Coordinate Geometry — NCERT Solutions

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

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

(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 $3x - y + 1 = 0$.)

(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 $r$: $(x-h)^2 + (y-k)^2 = r^2$.

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

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

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