Coordinate Geometry

Given: Two horizontal lines $y = 8$ and $y = -2$.

Concept: A line equidistant from two parallel lines lies exactly midway between them.

Working:
The required line is parallel to both given lines and passes through the midpoint of the perpendicular distance between them.

Midpoint of $y$-values:
$$y = \frac{8 + (-2)}{2} = \frac{6}{2} = 3$$

Answer: The required equation is $\boxed{y = 3}$.

Given: $A(1,4)$, $B(2,-3)$, $C(-1,-2)$.

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(i) Median through A:

The median from $A$ goes to the midpoint $D$ of $BC$.
$$D = \left(\frac{2+(-1)}{2},\, \frac{-3+(-2)}{2}\right) = \left(\frac{1}{2},\, -\frac{5}{2}\right)$$

Slope of $AD$:
$$m = \frac{-\frac{5}{2} - 4}{\frac{1}{2} - 1} = \frac{-\frac{13}{2}}{-\frac{1}{2}} = 13$$

Equation of median through $A(1,4)$:
$$y - 4 = 13(x - 1)$$
$$y - 4 = 13x - 13$$
$$\boxed{13x - y - 9 = 0}$$

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(ii) Altitude through A:

The altitude from $A$ is perpendicular to $BC$.

Slope of $BC$:
$$m_{BC} = \frac{-2-(-3)}{-1-2} = \frac{1}{-3} = -\frac{1}{3}$$

Slope of altitude from $A$ (perpendicular to $BC$):
$$m = 3$$

Equation through $A(1,4)$:
$$y - 4 = 3(x - 1)$$
$$y - 4 = 3x - 3$$
$$\boxed{3x - y + 1 = 0}$$

*(Note: The answer key lists $3x - y - 11 = 0$; the correct working gives $3x - y + 1 = 0$.)*

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(iii) Perpendicular bisector of BC:

Midpoint of $BC$: $D = \left(\frac{1}{2}, -\frac{5}{2}\right)$

Slope of $BC = -\frac{1}{3}$, so slope of perpendicular bisector $= 3$.

Equation through $D\left(\frac{1}{2}, -\frac{5}{2}\right)$:
$$y + \frac{5}{2} = 3\left(x - \frac{1}{2}\right)$$
$$y + \frac{5}{2} = 3x - \frac{3}{2}$$
$$y = 3x - \frac{3}{2} - \frac{5}{2} = 3x - 4$$
$$\boxed{3x - y - 4 = 0}$$

Concept: The angle bisectors of the coordinate axes (x-axis and y-axis) are the lines that make equal angles with both axes.

Working:
The x-axis has equation $y = 0$ and the y-axis has equation $x = 0$.

The bisectors of the angles between these two axes are the lines where $|x| = |y|$, i.e.,
$$y = x \quad \text{and} \quad y = -x$$

Answer: The equations of the bisectors are $\boxed{y = x}$ and $\boxed{y = -x}$, or equivalently $y = \pm x$.

Given: Line passes through $(2, 2)$; $x$-intercept $= a$, $y$-intercept $= b$, and $a + b = 9$.

Concept: Intercept form of a line: $\dfrac{x}{a} + \dfrac{y}{b} = 1$.

Working:
Since the line passes through $(2, 2)$:
$$\frac{2}{a} + \frac{2}{b} = 1 \quad \cdots (1)$$

Also, $b = 9 - a$ $\cdots (2)$

Substituting (2) into (1):
$$\frac{2}{a} + \frac{2}{9-a} = 1$$
$$2(9-a) + 2a = a(9-a)$$
$$18 - 2a + 2a = 9a - a^2$$
$$18 = 9a - a^2$$
$$a^2 - 9a + 18 = 0$$
$$(a-3)(a-6) = 0$$
$$a = 3 \quad \text{or} \quad a = 6$$

- If $a = 3$, then $b = 6$: $\dfrac{x}{3} + \dfrac{y}{6} = 1 \Rightarrow 2x + y = 6$
- If $a = 6$, then $b = 3$: $\dfrac{x}{6} + \dfrac{y}{3} = 1 \Rightarrow x + 2y = 6$

Answer: The required equations are $2x + y = 6$ and $\boxed{x + 2y = 6}$.

Given: Normal (perpendicular) distance from origin $= p = 3$; $\tan\alpha = \dfrac{5}{12}$.

Concept: Normal form of a line: $x\cos\alpha + y\sin\alpha = p$.

Working:
From $\tan\alpha = \dfrac{5}{12}$, we construct a right triangle with opposite $= 5$, adjacent $= 12$, hypotenuse $= \sqrt{25+144} = 13$.

$$\cos\alpha = \frac{12}{13}, \quad \sin\alpha = \frac{5}{13}$$

Substituting into the normal form:
$$x \cdot \frac{12}{13} + y \cdot \frac{5}{13} = 3$$
$$12x + 5y = 39$$

Answer: $\boxed{12x + 5y = 39}$

Given: $5x - 12y = 60$.

Step 1 – Intercept form:
Divide both sides by 60:
$$\frac{5x}{60} - \frac{12y}{60} = 1$$
$$\frac{x}{12} + \frac{y}{-5} = 1$$

So $x$-intercept $a = 12$ and $y$-intercept $b = -5$.

Step 2 – Length of intercept between axes:
The line meets the x-axis at $(12, 0)$ and the y-axis at $(0, -5)$.

$$\text{Length} = \sqrt{(12-0)^2 + (0-(-5))^2} = \sqrt{144 + 25} = \sqrt{169} = 13$$

Answer: Intercept form: $\dfrac{x}{12} + \dfrac{y}{-5} = 1$; Length of intercepted portion $= \boxed{13}$ units.

Given: $x + y - 2 = 0$, i.e., $x + y = 2$.

Concept: Normal form: $x\cos\alpha + y\sin\alpha = p$, obtained by dividing by $\sqrt{A^2+B^2}$.

Working:
Here $A = 1$, $B = 1$, $C = -2$.
$$\sqrt{A^2 + B^2} = \sqrt{1+1} = \sqrt{2}$$

Divide throughout by $\sqrt{2}$:
$$\frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$$

$$x\cos 45^\circ + y\sin 45^\circ = \sqrt{2}$$

Answer: $\boxed{x\cos 45^\circ + y\sin 45^\circ = \sqrt{2}}$

Given: Line $\dfrac{x}{3} + \dfrac{y}{4} = 1$, i.e., $4x + 3y - 12 = 0$. Points on x-axis have the form $(h, 0)$.

Concept: Distance from point $(h, 0)$ to line $4x + 3y - 12 = 0$:
$$d = \frac{|4h + 3(0) - 12|}{\sqrt{4^2+3^2}} = \frac{|4h - 12|}{5} = 4$$

Working:
$$|4h - 12| = 20$$
$$4h - 12 = 20 \quad \Rightarrow \quad 4h = 32 \quad \Rightarrow \quad h = 8$$
$$4h - 12 = -20 \quad \Rightarrow \quad 4h = -8 \quad \Rightarrow \quad h = -2$$

Answer: The required points on the x-axis are $\boxed{(8,\, 0)}$ and $\boxed{(-2,\, 0)}$.

Given: Two points on the linear cost function: $(q_1, C_1) = (30, 1500)$ and $(q_2, C_2) = (50, 2000)$.

Step 1 – Find the slope (variable cost per unit):
$$m = \frac{C_2 - C_1}{q_2 - q_1} = \frac{2000 - 1500}{50 - 30} = \frac{500}{20} = 25$$

Step 2 – Find the equation using point-slope form:
$$C - 1500 = 25(q - 30)$$
$$C - 1500 = 25q - 750$$
$$C = 25q + 750$$

Verification: At $q = 50$: $C = 25(50) + 750 = 1250 + 750 = 2000$ ✓

Answer: The cost equation is $\boxed{C = 25q + 750}$, where the fixed cost is Rs. 750 and the variable cost is Rs. 25 per shoe.

Concept: Standard equation of a circle with centre $(h, k)$ and radius $r$: $(x-h)^2 + (y-k)^2 = r^2$.

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(i) Centre $(0, 2)$, radius $= 2$:
$$(x-0)^2 + (y-2)^2 = 4$$
$$\boxed{x^2 + (y-2)^2 = 4}$$
Expanding: $x^2 + y^2 - 4y = 0$.

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(ii) Centre $(0, 0)$, radius $= 3$:
$$(x-0)^2 + (y-0)^2 = 9$$
$$\boxed{x^2 + y^2 = 9}$$

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(iii) Centre $(-a, -b)$, radius $= \sqrt{a^2 - b^2}$:
$$(x+a)^2 + (y+b)^2 = a^2 - b^2$$
$$\boxed{x^2 + y^2 + 2ax + 2by + 2b^2 = 0}$$

*(Expanding: $x^2+2ax+a^2+y^2+2by+b^2 = a^2-b^2 \Rightarrow x^2+y^2+2ax+2by+2b^2=0$.)*

Given: Rectangle with sides $x = 4$, $x = -5$, $y = 5$, $y = -1$.

Step 1 – Find the vertices (corners) of the rectangle:
The four corners are $(4, 5)$, $(-5, 5)$, $(-5, -1)$, $(4, -1)$.

Step 2 – Identify the diagonal endpoints:
Take diagonal from $(4, 5)$ to $(-5, -1)$ (or the other diagonal — both give the same circle).

Step 3 – Centre = midpoint of diagonal:
$$\text{Centre} = \left(\frac{4+(-5)}{2},\, \frac{5+(-1)}{2}\right) = \left(-\frac{1}{2},\, 2\right)$$

Step 4 – Radius = half the diagonal length:
$$\text{Diameter} = \sqrt{(4-(-5))^2+(5-(-1))^2} = \sqrt{81+36} = \sqrt{117}$$
$$r = \frac{\sqrt{117}}{2}, \quad r^2 = \frac{117}{4}$$

Step 5 – Equation:
$$\left(x+\frac{1}{2}\right)^2 + (y-2)^2 = \frac{117}{4}$$

Expanding:
$$x^2 + x + \frac{1}{4} + y^2 - 4y + 4 = \frac{117}{4}$$
$$x^2 + y^2 + x - 4y + \frac{1}{4} + 4 - \frac{117}{4} = 0$$
$$x^2 + y^2 + x - 4y + \frac{1 + 16 - 117}{4} = 0$$
$$\boxed{x^2 + y^2 + x - 4y - 25 = 0}$$

Concept: An equation $x^2+y^2+2gx+2fy+c=0$ represents a real circle if g^2+f^2-c > 0.

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(i) $3x^2 + 3y^2 + 6x - 4y = 1$:

Divide by 3:
$$x^2 + y^2 + 2x - \frac{4}{3}y - \frac{1}{3} = 0$$

Here $g = 1$, $f = -\dfrac{2}{3}$, $c = -\dfrac{1}{3}$.

g^2+f^2-c = 1 + \frac{4}{9} + \frac{1}{3} = \frac{9+4+3}{9} = \frac{16}{9} > 0

It represents a circle.
- Centre $= (-g, -f) = \left(-1,\, \dfrac{2}{3}\right)$
- Radius $= \sqrt{\dfrac{16}{9}} = \dfrac{4}{3}$

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(ii) $2x^2 + 2y^2 + 3y + 10 = 0$:

Divide by 2:
$$x^2 + y^2 + \frac{3}{2}y + 5 = 0$$

Here $g = 0$, $f = \dfrac{3}{4}$, $c = 5$.

g^2+f^2-c = 0 + \frac{9}{16} - 5 = \frac{9-80}{16} = -\frac{71}{16} < 0

It does NOT represent a real circle (imaginary circle).

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(iii) $x^2 + y^2 - 12x + 6y + 45 = 0$:

Here $g = -6$, $f = 3$, $c = 45$.

$$g^2+f^2-c = 36 + 9 - 45 = 0$$

Since $g^2+f^2-c = 0$, the radius is zero — this represents a point circle (degenerate circle) at the point $(6, -3)$.

*(Some textbooks classify this as a circle of radius 0.)*
- Centre $= (6, -3)$, Radius $= 0$.

Given: Circle: $x^2 + y^2 - 6x + 5y - 7 = 0$. One end of diameter: $A(-1, 3)$.

Step 1 – Find the centre of the circle:
Comparing with $x^2+y^2+2gx+2fy+c=0$:
$$2g = -6 \Rightarrow g = -3; \quad 2f = 5 \Rightarrow f = \frac{5}{2}$$

Centre $= (-g, -f) = \left(3,\, -\dfrac{5}{2}\right)$.

Step 2 – The centre is the midpoint of the diameter:
Let the other end be $B(x_2, y_2)$.
$$\frac{-1 + x_2}{2} = 3 \Rightarrow x_2 = 7$$
$$\frac{3 + y_2}{2} = -\frac{5}{2} \Rightarrow 3 + y_2 = -5 \Rightarrow y_2 = -8$$

Answer: The other end of the diameter is $\boxed{(7,\, -8)}$.

Given: Circle passes through $P(2, 3)$ and $Q(-1, 1)$; centre lies on $x - 3y - 11 = 0$.

Let the equation of the circle be $x^2+y^2+2gx+2fy+c=0$.

Condition 1 (passes through $(2,3)$):
$$4 + 9 + 4g + 6f + c = 0 \Rightarrow 4g + 6f + c = -13 \quad\cdots(1)$$

Condition 2 (passes through $(-1,1)$):
$$1 + 1 - 2g + 2f + c = 0 \Rightarrow -2g + 2f + c = -2 \quad\cdots(2)$$

Condition 3 (centre $(-g,-f)$ lies on $x-3y-11=0$):
$$-g - 3(-f) - 11 = 0 \Rightarrow -g + 3f = 11 \quad\cdots(3)$$

Subtract (2) from (1):
$$6g + 4f = -11 \quad\cdots(4)$$

From (3): $g = 3f - 11$. Substitute into (4):
$$6(3f-11) + 4f = -11$$
$$18f - 66 + 4f = -11$$
$$22f = 55 \Rightarrow f = \frac{5}{2}$$

$$g = 3\left(\frac{5}{2}\right) - 11 = \frac{15}{2} - 11 = -\frac{7}{2}$$

From (2): $c = -2 + 2g - 2f = -2 + 2(-\frac{7}{2}) - 2(\frac{5}{2}) = -2 - 7 - 5 = -14$

Equation of circle:
$$x^2 + y^2 + 2\left(-\frac{7}{2}\right)x + 2\left(\frac{5}{2}\right)y - 14 = 0$$
$$\boxed{x^2 + y^2 - 7x + 5y - 14 = 0}$$

Given: $x^2+y^2+8x+10y+p=0$; radius $= 7$.

Comparing with $x^2+y^2+2gx+2fy+c=0$:
$$g = 4,\quad f = 5,\quad c = p$$

Radius formula:
$$r = \sqrt{g^2+f^2-c} = 7$$
$$g^2+f^2-c = 49$$
$$16 + 25 - p = 49$$
$$41 - p = 49$$
$$p = -8$$

Answer: $\boxed{p = -8}$

Concept: Two circles are concentric if they have the same centre.

Circle 1: $x^2+y^2-3x+ky-5=0$
$$g_1 = -\frac{3}{2},\quad f_1 = \frac{k}{2}$$
Centre$_1 = \left(\dfrac{3}{2},\, -\dfrac{k}{2}\right)$

Circle 2: $4x^2+4y^2-12x-y-9=0$

Divide by 4:
$$x^2+y^2-3x-\frac{1}{4}y-\frac{9}{4}=0$$
$$g_2 = -\frac{3}{2},\quad f_2 = -\frac{1}{8}$$
Centre$_2 = \left(\dfrac{3}{2},\, \dfrac{1}{8}\right)$

For concentric circles, centres must be equal:
$$-\frac{k}{2} = \frac{1}{8}$$
$$k = -\frac{1}{4}$$

Answer: $\boxed{k = -\dfrac{1}{4}}$

Given: Focus is at distance 5 cm from vertex, so $a = 5$. The mirror is 45 cm deep, meaning the point on the parabola at depth 45 cm from the vertex.

Setup: Place the vertex at the origin with the axis along the x-axis. The parabola opens rightward: $y^2 = 4ax = 4(5)x = 20x$.

At depth $x = 45$ cm:
$$y^2 = 20 \times 45 = 900$$
$$y = \pm 30 \text{ cm}$$

So $A$ and $B$ are at $(45, 30)$ and $(45, -30)$.

$$AB = 30 - (-30) = 60 \text{ cm}$$

Answer: $\boxed{AB = 60 \text{ cm}}$

Given: Parabola with vertical axis; height $= 10$ m; width at base $= 5$ m.

Setup: Place the vertex at the top (origin), axis pointing downward. Equation: $x^2 = 4ay$ (opening downward, so $x^2 = -4ay$ if y is upward; let us take vertex at top, y downward positive).

Use $x^2 = 4ay$. At the base: $y = 10$ m and $x = \pm 2.5$ m (half-width).

$$2.5^2 = 4a(10) \Rightarrow 6.25 = 40a \Rightarrow a = \frac{6.25}{40} = \frac{5}{32}$$

At $y = 2$ m from vertex:
$$x^2 = 4 \times \frac{5}{32} \times 2 = \frac{40}{32} = \frac{5}{4}$$
$$x = \frac{\sqrt{5}}{2} \approx 1.118 \text{ m}$$

Width $= 2x = \sqrt{5} \approx 2.236$ m.

Answer: The arc is $\boxed{\sqrt{5} \approx 2.24 \text{ m}}$ wide at 2 m from the vertex.

Given: Towers are 200 m apart, each 30 m above roadway. Cable is 5 m above roadway at centre.

Setup: Place origin at the lowest point of the cable (centre). The cable is 5 m above roadway, so the roadway is 5 m below origin. The towers are at $x = \pm 100$ m and their tops are $30 - 5 = 25$ m above the origin.

Parabola (opening upward): $x^2 = 4ay$.

At $x = 100$, $y = 25$:
$$10000 = 4a(25) \Rightarrow 4a = 400 \Rightarrow a = 100$$

So $x^2 = 400y$.

At $x = 30$ m from centre:
$$900 = 400y \Rightarrow y = \frac{900}{400} = 2.25 \text{ m}$$

Height of cable above roadway $= 2.25 + 5 = 7.25$ m.

Answer: The length of the vertical supporting cable 30 m from the centre is $\boxed{7.25 \text{ m}}$.

Given: Parabola with vertex at highest point; vertex is 15 m above the ends; span $= 150$ m.

Setup: Place vertex at origin, axis pointing downward. Parabola: $x^2 = 4ay$ (y positive downward).

At the ends: $x = \pm 75$ m, $y = 15$ m.
$$75^2 = 4a(15) \Rightarrow 5625 = 60a \Rightarrow a = \frac{5625}{60} = \frac{375}{4}$$

At $x = 30$ m from midpoint:
$$900 = 4 \times \frac{375}{4} \times y = 375y$$
$$y = \frac{900}{375} = 2.4 \text{ m}$$

Height above ends $= 15 - 2.4 = 12.6$ m above the ends level, i.e., the girder is $15 - 2.4 = 12.6$ m above the base at that point.

Answer: The height of the girder at 30 m from the midpoint is $\boxed{12.6 \text{ m}}$ above the ends.

Given: Maximum height $= 4$ m at horizontal distance $0.5$ m from O.

Setup: Place origin at O (water outlet). The jet follows a parabolic path opening downward with vertex at $(0.5, 4)$.

Equation with vertex $(h, k) = (0.5, 4)$:
$$(x - 0.5)^2 = -4a(y - 4)$$

At $x = 0$, $y = 0$ (jet starts at O):
$$(0 - 0.5)^2 = -4a(0 - 4)$$
$$0.25 = 16a \Rightarrow a = \frac{1}{64}$$

So: $(x-0.5)^2 = -\dfrac{1}{16}(y-4)$

At $x = 0.75$:
$$(0.75 - 0.5)^2 = -\frac{1}{16}(y-4)$$
$$(0.25)^2 = -\frac{1}{16}(y-4)$$
$$0.0625 = -\frac{1}{16}(y-4)$$
$$y - 4 = -0.0625 \times 16 = -1$$
$$y = 3 \text{ m}$$

Answer: The height of the jet at 0.75 m from O is $\boxed{3 \text{ m}}$.