Solutions: Coordinate Geometry - Vardaan Learning Institute

1. The distance of the point P(2, 3) from the x-axis is:

Distance from x-axis is the absolute value of y-coordinate = |3| = 3.

Correct Option: (b) 3

2. The points (-1, -2), (1, 0), (-1, 2), (-3, 0) form a quadrilateral of type:

All sides are equal ($\sqrt{8}$). Diagonals are equal ($\sqrt{16}=4$). Hence, Square.

Correct Option: (a) Square

3. If the distance between the points (4, p) and (1, 0) is 5, then the value of p is:

$\sqrt{(4-1)^2 + p^2} = 5 \Rightarrow 9 + p^2 = 25 \Rightarrow p^2 = 16 \Rightarrow p = \pm 4$.

Correct Option: (b) $\pm 4$

4. The mid-point of the line segment joining the points A(-2, 8) and B(-6, -4) is:

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

Correct Option: (c) (-4, 2)

// Wait, in the options provided in file, option b was 2,6. Option c was -4,2. Let me check my memory of options. // In BonusQuestions, opt c is (-4, 2). Answer correct.

5. The point which divides the line segment joining the points (7, -6) and (3, 4) in the ratio 1:2 internally lies in the:

$x = \frac{1(3)+2(7)}{3} = \frac{17}{3}$ (+ve).
$y = \frac{1(4)+2(-6)}{3} = \frac{-8}{3}$ (-ve).
(+ve, -ve) is IV Quadrant.

Correct Option: (d) IV quadrant

6. If the points A(1, 2), B(0, 0) and C(a, b) are collinear, then:

Slope AB = Slope BC. $\frac{0-2}{0-1} = \frac{b-0}{a-0} \Rightarrow 2 = \frac{b}{a} \Rightarrow b = 2a$.

Correct Option: (c) 2a = b

// Option c is 2a = b. Correct.

7. The distance of the point P(-6, 8) from the origin is:

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

Correct Option: (c) 10

8. If the mid-point of the line segment joining the points P(6, b-2) and Q(-2, 4) is (2, -3), find the value of b.

y-coordinate of midpoint: $\frac{(b-2) + 4}{2} = -3$
$b + 2 = -6 \Rightarrow b = -8$.

Correct Option: (b) -8

9. The coordinates of the point which divides the line segment joining (-1, 3) and (4, -7) internally in the ratio 3:2 are:

$x = \frac{3(4) + 2(-1)}{5} = \frac{12-2}{5} = \frac{10}{5} = 2$.
$y = \frac{3(-7) + 2(3)}{5} = \frac{-21+6}{5} = \frac{-15}{5} = -3$.

Correct Option: (a) (2, -3)

10. Assertion (A): The point (0, 6) lies on the y-axis. Reason (R): The x-coordinate of every point on the y-axis is zero.

(0, 6) has x=0, so it is on y-axis. A is true.
Definition of y-axis is x=0. R is true and explains A.

Correct Option: (a) Both A and R are true and R is the correct explanation of A.

11. Find the relations between x and y such that the point (x, y) is equidistant from the points (7, 1) and (3, 5).

$(x-7)^2 + (y-1)^2 = (x-3)^2 + (y-5)^2$.
$x^2 - 14x + 49 + y^2 - 2y + 1 = x^2 - 6x + 9 + y^2 - 10y + 25$.
$-14x - 2y + 50 = -6x - 10y + 34$.
$-8x + 8y + 16 = 0 \Rightarrow -x + y + 2 = 0 \Rightarrow x - y = 2$.

Relation: x - y = 2

12. Find the ratio in which the y-axis divides the line segment joining the points (5, -6) and (-1, -4). Also find the point of intersection.

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

Ratio: 5:1; Point: (0, -13/3)

13. Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, -3) and B is (1, 4).

Midpoint of AB is Centre C(2, -3). Let A be $(x, y)$.
$\frac{x+1}{2} = 2 \Rightarrow x = 3$.
$\frac{y+4}{2} = -3 \Rightarrow y = -10$.

A(3, -10)

14. Check whether (5, -2), (6, 4) and (7, -2) are the vertices of an isosceles triangle.

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

Yes, Isosceles

15. Find the area of the rhombus if its vertices are (3, 0), (4, 5), (-1, 4) and (-2, -1) taken in order.

$d1 = \text{Distance bet. (3,0) and (-1,4)} = \sqrt{(-4)^2 + 4^2} = \sqrt{32} = 4\sqrt{2}$.
$d2 = \text{Distance bet. (4,5) and (-2,-1)} = \sqrt{(-6)^2 + (-6)^2} = \sqrt{72} = 6\sqrt{2}$.
Area = $\frac{1}{2} \times d1 \times d2 = \frac{1}{2} \times 4\sqrt{2} \times 6\sqrt{2} = \frac{1}{2} \times 24 \times 2 = 24$.

Area = 24 sq units

16. Find the coordinates of the points of trisection of the line segment joining (4, -1) and (-2, -3).

Case 1 (Ratio 1:2): $P(\frac{1(-2)+2(4)}{3}, \frac{1(-3)+2(-1)}{3}) = (\frac{6}{3}, \frac{-5}{3}) = (2, -5/3)$.
Case 2 (Ratio 2:1): $Q(\frac{2(-2)+1(4)}{3}, \frac{2(-3)+1(-1)}{3}) = (\frac{0}{3}, \frac{-7}{3}) = (0, -7/3)$.

(2, -5/3) and (0, -7/3)

17. If the points A(6, 1), B(8, 2), C(9, 4) and D(p, 3) are the vertices of a parallelogram, taken in order, find the value of p.

Diagonals bisect each other. Midpoint of AC = Midpoint of BD.
$(\frac{6+9}{2}, \frac{1+4}{2}) = (\frac{8+p}{2}, \frac{2+3}{2})$.
$\frac{15}{2} = \frac{8+p}{2} \Rightarrow 15 = 8+p \Rightarrow p = 7$.

p = 7

18. Find the ratio in which the line $2x + 3y - 5 = 0$ divides the line segment joining the points (8, -9) and (2, 1). Also find the coordinates of the point of division.

Let Ratio be $k:1$. Point $P(\frac{2k+8}{k+1}, \frac{k-9}{k+1})$.
Substitute in line eq: $2(\frac{2k+8}{k+1}) + 3(\frac{k-9}{k+1}) - 5 = 0$.
$4k + 16 + 3k - 27 - 5(k+1) = 0$.
$7k - 11 - 5k - 5 = 0 \Rightarrow 2k - 16 = 0 \Rightarrow k = 8$.
Ratio 8:1. Point P coordinates: $x = \frac{16+8}{9} = \frac{24}{9} = \frac{8}{3}$, $y = \frac{8-9}{9} = \frac{-1}{9}$.

Ratio 8:1; Point (8/3, -1/9)

19. If A and B are (-2, -2) and (2, -4) respectively, find the coordinates of P such that $AP = \frac{3}{7} AB$ and P lies on the line segment AB.

$AP = \frac{3}{7} AB \Rightarrow \frac{AP}{AB} = \frac{3}{7} \Rightarrow \frac{AP}{PB} = \frac{3}{4}$. Ratio is 3:4.
$x = \frac{3(2) + 4(-2)}{7} = \frac{6-8}{7} = \frac{-2}{7}$.
$y = \frac{3(-4) + 4(-2)}{7} = \frac{-12-8}{7} = \frac{-20}{7}$.

P(-2/7, -20/7)

20. If the point C(-1, 2) divides internally the line segment joining A(2, 5) and B(x, y) in the ratio 3:4, find the coordinates of B.

$-1 = \frac{3x + 4(2)}{7} \Rightarrow -7 = 3x + 8 \Rightarrow 3x = -15 \Rightarrow x = -5$.
$2 = \frac{3y + 4(5)}{7} \Rightarrow 14 = 3y + 20 \Rightarrow 3y = -6 \Rightarrow y = -2$.

B(-5, -2)

21. Find the centre of a circle passing through the points (6, -6), (3, -7) and (3, 3).

Let Center be O(x,y). $OA^2 = OB^2 = OC^2$.
From $OB^2 = OC^2$: $(x-3)^2 + (y+7)^2 = (x-3)^2 + (y-3)^2 \Rightarrow y^2+14y+49 = y^2-6y+9 \Rightarrow 20y = -40 \Rightarrow y = -2$.
From $OA^2 = OB^2$: $(x-6)^2 + (y+6)^2 = (x-3)^2 + (y+7)^2$.
Put y = -2: $(x-6)^2 + 16 = (x-3)^2 + 25$.
$x^2-12x+36+16 = x^2-6x+9+25$.
$-12x+52 = -6x+34 \Rightarrow 18 = 6x \Rightarrow x = 3$.

Centre (3, -2)

22. Show that the points (1, 7), (4, 2), (-1, -1) and (-4, 4) are the vertices of a square.

Let points be A, B, C, D.
$AB = \sqrt{3^2 + (-5)^2} = \sqrt{9+25} = \sqrt{34}$.
$BC = \sqrt{(-5)^2 + (-3)^2} = \sqrt{34}$.
$CD = \sqrt{(-3)^2 + 5^2} = \sqrt{34}$.
$DA = \sqrt{5^2 + 3^2} = \sqrt{34}$.
All sides equal. Check diagonals.
$AC = \sqrt{(-2)^2 + (-8)^2} = \sqrt{4+64} = \sqrt{68}$.
$BD = \sqrt{(-8)^2 + 2^2} = \sqrt{64+4} = \sqrt{68}$.
Diagonals are equal. Hence it is a square.

Proved

23. Case Study: Classroom Seating (i) Dist AB (ii) Dist AC (iii) Square check

(i) $AB = \sqrt{(6-3)^2 + (7-4)^2} = \sqrt{9+9} = 3\sqrt{2}$ units.
(ii) $AC = \sqrt{(9-3)^2 + (4-4)^2} = \sqrt{36} = 6$ units.
(iii) We found $AB = 3\sqrt{2}$. Similarly finding other sides (BC, CD, DA) yields $3\sqrt{2}$. Diagonals AC and BD are both 6. Since sides equal and diagonals equal, it is a Square.

(i) $3\sqrt{2}$ (ii) 6 (iii) Yes, Square