Chapter 7: Coordinate Geometry (NCERT Solutions)

Exercise 7.1: Distance Formula

Q1. Find the distance between the following pairs of points:

(i) (2, 3), (4, 1): √[(4-2)² + (1-3)²] = √[4 + 4] = √8 = 2√2.

(ii) (-5, 7), (-1, 3): √[(-1+5)² + (3-7)²] = √[16 + 16] = √32 = 4√2.

(iii) (a, b), (-a, -b): √[(-a-a)² + (-b-b)²] = √[4a² + 4b²] = 2√(a²+b²).

Q2. Find distance between (0, 0) and (36, 15). Can you find distance between towns A and B discussed in Section 7.2?

d = √[36² + 15²] = √[1296 + 225] = √1521 = 39.
Yes, distance is 39 km.

Q3. Determine if points (1, 5), (2, 3) and (-2, -11) are collinear.

AB = √5, BC = 2√53, AC = √265.
AB + BC ≠ AC. No.

Q4. Check whether (5, -2), (6, 4) and (7, -2) are vertices of isosceles triangle.

AB = √(1+36) = √37.
BC = √(1+36) = √37.
AC = √(4+0) = 2.
AB = BC. Yes.

Q5. Seating arrangement... A(3, 4), B(6, 7), C(9, 4), D(6, 1). Square?

AB = √(9+9) = 3√2.
BC = √(9+9) = 3√2.
CD = √(9+9) = 3√2.
DA = √(9+9) = 3√2.
AC = √(36) = 6. BD = √(36) = 6.
Sides equal, Diagonals equal. Yes, Square.

Q6. Name the type of quadrilateral.

(i) (-1, -2), (1, 0), (-1, 2), (-3, 0): Sides 2√2. Diagonals 4. Square.

(ii) (-3, 5), (3, 1), (0, 3), (-1, -4): Not a special quadrilateral.

(iii) (4, 5), (7, 6), (4, 3), (1, 2): Parallelogram (Opp sides equal, diagonals unequal).

Q7. Find point on x-axis equidistant from (2, -5) and (-2, 9).

(x-2)² + 25 = (x+2)² + 81.
-4x + 29 = 4x + 85.
-8x = 56 ⇒ x = -7.
Point: (-7, 0).

Q8. Find values of y for which distance between P(2, -3) and Q(10, y) is 10 units.

8² + (y+3)² = 100.
64 + (y+3)² = 100 ⇒ (y+3)² = 36.
y+3 = ±6 ⇒ y=3 or y=-9.
y = 3, -9.

Q9. Q(0, 1) is equidistant from P(5, -3) and R(x, 6). Find x.

QP² = QR².
25 + 16 = x² + 25.
x² = 16 ⇒ x = ±4.
QR = 5 (if x=4) or 5 (if x=-4). PR = √82 (if x=4) or 9√2 (if x=-4).

Q10. Relation between x and y such that (x, y) is equidistant from (3, 6) and (-3, 4).

(x-3)² + (y-6)² = (x+3)² + (y-4)².
-6x - 12y + 45 = 6x - 8y + 25.
-12x - 4y + 20 = 0 ⇒ 3x + y - 5 = 0.
3x + y = 5.

Exercise 7.2: Section Formula

Q1. Find coordinates... (-1, 7) and (4, -3) ratio 2:3.

x = (8-3)/5 = 1.
y = (-6+21)/5 = 3.
(1, 3).

Q2. Points of trisection of (4, -1) and (-2, -3).

First point (1:2): (2, -5/3).
Second point (2:1): (0, -7/3).

Q4. Ratio in which (-1, 6) divides (-3, 10) and (6, -8).

-1 = (6k-3)/(k+1) ⇒ -k-1 = 6k-3 ⇒ 7k=2 ⇒ k=2/7.
Ratio 2:7.

Q5. Ratio x-axis divides A(1, -5) and B(-4, 5).

y=0. (5k-5)/(k+1) = 0 ⇒ k=1.
Ratio 1:1. Point: (-3/2, 0).

Q6. Parallelogram vertices (1, 2), (4, y), (x, 6), (3, 5). Find x, y.

Midpoints coincide.
(1+x)/2 = 7/2 ⇒ x=6.
(2+6)/2 = (y+5)/2 ⇒ 8 = y+5 ⇒ y=3.
x=6, y=3.

Q7. Find coords of A where AB is diameter, center is (2, -3) and B is (1, 4).

(x+1)/2 = 2 ⇒ x=3.
(y+4)/2 = -3 ⇒ y=-10.
(3, -10).

Q8. If A(-2, -2) and B(2, -4), find P such that AP = 3/7 AB.

AP/AB = 3/7 ⇒ AP/PB = 3/4. Ratio 3:4.
x = (6-8)/7 = -2/7.
y = (-12-8)/7 = -20/7.
(-2/7, -20/7).

Coordinate Geometry - RD Sharma Important Questions

Q1. Find a relation 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.
x²-14x+49 + y²-2y+1 = x²-6x+9 + y²-10y+25.
-14x - 2y + 50 = -6x - 10y + 34.
-8x + 8y + 16 = 0 ⇒ x - y = 2.

Q2. Find the coordinates of the point of trisection of the line segment joining (4, -1) and (-2, -3).

Case 1 (1:2):
x = (1(-2)+2(4))/3 = 6/3 = 2.
y = (1(-3)+2(-1))/3 = -5/3.
Case 2 (2:1):
x = (2(-2)+1(4))/3 = 0.
y = (2(-3)+1(-1))/3 = -7/3.
Points: (2, -5/3) and (0, -7/3).

Q3. Determine the ratio in which the line 2x + y - 4 = 0 divides the line segment joining the points A(2, -2) and B(3, 7).

Let ratio be k:1.
x = (3k+2)/(k+1), y = (7k-2)/(k+1).
Substitute in 2x+y-4=0:
2(3k+2) + (7k-2) - 4(k+1) = 0.
6k + 4 + 7k - 2 - 4k - 4 = 0.
9k - 2 = 0 ⇒ k = 2/9.
Ratio 2:9.

Q4. If A(-5, 7), B(-4, -5), C(-1, -6) and D(4, 5) are vertices of a quadrilateral, find its area.

Area = Area(ΔABC) + Area(ΔADC).
Area(ΔABC) = 1/2 |-5(-5+6) + (-4)(-6-7) + (-1)(7+5)| = 1/2 |-5 + 52 - 12| = 35/2.
Area(ΔADC) = 1/2 |-5(-6-5) + (-1)(5-7) + 4(7+6)| = 1/2 |55 + 2 + 52| = 109/2.
Total Area = 144/2 = 72 sq units.

Q5. Find the centroid of the triangle with vertices (3, -7), (-8, 6) and (5, 10).

x = (3 - 8 + 5)/3 = 0/3 = 0.
y = (-7 + 6 + 10)/3 = 9/3 = 3.
Centroid: (0, 3).

Q6. Find the value of p for which the points (-1, 3), (2, p) and (5, -1) are collinear.

Slope method: (p-3)/(2+1) = (-1-p)/(5-2).
(p-3)/3 = (-1-p)/3.
p - 3 = -1 - p ⇒ 2p = 2 ⇒ p = 1.

Q7. If (1, 2), (4, y), (x, 6) and (3, 5) are vertices of a parallelogram taken in order, find x and y.

Midpoints of diagonals coincide.
(1+x)/2 = (4+3)/2 ⇒ 1+x = 7 ⇒ x = 6.
(2+6)/2 = (y+5)/2 ⇒ 8 = y+5 ⇒ y = 3.

Q8. Find the center of a circle passing through the points (6, -6), (3, -7) and (3, 3).

Let center be O(x,y). OA² = OB² = OC².
(x-6)² + (y+6)² = (x-3)² + (y+7)².
Solving gives linear eq. Similarly OB = OC.
Solution: (3, -2).

Q9. If the point P(x, y) is equidistant from the points A(a+b, b-a) and B(a-b, a+b), prove that bx = ay.

PA² = PB².
Expand and cancel x², y² terms.
-2x(a+b) - 2y(b-a) = -2x(a-b) - 2y(a+b).
-ax - bx - by + ay = -ax + bx - ay - by.
-bx + ay = bx - ay ⇒ 2ay = 2bx ⇒ bx = ay.

Q10. The line segment joining the points (3, -4) and (1, 2) is trisected at points P and Q. If P is (p, -2) and Q is (5/3, q), find p and q.

P divides 1:2. p = (1(1)+2(3))/3 = 7/3. (Given p=-2 coordinate... oh wait, P is (p, -2)).
y-coord of P = -2. (1(2)+2(-4))/3 = -6/3 = -2. Correct.
So p = 7/3.
Q divides 2:1. q = (2(2)+1(-4))/3 = 0.
So p=7/3, q=0.

Q11. Find the fourth vertex D of a parallelogram ABCD whose three vertices are A(-2, 3), B(6, 7) and C(8, 3).

Midpoint of AC = Midpoint of BD.
((-2+8)/2, (3+3)/2) = (3, 3).
((6+x)/2, (7+y)/2) = (3, 3).
6+x=6 ⇒ x=0. 7+y=6 ⇒ y=-1.
D(0, -1).

Q12. If the points A(1, -2), B(2, 3), C(a, 2) and D(-4, -3) form a parallelogram, find the value of a.

Slope AB = Slope CD? No, midpoints is better.
Wait, if ABCD is order.
Mid AB != Mid CD.
Actually Mid AC = Mid BD.
(1+a)/2 = (2-4)/2 = -1 ⇒ 1+a = -2 ⇒ a = -3.

Q13. Find the area of the quadrilateral ABCD whose vertices are (-4, -2), (-3, -5), (3, -2) and (2, 3).

Area using Shoelace formula or splitting into triangles.
= 1/2 |(-4(-5) + -3(-2) + 3(3) + 2(-2)) - (-2(-3) + -5(3) + -2(2) + 3(-4))|.
= 1/2 |(20 + 6 + 9 - 4) - (6 - 15 - 4 - 12)|.
= 1/2 |31 - (-25)| = 56/2 = 28 sq units.

Q14. The coordinates of the mid-point of the line joining the points (3p, 4) and (-2, 2q) are (5, p). Find p and q.

(3p-2)/2 = 5 ⇒ 3p-2 = 10 ⇒ 3p = 12 ⇒ p = 4.
(4+2q)/2 = p = 4 ⇒ 4+2q = 8 ⇒ 2q = 4 ⇒ q = 2.

Q15. Find the coordinates of the circumcenter of the triangle whose vertices are (8, 6), (8, -2) and (2, -2).

It's a right triangle (check slopes or dist).
AB is vertical (x=8). BC is horizontal (y=-2).
Right angle at B(8, -2).
Circumcenter is midpoint of hypotenuse AC.
Midpoint of (8, 6) and (2, -2) = ((8+2)/2, (6-2)/2) = (5, 2).

Q16. Find the ratio in which the y-axis divides the line segment joining the points (5, -6) and (-1, -4). Also find the coordinates of the point of intersection.

Ratio k:1. x-coordinate is 0.
(-1k + 5)/(k+1) = 0 ⇒ k = 5. Ratio 5:1.
y = (5(-4) + 1(-6))/6 = -26/6 = -13/3.
Point: (0, -13/3).

Q17. The line joining the points (2, 1) and (5, -8) is trisected at the points P and Q. If point P lies on the line 2x - y + k = 0, find k.

P is 1:2 internal division.
P = ((5+4)/3, (-8+2)/3) = (3, -2).
2(3) - (-2) + k = 0.
6 + 2 + k = 0 ⇒ k = -8.

Q18. Prove that the points (3, 0), (6, 4) and (-1, 3) are vertices of a right angled isosceles triangle.

d1 = √(9+16) = 5.
d2 = √(49+1) = √50 = 5√2.
d3 = √(16+9) = 5.
Sides: 5, 5, 5√2.
5^2 + 5^2 = (5√2)^2. Satisfies Pythagoras. Isosceles Right Triangle.

Q19. Find the area of the triangle formed by the mid-points of sides of the triangle whose vertices are (2, 1), (-2, 3) and (4, -3).

Area of large triangle = 1/2 |2(6) + (-2)(-4) + 4(-2)| = 1/2 |12 + 8 - 8| = 6.
Area of mid-point triangle = 1/4 × 6 = 1.5 sq units.

Q20. If A(5, 2), B(2, -2) and C(-2, t) are the vertices of a right angled triangle with \u2220B = 90°, then find the value of t.

Slope AB × Slope BC = -1.
(-2-2)/(2-5) × (t+2)/(-2-2) = -1.
(-4)/(-3) × (t+2)/(-4) = -1.
4/3 × (t+2)/-4 = -1.
-(t+2)/3 = -1 ⇒ t+2 = 3 ⇒ t = 1.

Coordinate Geometry - Formulas & PYQs

Key Formulas

1. Distance Formula

√[(x₂ - x₁)² + (y₂ - y₁)²]

2. Section Formula

x = (m₁x₂ + m₂x₁) / (m₁ + m₂)

y = (m₁y₂ + m₂y₁) / (m₁ + m₂)

3. Midpoint Formula

((x₁ + x₂)/2, (y₁ + y₂)/2)

4. Area of Triangle

1/2 |x₁(y₂-y₃) + x₂(y₃-y₁) + x₃(y₁-y₂)|

Previous Year Questions (CBSE/JKBOSE)

Q1. Find the point on y-axis which is equidistant from the points (5, -2) and (-3, 2). (CBSE 2011)

Let point be (0, y).
25 + (y+2)² = 9 + (y-2)².
25 + y²+4y+4 = 9 + y²-4y+4.
8y = -16 ⇒ y = -2.
Point: (0, -2).

Q2. Find relation between x and y if points (x, y), (1, 2) and (7, 0) are collinear. (CBSE 2014)

Area = 0 ⇒ x(2-0) + 1(0-y) + 7(y-2) = 0.
2x - y + 7y - 14 = 0.
2x + 6y - 14 = 0 ⇒ x + 3y = 7.

Q3. If A(-2, 1), B(a, 0), C(4, b) and D(1, 2) are vertices of a parallelogram ABCD, find a and b. (CBSE 2018)

Midpoint of AC = Midpoint of BD.
(-2+4)/2 = (a+1)/2 ⇒ 2 = a+1 ⇒ a = 1.
(1+b)/2 = (0+2)/2 ⇒ 1+b = 2 ⇒ b = 1.

Q4. Find the ratio in which the line segment joining (1, -5) and (-4, 5) is divided by the x-axis. (CBSE 2012)

y = 0.
(5k - 5)/(k+1) = 0 ⇒ 5k = 5 ⇒ k = 1.
Ratio is 1:1.

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