Solutions: Coordinate Geometry - Test 2

Q1. (a) $(7, 3)$

$x = \frac{3(8)+1(4)}{3+1} = \frac{28}{4} = 7$, $y = \frac{3(5)+1(-3)}{3+1} = \frac{12}{4} = 3$.

Q2. (b) $-1$

$k = \frac{1(-7) + 2(2)}{1+2} = \frac{-3}{3} = -1$.

Q3. (b) $(0, 3)$

Centroid = $(\frac{3-8+5}{3}, \frac{-7+6+10}{3}) = (\frac{0}{3}, \frac{9}{3}) = (0, 3)$.

Q4. (a) Both A and R are true and R is the correct explanation of A.

Using the formula in R, $x = \frac{2(3)+7(-6)}{9} = \frac{6-42}{9} = -4$. Correct.

Q5. Midpoint = $(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$.
$M = (\frac{-5-1}{2}, \frac{7+3}{2}) = (\frac{-6}{2}, \frac{10}{2}) = (-3, 5)$. [2]

Q6. Let ratio be $k:1$. Using x-coordinate: $-1 = \frac{k(6) + 1(-3)}{k+1}$.
$-k - 1 = 6k - 3 \Rightarrow 7k = 2 \Rightarrow k = 2/7$.
Ratio is $2:7$. [2]

Q7. Let $A = (x, y)$. Centre $O(2, -3)$ is the midpoint of AB.
$\frac{x+1}{2} = 2 \Rightarrow x+1 = 4 \Rightarrow x = 3$.
$\frac{y+4}{2} = -3 \Rightarrow y+4 = -6 \Rightarrow y = -10$.
Coordinates of A are $(3, -10)$. [2]

Q8. Let points be P and Q. P divides in 1:2, Q divides in 2:1.
P: $(\frac{1(-7)+2(2)}{3}, \frac{1(4)+2(-2)}{3}) = (\frac{-3}{3}, \frac{0}{3}) = (-1, 0)$. [1.5]
Q: $(\frac{2(-7)+1(2)}{3}, \frac{2(4)+1(-2)}{3}) = (\frac{-12}{3}, \frac{6}{3}) = (-4, 2)$. [1.5]

Q9. Let ratio be $k:1$. Point on y-axis is $(0, y)$.
$0 = \frac{k(-1) + 1(5)}{k+1} \Rightarrow -k + 5 = 0 \Rightarrow k = 5$. Ratio is $5:1$. [1.5]
$y = \frac{5(-4) + 1(-6)}{5+1} = \frac{-20-6}{6} = \frac{-26}{6} = -\frac{13}{3}$.
Point of intersection is $(0, -13/3)$. [1.5]

Q10. Given $AP = \frac{3}{7} AB$. This implies $P$ divides $AB$ in the ratio $3:4$.
$m_1 = 3, m_2 = 4$. Points $A(-2, -2), B(2, -4)$. [1]
$x = \frac{3(2) + 4(-2)}{3+4} = \frac{6-8}{7} = -\frac{2}{7}$. [2]
$y = \frac{3(-4) + 4(-2)}{3+4} = \frac{-12-8}{7} = -\frac{20}{7}$. [2]
Coordinates of P are $(-2/7, -20/7)$.

(i) Green flag is on 2nd line at 1/4 of distance AD (100m).
$x = 2, y = \frac{1}{4} \times 100 = 25$. Coordinates: $(2, 25)$. [1]

(ii) Red flag is on 8th line at 1/5 of distance AD (100m).
$x = 8, y = \frac{1}{5} \times 100 = 20$. Coordinates: $(8, 20)$. [1]

(iii) Blue flag is at the midpoint of $(2, 25)$ and $(8, 20)$.
$x = \frac{2+8}{2} = 5$.
$y = \frac{25+20}{2} = \frac{45}{2} = 22.5$.
Position: 5th line, at distance 22.5m. [2]