Co Ordinate Geometry — RRB NTPC Quantitative Aptitude

• Given: Point $P(-4, 5)$ and reflection axis is x-axis.

• Solution:
Step 1: When a point $(x, y)$ is reflected across the x-axis, its x-coordinate remains the same, but the sign of its y-coordinate changes.
Step 2: Therefore, the image of $(-4, 5)$ becomes $(-4, -5)$.

• Exam Trick: x-axis mirror $\rightarrow$ change y-sign. y-axis mirror $\rightarrow$ change x-sign. Origin mirror $\rightarrow$ change both signs. Here, x-axis means $y$ goes from $5$ to $-5$.

• Final Answer: $(-4, -5)$ (Option 1)

• Given: Points $(a+b, a-b)$ and $(a-b, a+b)$. Distance = ?

• Solution:
Step 1: Use the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.
Step 2: $x_2 - x_1 = (a-b) - (a+b) = -2b$.
Step 3: $y_2 - y_1 = (a+b) - (a-b) = 2b$.
Step 4: $d = \sqrt{(-2b)^2 + (2b)^2} = \sqrt{4b^2 + 4b^2} = \sqrt{8b^2}$.
Step 5: Simplify to get $2\sqrt{2}b$.

• Exam Trick: Difference in x is $2b$, difference in y is $2b$. Distance is hypotenuse of right triangle with legs $2b, 2b$. Thus, $d = 2b\sqrt{2}$.

• Final Answer: $2\sqrt{2}b$ (Option 1)

• Given: Point $(-15, 20)$ and Origin $(0,0)$.

• Solution:
Step 1: Distance from origin is $d = \sqrt{x^2 + y^2}$.
Step 2: $d = \sqrt{(-15)^2 + 20^2} = \sqrt{225 + 400}$.
Step 3: $d = \sqrt{625} = 25$.

• Exam Trick: Recognize the Pythagorean triplet $(15, 20, x) \rightarrow 5 \times (3, 4, 5) \rightarrow 5 \times 5 = 25$. So distance is 25 without full calculation.

• Final Answer: $25$ (Option 1)

• Given: Points $(2, 4)$ and $(7, 9)$, divided internally in $m:n = 2:3$.

• Solution:
Step 1: Section formula: $x = \frac{mx_2 + nx_1}{m+n}, y = \frac{my_2 + ny_1}{m+n}$.
Step 2: $x = \frac{2(7) + 3(2)}{2+3} = \frac{14+6}{5} = \frac{20}{5} = 4$.
Step 3: $y = \frac{2(9) + 3(4)}{2+3} = \frac{18+12}{5} = \frac{30}{5} = 6$.

• Exam Trick: Difference in x is $7-2 = 5$. A $2:3$ ratio means moving $2/5$ of the total distance. $2 + \frac{2}{5}(5) = 4$. Similarly, difference in y is $9-4 = 5$. $4 + \frac{2}{5}(5) = 6$. So, $(4, 6)$.

• Final Answer: $(4, 6)$ (Option 1)

• Given: Midpoint $M(3, -2)$ and one endpoint $A(7, 4)$. Other endpoint $B(x,y) = ?$

• Solution:
Step 1: Midpoint formula: $\frac{x_1 + x_2}{2} = x_m$ and $\frac{y_1 + y_2}{2} = y_m$.
Step 2: $\frac{7 + x}{2} = 3 \implies 7 + x = 6 \implies x = -1$.
Step 3: $\frac{4 + y}{2} = -2 \implies 4 + y = -4 \implies y = -8$.

• Exam Trick: Other Point = $2 \times \text{Midpoint} - \text{Given Point}$.
$x = 2(3) - 7 = -1$.
$y = 2(-2) - 4 = -8$. Answer is $(-1, -8)$.

• Final Answer: $(-1, -8)$ (Option 1)

• Given: Points $(-3, 4)$ and $(5, -2)$, divided externally in $m:n = 3:2$.

• Solution:
Step 1: External section formula: $x = \frac{mx_2 - nx_1}{m-n}, y = \frac{my_2 - ny_1}{m-n}$.
Step 2: $x = \frac{3(5) - 2(-3)}{3-2} = \frac{15 + 6}{1} = 21$.
Step 3: $y = \frac{3(-2) - 2(4)}{3-2} = \frac{-6 - 8}{1} = -14$.

• Exam Trick: Write ratio as $3:(-2)$. Then apply standard internal division logic: $x = \frac{3(5) + (-2)(-3)}{3-2} = 15+6 = 21$. This prevents formula confusion.

• Final Answer: $(21, -14)$ (Option 1)

• Given: Vertices $(1, 2), (-4, -3), (9, 7)$. Centroid = ?

• Solution:
Step 1: Centroid formula: $x = \frac{x_1 + x_2 + x_3}{3}, y = \frac{y_1 + y_2 + y_3}{3}$.
Step 2: $x = \frac{1 + (-4) + 9}{3} = \frac{6}{3} = 2$.
Step 3: $y = \frac{2 + (-3) + 7}{3} = \frac{6}{3} = 2$.

• Exam Trick: Centroid is simply the average of the coordinates. $1 - 4 + 9 = 6 ightarrow 6/3 = 2$. Just doing x-coord gives 2, eliminating most options instantly.

• Final Answer: $(2, 2)$ (Option 1)

• Given: Line $L_1$: $5x + 3y - 7 = 0$. Perp line slope $m_2 = ?$

• Solution:
Step 1: Slope of line $ax + by + c = 0$ is $m_1 = -a/b$.
Step 2: $m_1 = -5/3$.
Step 3: For perpendicular lines, $m_1 \times m_2 = -1$.
Step 4: $m_2 = -1 / (-5/3) = 3/5$.

• Exam Trick: If a line is $Ax + By + C = 0$, its perpendicular line has the form $Bx - Ay + K = 0$. The slope of $Bx - Ay + K = 0$ is $B/A$. Here $B=3, A=5$, so slope is $3/5$.

• Final Answer: $3/5$ (Option 1)

• Given: Point $(2, -3)$, slope $m = 1/2$.

• Solution:
Step 1: Point-slope form is $y - y_1 = m(x - x_1)$.
Step 2: $y - (-3) = \frac{1}{2}(x - 2)$.
Step 3: $2(y + 3) = x - 2$.
Step 4: $2y + 6 = x - 2 \implies x - 2y = 8$.

• Exam Trick: Put $(2, -3)$ into the options.
Opt 1: $2 - 2(-3) = 2 + 6 = 8$ (Matches)
Opt 2: $2(2) - (-3) = 4 + 3 = 7$ (Matches, but slope is 2, not 1/2)
Only Option 1 has slope $-A/B = -1/(-2) = 1/2$.

• Final Answer: $x - 2y = 8$ (Option 1)

• Given: Line passes through $(1, -2)$ and $(3, 4)$.

• Solution:
Step 1: Find slope $m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{4 - (-2)}{3 - 1} = \frac{6}{2} = 3$.
Step 2: Use point-slope form $y - y_1 = m(x - x_1)$.
Step 3: $y - 4 = 3(x - 3)$.
Step 4: $y - 4 = 3x - 9 \implies 3x - y = 5$.

• Exam Trick: Just plug the point $(3,4)$ into options.
Opt 1: $3(3) - 4 = 5$ (Matches)
Opt 2: $3(3) + 4 = 13 \neq 1$ (False)
Opt 3: $3 - 3(4) = -9 \neq 7$ (False)
Opt 4: $2(3) - 4 = 2 \neq 4$ (False). Option elimination is 5 seconds.

• Final Answer: $3x - y = 5$ (Option 1)