Case Study of Chapter 3 Coordinate Geometry Class 9th

Case Study of Chapter 3 Coordinate Geometry Class 9th

Case Study of Chapter 3 Coordinate Geometry Class 9th | Free Online Test

Case Study of Chapter 3 Coordinate Geometry Class 9th

Students preparing for exams often search for Case Study math questions for class 9. These exercises help strengthen concepts in the coordinate geometry. Our online tests include interactive math case study questions class 9 that focus on real-life applications of numbers. Practicing these questions regularly improves accuracy and speed. Additionally, students develop problem-solving skills while applying formulas in practical situations.

Importance of Case Study of Chapter 3 Coordinate Geometry Class 9th

Math case study questions class 9 encourage analytical thinking and logical reasoning. For instance, questions on rational and irrational numbers allow deeper understanding. Furthermore, solving these problems enhances critical thinking. Short exercises help reinforce key formulas. Therefore, students gain confidence and clarity in the coordinate geometry through consistent practice.

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Case Study 3: Coordinate Geometry

A navigation team is analyzing a small island map represented on the Cartesian plane to plan rescue routes. Several key points are marked: a lookout tower at $P(1,2)$, a supply depot at $Q(4,8)$, an emergency beacon at $R(7,14)$, a stranded hut at $S(-3,-7)$, and two temporary posts at $A(2,1)$ and $B(-1,4)$. The team must: (i) check whether three points lie on the same straight path (collinearity), (ii) compute precise straight-line distances for rope-laying, (iii) find reflections across axes for mirror-based signaling, (iv) determine intersection coordinates of proposed radio mast lines, and (v) compute the area of a triangular search-sector to estimate coverage. Use standard formulas: distance between $(x_1,y_1)$ and $(x_2,y_2)$, \[ d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}, \] slope of the line joining two points, \[ m=\frac{y_2-y_1}{x_2-x_1}\quad (x_1\neq x_2), \] collinearity test via slopes ($m_{12}=m_{23}$) or determinant method: \[ \text{Points }(x_1,y_1),(x_2,y_2),(x_3,y_3)\text{ are collinear iff } \begin{vmatrix} x_1 & y_1 & 1\\[4pt] x_2 & y_2 & 1\\[4pt] x_3 & y_3 & 1 \end{vmatrix}=0, \] and area of triangle with vertices $(x_1,y_1),(x_2,y_2),(x_3,y_3)$: \[ \text{Area}=\frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|. \] Apply these precisely to answer the questions below.

MCQ Questions

1. Are the points $P(1,2)$, $Q(4,8)$ and $R(7,14)$ collinear?

  • A) Yes, they are collinear.
  • B) No, they are not collinear.
  • C) They are collinear only if shifted by $(1,1)$.
  • D) Only $P$ and $Q$ are collinear with the origin.
Answer: A) Yes, they are collinear.
Solution: Compute slope $m_{PQ}=\dfrac{8-2}{4-1}=\dfrac{6}{3}=2$ and slope $m_{QR}=\dfrac{14-8}{7-4}=\dfrac{6}{3}=2$. Since $m_{PQ}=m_{QR}$ the three points lie on the same straight line. Alternatively the determinant \[ \begin{vmatrix} 1 & 2 & 1\\[4pt] 4 & 8 & 1\\[4pt] 7 & 14 & 1 \end{vmatrix}=0, \] confirming collinearity.

2. The straight-line distance between $P(1,2)$ and $R(7,14)$ is:

  • A) $6\sqrt{5}$
  • B) $\sqrt{1800}$
  • C) $12$
  • D) $8\sqrt{2}$
Answer: A) $6\sqrt{5}$
Solution: Use distance formula: \[ d=\sqrt{(7-1)^2+(14-2)^2}=\sqrt{6^2+12^2}=\sqrt{36+144}=\sqrt{180}=6\sqrt{5}. \] Note option (b) is $\sqrt{1800}=10\sqrt{18}$ which is not equal to $\sqrt{180}$.

3. The reflection of the hut $S(-3,-7)$ across the $x$-axis has coordinates:

  • A) $(-3,7)$
  • B) $(3,-7)$
  • C) $(-3,-7)$
  • D) $(3,7)$
Answer: A) $(-3,7)$
Solution: Reflection across the $x$-axis maps $(x,y)\mapsto (x,-y)$. Thus $S(-3,-7)\mapsto (-3,7)$.

4. Lines considered for radio masts: line $\ell_1$ through $A(2,1)$ with slope $3$, and line $\ell_2$ through $B(-1,4)$ with slope $-\tfrac{1}{3}$. Their intersection point lies in which quadrant?

  • A) First Quadrant
  • B) Second Quadrant
  • C) Third Quadrant
  • D) Fourth Quadrant
Answer: A) First Quadrant
Solution: Equation of $\ell_1$: $y-1=3(x-2)\Rightarrow y=3x-5$. Equation of $\ell_2$: $y-4=-\tfrac{1}{3}(x+1)\Rightarrow y=-\tfrac{1}{3}x+\tfrac{11}{3}$. Solve: \[ 3x-5=-\tfrac{1}{3}x+\tfrac{11}{3}\quad\Rightarrow\quad 9x-15=-x+11\quad\Rightarrow\quad 10x=26, \] so $x=\dfrac{26}{10}=\dfrac{13}{5}>0$. Then \[ y=3\cdot \dfrac{13}{5}-5=\dfrac{39}{5}-\dfrac{25}{5}=\dfrac{14}{5}>0. \] Both coordinates positive $\Rightarrow$ First Quadrant.

5. The area of the triangle with vertices $O(0,0)$, $B(4,0)$ and $T(1,3)$ is:

  • A) $6$ square units
  • B) $\dfrac{9}{2}$ square units
  • C) $4$ square units
  • D) $3$ square units
Answer: A) $6$ square units
Solution: Using the determinant (shoelace) formula: \[ \text{Area}=\frac{1}{2}\left|0(0-3)+4(3-0)+1(0-0)\right|=\frac{1}{2}\left|0+12+0\right|=6. \]

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