Exercise 7.2 Practice
Mastering Section Formula & Mid-Point Formula
Q1: Section Formula Basic
Find the coordinates of the point which divides the join of (-2, 5) and (3, -5) in the ratio 2 : 3.
Given: Points $A(-2, 5)$ and $B(3, -5)$. Ratio $m_1:m_2 = 2:3$.
Formula: $P(x, y) = \left(\frac{m_1x_2 + m_2x_1}{m_1 + m_2}, \frac{m_1y_2 + m_2y_1}{m_1 + m_2}\right)$
For x: $x = \frac{2(3) + 3(-2)}{2+3} = \frac{6 - 6}{5} = 0$
For y: $y = \frac{2(-5) + 3(5)}{2+3} = \frac{-10 + 15}{5} = \frac{5}{5} = 1$
The required point is (0, 1)
Q2: Trisection
Find the coordinates of the points of trisection of the line segment joining (6, -2) and (-3, 4).
Points of trisection divide the line into 3 equal parts. Let points be P (1:2) and Q (2:1).
Point P (1:2):
$x = \frac{1(-3) + 2(6)}{3} = \frac{9}{3} = 3$
$y = \frac{1(4) + 2(-2)}{3} = \frac{0}{3} = 0$
P is (3, 0).
$x = \frac{1(-3) + 2(6)}{3} = \frac{9}{3} = 3$
$y = \frac{1(4) + 2(-2)}{3} = \frac{0}{3} = 0$
P is (3, 0).
Point Q (2:1):
$x = \frac{2(-3) + 1(6)}{3} = \frac{0}{3} = 0$
$y = \frac{2(4) + 1(-2)}{3} = \frac{6}{3} = 2$
Q is (0, 2).
$x = \frac{2(-3) + 1(6)}{3} = \frac{0}{3} = 0$
$y = \frac{2(4) + 1(-2)}{3} = \frac{6}{3} = 2$
Q is (0, 2).
Points of trisection are (3, 0) and (0, 2)
Q3: Activity Based
In a rectangular sports field, lines are drawn 1m apart. 80 flower pots are placed 1m apart along AD.
Student A runs $\frac{1}{4}$th of AD on line 2. Student B runs $\frac{1}{5}$th of AD on line 8.
Find the distance between their flags. If Student C posts a flag exactly halfway between them, where should it be?
Coordinates:
Total AD = 80m.
Green Flag (A): Line 2, $\frac{1}{4} \times 80 = 20$. Point G(2, 20).
Red Flag (B): Line 8, $\frac{1}{5} \times 80 = 16$. Point R(8, 16).
Total AD = 80m.
Green Flag (A): Line 2, $\frac{1}{4} \times 80 = 20$. Point G(2, 20).
Red Flag (B): Line 8, $\frac{1}{5} \times 80 = 16$. Point R(8, 16).
Distance:
$D = \sqrt{(8-2)^2 + (16-20)^2} = \sqrt{36 + 16} = \sqrt{52} \approx 7.21$ m.
$D = \sqrt{(8-2)^2 + (16-20)^2} = \sqrt{36 + 16} = \sqrt{52} \approx 7.21$ m.
Midpoint (Blue Flag):
$x = \frac{2+8}{2} = 5$, $y = \frac{20+16}{2} = 18$.
$x = \frac{2+8}{2} = 5$, $y = \frac{20+16}{2} = 18$.
Blue flag is on the 5th line, 18m away.
Q4: Finding Ratio
Find the ratio in which the line segment joining the points (-2, 5) and (4, -4) is divided by the point (-1, 3.5).
Let the ratio be $k:1$. Using x-coordinate of division point (-1):
$-1 = \frac{k(4) + 1(-2)}{k+1}$
$-k - 1 = 4k - 2$
$1 = 5k \implies k = 1/5$
$1 = 5k \implies k = 1/5$
The required ratio is 1:5
Q5: Division by Axis
Find the ratio in which the line segment joining A(2, -4) and B(-3, 6) is divided by the x-axis. Also find the coordinates of the point of division.
Point on x-axis is of form $(x, 0)$. Let ratio be $k:1$.
Using y-coordinate (which is 0):
$0 = \frac{k(6) + 1(-4)}{k+1} \implies 6k - 4 = 0 \implies k = 2/3$.
$0 = \frac{k(6) + 1(-4)}{k+1} \implies 6k - 4 = 0 \implies k = 2/3$.
Find x-coordinate:
$x = \frac{2(-3) + 3(2)}{2+3} = \frac{-6+6}{5} = 0$.
$x = \frac{2(-3) + 3(2)}{2+3} = \frac{-6+6}{5} = 0$.
Ratio is 2:3 and Point is (0, 0)
Q6: Parallelogram
If (2, 3), (5, y), (x, 7) and (4, 6) are the vertices of a parallelogram taken in order, find x and y.
Diagonals of a parallelogram bisect each other. Midpoint of AC = Midpoint of BD.
Midpoint AC: $(\frac{2+x}{2}, \frac{3+7}{2}) = (\frac{2+x}{2}, 5)$
Midpoint BD: $(\frac{5+4}{2}, \frac{y+6}{2}) = (4.5, \frac{y+6}{2})$
Equating coordinates:
$2+x = 9 \implies x=7$
$10 = y+6 \implies y=4$
$2+x = 9 \implies x=7$
$10 = y+6 \implies y=4$
x = 7, y = 4
Q7: Circle Diameter
Find the coordinates of point A, where AB is the diameter of a circle whose centre is (3, -2) and B is (2, 5).
Let A be $(x, y)$. Centre (3, -2) is the midpoint of AB.
$\frac{x+2}{2} = 3 \implies x = 6 - 2 = 4$
$\frac{y+5}{2} = -2 \implies y = -4 - 5 = -9$
Coordinates of A are (4, -9)
Q8: Line Segment
If A and B are (-2, -2) and (2, -4) respectively, find the coordinates of P such that $AP = \frac{4}{7}AB$ and P lies on the line segment AB.
If $AP = \frac{4}{7}AB$, then $AP:AB = 4:7$.
This means $AP:PB = 4:3$.
This means $AP:PB = 4:3$.
Section Formula (4:3):
$x = \frac{4(2) + 3(-2)}{7} = \frac{8-6}{7} = \frac{2}{7}$
$y = \frac{4(-4) + 3(-2)}{7} = \frac{-16-6}{7} = \frac{-22}{7}$
$x = \frac{4(2) + 3(-2)}{7} = \frac{8-6}{7} = \frac{2}{7}$
$y = \frac{4(-4) + 3(-2)}{7} = \frac{-16-6}{7} = \frac{-22}{7}$
P is $(\frac{2}{7}, \frac{-22}{7})$
Q9: Four Equal Parts
Find the coordinates of the points which divide the line segment joining A(-4, 0) and B(4, 8) into four equal parts.
Midpoint M (of AB): $(\frac{-4+4}{2}, \frac{0+8}{2}) = (0, 4)$
Midpoint P1 (of AM): $(\frac{-4+0}{2}, \frac{0+4}{2}) = (-2, 2)$
Midpoint P3 (of MB): $(\frac{0+4}{2}, \frac{4+8}{2}) = (2, 6)$
Points are (-2, 2), (0, 4), and (2, 6)
Q10: Area of Rhombus
Find the area of a rhombus if its vertices are (4, 0), (5, 6), (-2, 5) and (-3, -1) taken in order.
Area = $\frac{1}{2} \times d_1 \times d_2$. Calculate diagonals AC and BD.
$d_1 (AC) = \sqrt{(-2-4)^2 + (5-0)^2} = \sqrt{36+25} = \sqrt{61}$
$d_2 (BD) = \sqrt{(-3-5)^2 + (-1-6)^2} = \sqrt{64+49} = \sqrt{113}$
Area = $\frac{1}{2}\sqrt{6893}$ sq units (approx 41.5)