Exercise 7.2, Question 1
Find the point dividing the join of \((-1, 7)\) and \((4, -3)\) in the ratio \(2 : 3\).
Show solution
- \(x = \dfrac{2(4) + 3(-1)}{5} = 1\), \(y = \dfrac{2(-3) + 3(7)}{5} = 3\).
Answer: \((1, 3)\)
NCERT Solutions for Class 10 Maths Chapter 7 Exercise 7.2
Exercise 7.2: Section formula. The point dividing the join of \((x_1, y_1)\) and \((x_2, y_2)\) internally in the ratio \(m_1 : m_2\) is \(\left(\dfrac{m_1x_2 + m_2x_1}{m_1 + m_2}, \dfrac{m_1y_2 + m_2y_1}{m_1 + m_2}\right)\); the midpoint is the case \(1 : 1\). For an unknown ratio, call it \(k : 1\).
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Exercise 7.2 questions and solutions
Exercise 7.2, Question 2
Find the points of trisection of the join of \((4, -1)\) and \((-2, -3)\).
Show solution
- Ratio \(1 : 2\): \[\left(\dfrac{-2 + 8}{3}, \dfrac{-3 - 2}{3}\right) = \left(2, -\tfrac53\right)\]
- Ratio \(2 : 1\): \[\left(\dfrac{-4 + 4}{3}, \dfrac{-6 - 1}{3}\right) = \left(0, -\tfrac73\right)\]
Answer: \(\left(2, -\tfrac53\right)\) and \(\left(0, -\tfrac73\right)\)
Exercise 7.2, Question 3
On a sports ground, lines are 1 m apart and flower pots are 1 m apart along AD (100 pots). One runner posts a green flag \(\tfrac14\) of the way along line 2; another posts a red flag \(\tfrac15\) of the way along line 8. How far apart are the flags? Where should a blue flag go exactly halfway between them?
Show solution
- Take A as origin, lines along the x-axis and AD along the y-axis: green \((2, 25)\), red \((8, 20)\).
- Distance \(= \sqrt{36 + 25} = \sqrt{61}\) m.
- Midpoint: \(\left(5, \tfrac{45}{2}\right)\), i.e. on line 5 at 22.5 m.
Answer: \(\sqrt{61}\) m apart; blue flag on the 5th line, 22.5 m along
Exercise 7.2, Question 4
In what ratio does \((-1, 6)\) divide the join of \((-3, 10)\) and \((6, -8)\)?
Show solution
- Let the ratio be \(k : 1\): \[\begin{aligned}&\dfrac{6k - 3}{k + 1} = -1 \\ \Rightarrow\ &7k = 2 \\ \Rightarrow\ &k = \tfrac27\end{aligned}\]
- Check with \(y\): \[\begin{aligned}\dfrac{-8(\frac27) + 10}{\frac97} &= \dfrac{54/7}{9/7} \\ &= 6\end{aligned}\] ✓
Answer: \(2 : 7\)
Exercise 7.2, Question 5
Find the ratio in which the x-axis divides the join of \(A(1, -5)\) and \(B(-4, 5)\), and the point of division.
Show solution
- Ratio \(k : 1\); on the x-axis \(y = 0\): \[\begin{aligned}&\dfrac{5k - 5}{k + 1} = 0 \\ \Rightarrow\ &k = 1\end{aligned}\]
- Point: \[\left(\dfrac{-4 + 1}{2}, 0\right) = \left(-\tfrac32, 0\right)\]
Answer: \(1 : 1\), at \(\left(-\tfrac32, 0\right)\)
Exercise 7.2, Question 6
\((1, 2),\ (4, y),\ (x, 6),\ (3, 5)\) are the vertices of a parallelogram taken in order. Find \(x\) and \(y\).
Show solution
- Diagonals of a parallelogram bisect each other: midpoint of \((1, 2)\) and \((x, 6)\) = midpoint of \((4, y)\) and \((3, 5)\).
- \[\begin{aligned}&\dfrac{1 + x}{2} = \dfrac72 \\ \Rightarrow\ &x = 6\end{aligned}\]; \[\begin{aligned}&\dfrac{8}{2} = \dfrac{y + 5}{2} \\ \Rightarrow\ &y = 3\end{aligned}\]
Answer: \(x = 6,\ y = 3\)
Exercise 7.2, Question 7
AB is a diameter of a circle with centre \((2, -3)\) and \(B = (1, 4)\). Find A.
Show solution
- The centre is the midpoint of AB: \(\dfrac{x + 1}{2} = 2\), \(\dfrac{y + 4}{2} = -3\).
Answer: \(A(3, -10)\)
Exercise 7.2, Question 8
\(A(-2, -2)\), \(B(2, -4)\). Find P on AB with \(AP = \tfrac37 AB\).
Show solution
- \(AP : PB = 3 : 4\).
- \[\begin{aligned}P &= \left(\dfrac{3(2) + 4(-2)}{7}, \dfrac{3(-4) + 4(-2)}{7}\right) \\ &= \left(-\tfrac27, -\tfrac{20}{7}\right)\end{aligned}\]
Answer: \(\left(-\tfrac27, -\tfrac{20}{7}\right)\)
Exercise 7.2, Question 9
Find the points dividing the join of \(A(-2, 2)\) and \(B(2, 8)\) into four equal parts.
Show solution
- The middle point is the midpoint \((0, 5)\).
- The others are midpoints of \(A\) and \((0, 5)\), and of \((0, 5)\) and \(B\): \(\left(-1, \tfrac72\right)\) and \(\left(1, \tfrac{13}{2}\right)\).
Answer: \[\begin{gathered}\left(-1, \tfrac72\right), \\ (0, 5), \\ \left(1, \tfrac{13}{2}\right)\end{gathered}\]
Exercise 7.2, Question 10
Find the area of the rhombus with vertices \((3, 0),\ (4, 5),\ (-1, 4),\ (-2, -1)\) in order.
Show solution
- Diagonals: \[\begin{aligned}d_1 &= \sqrt{(3 + 1)^2 + (0 - 4)^2} \\ &= 4\sqrt2\end{aligned}\], \[\begin{aligned}d_2 &= \sqrt{(4 + 2)^2 + (5 + 1)^2} \\ &= 6\sqrt2\end{aligned}\]
- Area \[\begin{aligned}= \tfrac12 d_1 d_2 &= \tfrac12 \times 4\sqrt2 \times 6\sqrt2 \\ &= 24\end{aligned}\]
Answer: 24 square units
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