0)$, $(0,-1)$, $(-1,0)$ — a rhombus with diagonals $2$ and $2$. Area $=\dfrac{d_{1}d_{2}}{2}=\dfrac{2\times2}{2}=2$.

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Solution

Vertices: $A(0,0),\ B(4,0),\ C(3,4)$ Slope of $BC = \dfrac{4 - 0}{3 - 4} = -4$ Equation of altitude from $A$: perpendicular to BC → slope $\dfrac{1}{4}$ $\Rightarrow y = \dfrac{1}{4}x$ … (1) Slope of $AC = \dfrac{4 - 0}{3 - 0} = \dfrac{4}{3}$ Equation of altitude from $B$: perpendicular slope $-\dfrac{3}{4}$, passes through $(4,0)$ $\Rightarrow y - 0 = -\dfrac{3}{4}(x - 4)$ $\Rightarrow y = -\dfrac{3}{4}x + 3$ … (2) Solving (1) and (2): $\dfrac{x}{4} = -\dfrac{3x}{4} + 3 \Rightarrow x = 3,\ y = \dfrac{3}{4}$

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Solution

Let $A = (x,0)$. Slope of incident ray $= \dfrac{2 - 0}{1 - x} = \dfrac{2}{1 - x}$ Slope of reflected ray $= \dfrac{3 - 0}{5 - x} = \dfrac{3}{5 - x}$ For reflection from $X$–axis: angle of incidence = angle of reflection. Hence, their slopes are negatives of each other: $\dfrac{2}{1 - x} = -\dfrac{3}{5 - x}$ $\Rightarrow 2(5 - x) = -3(1 - x)$ $\Rightarrow 10 - 2x = -3 + 3x$ $\Rightarrow 5x = 13 \Rightarrow x = \dfrac{13}{5}$ Thus, $A\left(\dfrac{13}{5}, 0\right)$.

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Solution

For a set of $n$ elements, power set size $= 2^n = 2^3 = 8.$