Coordinate Geometry — NCERT Solutions

Given: $Q(0,1)$, $P(5,-3)$, $R(x,6)$, and $QP = QR$.

Step 1: Find QP and QR.
$$QP = \sqrt{(5-0)^2+(-3-1)^2} = \sqrt{25+16} = \sqrt{41}$$
$$QR = \sqrt{(x-0)^2+(6-1)^2} = \sqrt{x^2+25}$$

Step 2: Set QP = QR and solve.
$$\sqrt{x^2+25} = \sqrt{41}$$
$$x^2+25 = 41$$
$$x^2 = 16$$
$$x = \pm 4$$

Step 3: Find QR.
$$QR = \sqrt{41} \text{ units}$$

Step 4: Find PR for both values of $x$.

For $x = 4$, $R(4,6)$:
$$PR = \sqrt{(4-5)^2+(6-(-3))^2} = \sqrt{1+81} = \sqrt{82}$$

For $x = -4$, $R(-4,6)$:
$$PR = \sqrt{(-4-5)^2+(6-(-3))^2} = \sqrt{81+81} = 9\sqrt{2}$$

Answer: $x = 4$ or $x = -4$; $QR = \sqrt{41}$ units; $PR = \sqrt{82}$ units (when $x=4$) or $PR = 9\sqrt{2}$ units (when $x=-4$).

Given: Point $P(x,y)$ is equidistant from $A(3,6)$ and $B(-3,4)$.

Condition: $PA = PB$

Working:
$$PA = \sqrt{(x-3)^2+(y-6)^2}$$
$$PB = \sqrt{(x+3)^2+(y-4)^2}$$

Setting $PA = PB$ and squaring:
$$(x-3)^2+(y-6)^2 = (x+3)^2+(y-4)^2$$
$$x^2-6x+9+y^2-12y+36 = x^2+6x+9+y^2-8y+16$$
$$-6x-12y+45 = 6x-8y+25$$
$$-12x-4y+20 = 0$$
$$3x+y = 5$$

Answer: The required relation is $3x + y = 5$.

Given: $A(-1,7)$, $B(4,-3)$, ratio $m_1:m_2 = 2:3$.

Formula (Section Formula):
$$P(x,y) = \left(\frac{m_1 x_2+m_2 x_1}{m_1+m_2},\; \frac{m_1 y_2+m_2 y_1}{m_1+m_2}\right)$$

Working:
$$x = \frac{2(4)+3(-1)}{2+3} = \frac{8-3}{5} = \frac{5}{5} = 1$$
$$y = \frac{2(-3)+3(7)}{2+3} = \frac{-6+21}{5} = \frac{15}{5} = 3$$

Answer: The required point is $(1, 3)$.

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

Concept: Trisection means dividing into three equal parts. Let $P$ and $Q$ be the two trisection points. Then $P$ divides $AB$ in ratio $1:2$ and $Q$ divides $AB$ in ratio $2:1$.

For point P (ratio 1:2):
$$x_P = \frac{1(-2)+2(4)}{1+2} = \frac{-2+8}{3} = \frac{6}{3} = 2$$
$$y_P = \frac{1(-3)+2(-1)}{1+2} = \frac{-3-2}{3} = \frac{-5}{3}$$

For point Q (ratio 2:1):
$$x_Q = \frac{2(-2)+1(4)}{2+1} = \frac{-4+4}{3} = 0$$
$$y_Q = \frac{2(-3)+1(-1)}{2+1} = \frac{-6-1}{3} = \frac{-7}{3}$$

Answer: The two trisection points are $\left(2,\,-\dfrac{5}{3}\right)$ and $\left(0,\,-\dfrac{7}{3}\right)$.

Given: AD has 100 flower pots at 1 m apart, so $AD = 100$ m. Lines are 1 m apart.

Position of Green Flag (Niharika):
Line 2 from AD, distance $= \frac{1}{4} \times 100 = 25$ m along AD.
Coordinates of green flag: $G(2, 25)$.

Position of Red Flag (Preet):
Line 8 from AD, distance $= \frac{1}{5} \times 100 = 20$ m along AD.
Coordinates of red flag: $R(8, 20)$.

Distance between the flags:
$$GR = \sqrt{(8-2)^2+(20-25)^2} = \sqrt{36+25} = \sqrt{61} \text{ m}$$

Position of Blue Flag (Rashmi) — midpoint of GR:
$$x = \frac{2+8}{2} = 5, \quad y = \frac{25+20}{2} = \frac{45}{2} = 22.5$$

Answer: The distance between the flags is $\sqrt{61}$ m. Rashmi should post the blue flag on the 5th line at a distance of 22.5 m from AD.

Given: $A(-3,10)$, $B(6,-8)$, point $P(-1,6)$ divides $AB$ in ratio $k:1$.

Using Section Formula:
$$x = \frac{k \cdot x_2 + 1 \cdot x_1}{k+1}$$
$$-1 = \frac{k(6)+1(-3)}{k+1} = \frac{6k-3}{k+1}$$
$$-k-1 = 6k-3$$
$$2 = 7k$$
$$k = \frac{2}{7}$$

Verification with $y$-coordinate:
$$y = \frac{\frac{2}{7}(-8)+10}{\frac{2}{7}+1} = \frac{\frac{-16}{7}+10}{\frac{9}{7}} = \frac{\frac{54}{7}}{\frac{9}{7}} = 6 \checkmark$$

Answer: The point $(-1,6)$ divides the segment in the ratio $\dfrac{2}{7}:1 = \mathbf{2:7}$.

Given: $A(1,-5)$, $B(-4,5)$. The $x$-axis has $y = 0$.

Let the ratio be $k:1$. Using the $y$-coordinate of the section formula:
$$0 = \frac{k(5)+1(-5)}{k+1}$$
$$0 = 5k-5$$
$$k = 1$$

So the ratio is $1:1$.

Finding $x$-coordinate:
$$x = \frac{1(-4)+1(1)}{1+1} = \frac{-4+1}{2} = \frac{-3}{2}$$

Answer: The $x$-axis divides $AB$ in the ratio $\mathbf{1:1}$ and the point of division is $\left(-\dfrac{3}{2},\, 0\right)$.

Given: Vertices of parallelogram $A(1,2)$, $B(4,y)$, $C(x,6)$, $D(3,5)$.

Concept: Diagonals of a parallelogram bisect each other, so midpoint of $AC$ = midpoint of $BD$.

Midpoint of AC:
$$\left(\frac{1+x}{2},\; \frac{2+6}{2}\right) = \left(\frac{1+x}{2},\; 4\right)$$

Midpoint of BD:
$$\left(\frac{4+3}{2},\; \frac{y+5}{2}\right) = \left(\frac{7}{2},\; \frac{y+5}{2}\right)$$

Equating:
$$\frac{1+x}{2} = \frac{7}{2} \Rightarrow 1+x = 7 \Rightarrow x = 6$$
$$4 = \frac{y+5}{2} \Rightarrow y+5 = 8 \Rightarrow y = 3$$

Answer: $x = 6$ and $y = 3$.

Given: Centre $O(2,-3)$, one end of diameter $B(1,4)$, other end $A(x,y)$.

Concept: The centre is the midpoint of the diameter.

Working:
$$\frac{x+1}{2} = 2 \Rightarrow x+1 = 4 \Rightarrow x = 3$$
$$\frac{y+4}{2} = -3 \Rightarrow y+4 = -6 \Rightarrow y = -10$$

Answer: The coordinates of point $A$ are $(3, -10)$.

Given: $A(-2,-2)$, $B(2,-4)$, $AP = \dfrac{3}{7}AB$.

Finding the ratio AP:PB:
$$AP = \frac{3}{7}AB \Rightarrow PB = AB - AP = AB - \frac{3}{7}AB = \frac{4}{7}AB$$
$$\frac{AP}{PB} = \frac{3/7}{4/7} = \frac{3}{4}$$

So P divides AB in ratio $3:4$.

Using Section Formula:
$$x = \frac{3(2)+4(-2)}{3+4} = \frac{6-8}{7} = \frac{-2}{7}$$
$$y = \frac{3(-4)+4(-2)}{3+4} = \frac{-12-8}{7} = \frac{-20}{7}$$

Answer: The coordinates of P are $\left(-\dfrac{2}{7},\,-\dfrac{20}{7}\right)$.

Given: $A(-2,2)$ and $B(2,8)$.

Concept: Let $P_1$, $P_2$, $P_3$ be the three points dividing $AB$ into four equal parts.

$P_2$ is the midpoint of AB:
$$P_2 = \left(\frac{-2+2}{2},\frac{2+8}{2}\right) = (0, 5)$$

$P_1$ is the midpoint of $A$ and $P_2$:
$$P_1 = \left(\frac{-2+0}{2},\frac{2+5}{2}\right) = \left(-1,\frac{7}{2}\right)$$

$P_3$ is the midpoint of $P_2$ and $B$:
$$P_3 = \left(\frac{0+2}{2},\frac{5+8}{2}\right) = \left(1,\frac{13}{2}\right)$$

Answer: The three points are $\left(-1,\,\dfrac{7}{2}\right)$, $(0,\,5)$, and $\left(1,\,\dfrac{13}{2}\right)$.

Given: Vertices $A(3,0)$, $B(4,5)$, $C(-1,4)$, $D(-2,-1)$.

Concept: Area of rhombus $= \dfrac{1}{2} \times d_1 \times d_2$, where $d_1$ and $d_2$ are the diagonals.

Diagonal AC:
$$AC = \sqrt{(-1-3)^2+(4-0)^2} = \sqrt{16+16} = \sqrt{32} = 4\sqrt{2}$$

Diagonal BD:
$$BD = \sqrt{(-2-4)^2+(-1-5)^2} = \sqrt{36+36} = \sqrt{72} = 6\sqrt{2}$$

Area:
$$\text{Area} = \frac{1}{2} \times 4\sqrt{2} \times 6\sqrt{2} = \frac{1}{2} \times 24 \times 2 = 24 \text{ sq. units}$$

Answer: The area of the rhombus is $\mathbf{24}$ square units.