Solutions: Coordinate Geometry - Test 1

Q1. (b) 3

The distance of a point $(x, y)$ from the x-axis is $|y|$. Here $|3| = 3$.

Q2. (c) 10

Distance from origin = $\sqrt{(-6)^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$.

Q3. (c) $(-4, 2)$

Midpoint = $(\frac{-2-6}{2}, \frac{8-4}{2}) = (\frac{-8}{2}, \frac{4}{2}) = (-4, 2)$.

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

Since the x-coordinate is 0, the point lies on the y-axis.

Q5. $PQ = 10 \Rightarrow PQ^2 = 100$.
$(10-2)^2 + (y+3)^2 = 100 \Rightarrow 64 + (y+3)^2 = 100$.
$(y+3)^2 = 36 \Rightarrow y+3 = \pm 6$.
$y = 3$ or $y = -9$. [2]

Q6. Let the ratio be $k:1$. The x-coordinate of a point on y-axis is 0.
$x = \frac{m_1x_2 + m_2x_1}{m_1+m_2} \Rightarrow 0 = \frac{k(-1) + 1(5)}{k+1}$.
$-k + 5 = 0 \Rightarrow k = 5$. Ratio is 5:1. [2]

Q7. Diagonals of a parallelogram bisect each other.
Midpoint of AC = Midpoint of BD.
$(\frac{1+x}{2}, \frac{2+6}{2}) = (\frac{4+3}{2}, \frac{y+5}{2})$.
$\frac{1+x}{2} = \frac{7}{2} \Rightarrow 1+x=7 \Rightarrow x=6$.
$4 = \frac{y+5}{2} \Rightarrow 8 = y+5 \Rightarrow y=3$. [2]

Q8. Let points be P and Q. P divides in 1:2, Q divides in 2:1.
P: $(\frac{1(-2)+2(4)}{3}, \frac{1(-3)+2(-1)}{3}) = (\frac{6}{3}, \frac{-5}{3}) = (2, -5/3)$.
Q: $(\frac{2(-2)+1(4)}{3}, \frac{2(-3)+1(-1)}{3}) = (\frac{0}{3}, \frac{-7}{3}) = (0, -7/3)$. [3]

Q9. Let point be $P(x, 0)$. $PA = PB \Rightarrow PA^2 = PB^2$.
$(x-2)^2 + (0+5)^2 = (x+2)^2 + (0-9)^2$.
$x^2 - 4x + 4 + 25 = x^2 + 4x + 4 + 81$.
$-4x + 29 = 4x + 85 \Rightarrow -8x = 56 \Rightarrow x = -7$.
Point is $(-7, 0)$. [3]

Q10. Vertices $A(3, 0), B(4, 5), C(-1, 4), D(-2, -1)$.
Diagonal $AC = \sqrt{(-1-3)^2 + (4-0)^2} = \sqrt{(-4)^2 + 4^2} = \sqrt{16+16} = \sqrt{32} = 4\sqrt{2}$. [2]
Diagonal $BD = \sqrt{(-2-4)^2 + (-1-5)^2} = \sqrt{(-6)^2 + (-6)^2} = \sqrt{36+36} = \sqrt{72} = 6\sqrt{2}$. [2]
Area = $\frac{1}{2} \times d_1 \times d_2 = \frac{1}{2} \times 4\sqrt{2} \times 6\sqrt{2} = \frac{1}{2} \times 24 \times 2 = 24$ sq units. [1]

(i) $AB = \sqrt{(6-3)^2 + (7-4)^2} = \sqrt{3^2 + 3^2} = \sqrt{18} = 3\sqrt{2}$. [1]

(ii) $BC = \sqrt{(9-6)^2 + (4-7)^2} = \sqrt{3^2 + (-3)^2} = \sqrt{18} = 3\sqrt{2}$. [1]

(iii) $CD = 3\sqrt{2}, DA = 3\sqrt{2}$. All sides are equal.
Diagonal $AC = \sqrt{(9-3)^2 + (4-4)^2} = 6$.
Diagonal $BD = \sqrt{(6-6)^2 + (1-7)^2} = 6$.
Since sides are equal and diagonals are equal, ABCD is a square. [2]