Solutions: Coordinate Geometry - Test 3

Q1. (c) $\sqrt{a^2 + b^2}$

$D = \sqrt{(0 - (a\cos\theta + b\sin\theta))^2 + (a\sin\theta - b\cos\theta - 0)^2}$
$= \sqrt{a^2\cos^2\theta + b^2\sin^2\theta + 2ab\sin\theta\cos\theta + a^2\sin^2\theta + b^2\cos^2\theta - 2ab\sin\theta\cos\theta}$
$= \sqrt{a^2(\cos^2\theta+\sin^2\theta) + b^2(\sin^2\theta+\cos^2\theta)} = \sqrt{a^2+b^2}$.

Q2. (b) $AP = PB$

Check if P is midpoint. Midpoint of AB = $(\frac{4+8}{2}, \frac{2+4}{2}) = (6, 3)$. Wait, P is $(2, 1)$.
Actually, points are $A(4, 2), B(8, 4)$. $P(2, 1)$ is outside? No, let's check slope. Slope AB = $2/4 = 1/2$. Slope AP = $(1-2)/(2-4) = -1/-2 = 1/2$. Points are collinear.
Distance $AP = \sqrt{(4-2)^2 + (2-1)^2} = \sqrt{4+1} = \sqrt{5}$.
Distance $AB = \sqrt{(8-4)^2 + (4-2)^2} = \sqrt{16+4} = \sqrt{20} = 2\sqrt{5}$.
Wait, $P(2,1)$ is not between A and B. $A$ is between $P$ and $B$. $PA = \sqrt{5}, AB = 2\sqrt{5}$. So $PB = 3\sqrt{5}$.
The question implies P lies on segment. Let's re-read. "P(2, 1) lies on line segment joining A(4, 2) and B(8, 4)". This is geometrically impossible as $2 < 4 < 8$. P is outside.
Correction: If the question meant $P$ divides $AB$ externally, or maybe I miscalculated. Let's assume the question meant specific ratio. But based on options, if $P$ was $(6, 3)$, it would be midpoint. Let's assume the question is valid and check options. $AP = \sqrt{5}, AB = 2\sqrt{5}$. So $AB = 2AP$ or $AP = 1/2 AB$. Option (d).

Q3. (b) 12

Vertices $(0, 4), (0, 0), (3, 0)$. It's a right triangle with legs 3 and 4.
Hypotenuse = $\sqrt{3^2 + 4^2} = 5$. Perimeter = $3 + 4 + 5 = 12$.

Q4. (b) -12

Midpoint x-coord: $\frac{-6 + (-2)}{2} = \frac{-8}{2} = -4$.
Given $a/3 = -4 \Rightarrow a = -12$.

Q5. (a) Both A and R are true and R is the correct explanation of A.

Distance of $(0, 2)$ from $(-1, 1)$ is $\sqrt{1^2 + 1^2} = \sqrt{2}$.
Distance of $(0, 2)$ from $(3, 3)$ is $\sqrt{3^2 + 1^2} = \sqrt{10}$. Not equidistant.
Wait, let's check calculation. $P(0, 2), A(-1, 1)$. $PA^2 = (0+1)^2 + (2-1)^2 = 1+1=2$.
$P(0, 2), B(3, 3)$. $PB^2 = (0-3)^2 + (2-3)^2 = 9+1=10$.
Assertion is False. So answer should be (d). But wait, did I misread? "Intersection of y-axis and perpendicular bisector".
Let point on y-axis be $(0, y)$. Equidistant from $(-1, 1)$ and $(3, 3)$.
$(0+1)^2 + (y-1)^2 = (0-3)^2 + (y-3)^2 \Rightarrow 1 + y^2 - 2y + 1 = 9 + y^2 - 6y + 9$.
$-2y + 2 = -6y + 18 \Rightarrow 4y = 16 \Rightarrow y = 4$.
So point is $(0, 4)$. Assertion says $(0, 2)$. So A is False. Answer (d).

Q6. Let $P(x, y)$ be equidistant from $A(7, 1)$ and $B(3, 5)$.
$PA^2 = PB^2 \Rightarrow (x-7)^2 + (y-1)^2 = (x-3)^2 + (y-5)^2$.
$x^2 - 14x + 49 + y^2 - 2y + 1 = x^2 - 6x + 9 + y^2 - 10y + 25$.
$-14x - 2y + 50 = -6x - 10y + 34$.
$-8x + 8y + 16 = 0 \Rightarrow -x + y + 2 = 0$ or $x - y = 2$. [2]

Q7. Let $A(5, -2), B(6, 4), C(7, -2)$.
$AB = \sqrt{(6-5)^2 + (4+2)^2} = \sqrt{1+36} = \sqrt{37}$.
$BC = \sqrt{(7-6)^2 + (-2-4)^2} = \sqrt{1+36} = \sqrt{37}$.
$AC = \sqrt{(7-5)^2 + (-2+2)^2} = \sqrt{4} = 2$.
Since $AB = BC$, it is an isosceles triangle. [2]

Q8. Midpoint of diagonals coincide.
Midpoint of AC: $(\frac{6+9}{2}, \frac{1+4}{2}) = (7.5, 2.5)$.
Midpoint of BD: $(\frac{8+p}{2}, \frac{2+3}{2}) = (\frac{8+p}{2}, 2.5)$.
$\frac{8+p}{2} = 7.5 \Rightarrow 8+p = 15 \Rightarrow p = 7$. [2]

Q9. Points $P, Q, R$ divide $AB$ into 4 equal parts.
$Q$ is midpoint of $AB$: $(\frac{-2+2}{2}, \frac{2+8}{2}) = (0, 5)$. [1]
$P$ is midpoint of $AQ$: $(\frac{-2+0}{2}, \frac{2+5}{2}) = (-1, 3.5)$. [1]
$R$ is midpoint of $QB$: $(\frac{0+2}{2}, \frac{5+8}{2}) = (1, 6.5)$. [1]

Q10. Let center be $O(x, y)$. $OA = OB = OC = R$.
$OA^2 = OB^2 \Rightarrow (x-6)^2 + (y+6)^2 = (x-3)^2 + (y+7)^2$.
$-12x + 36 + 12y + 36 = -6x + 9 + 14y + 49$.
$-6x - 2y + 14 = 0 \Rightarrow 3x + y = 7$ ... (i)
$OB^2 = OC^2 \Rightarrow (x-3)^2 + (y+7)^2 = (x-3)^2 + (y-3)^2$.
$(y+7)^2 = (y-3)^2 \Rightarrow y^2 + 14y + 49 = y^2 - 6y + 9$.
$20y = -40 \Rightarrow y = -2$.
Put $y = -2$ in (i): $3x - 2 = 7 \Rightarrow 3x = 9 \Rightarrow x = 3$.
Center is $(3, -2)$. [3]

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

(ii) $BC = \sqrt{(8-6)^2 + (1-4)^2} = \sqrt{2^2 + (-3)^2} = \sqrt{4+9} = \sqrt{13}$ units. [1]

(iii) $AC = \sqrt{(8-3)^2 + (1-1)^2} = \sqrt{5^2} = 5$.
$AB^2 = 18, BC^2 = 13, AC^2 = 25$.
Since $AB^2 + BC^2 = 18 + 13 = 31 \neq AC^2$, it is not a right-angled triangle. [2]

(i) House $H(2, 4)$, Bank $B(5, 8)$.
$HB = \sqrt{(5-2)^2 + (8-4)^2} = \sqrt{3^2 + 4^2} = \sqrt{9+16} = 5$ units. [1]

(ii) Bank to School $S(13, 14)$: $BS = \sqrt{(13-5)^2 + (14-8)^2} = \sqrt{8^2 + 6^2} = 10$.
School to Office $O(13, 26)$: $SO = \sqrt{(13-13)^2 + (26-14)^2} = 12$.
Total distance = $5 + 10 + 12 = 27$ units. [2]

(iii) Direct House to Office: $HO = \sqrt{(13-2)^2 + (26-4)^2} = \sqrt{11^2 + 22^2} = \sqrt{121 + 484} = \sqrt{605} \approx 24.6$.
Distance saved = $27 - 24.6 = 2.4$ units (approx). [1]