CBSE Important Questions Class 10 Maths Ch7 Coordinate Geometry 2025-26

The chapter Coordinate Geometry is one of the most interesting and high-scoring topics in Class 10 Mathematics Syllabus. It connects algebra and geometry by helping you locate points on a plane and calculate distances, midpoints, and areas using coordinate values. 

It builds the foundation for advanced geometry and analytical concepts in higher classes. Since most questions are formula-based, accuracy and consistent revision are key to scoring full marks in this topic.

The Coordinate Geometry Important Questions for Class 10 Mathematics help you:

• Understand and apply basic coordinate formulas.
• Strengthen your visualization and calculation skills.
• It’s also helpful to refer to the NCERT Book Class 10 Maths to revise the core concepts clearly.
• Prepare efficiently for board exams through concept-based and numerical question

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IMPORTANT QUESTIONS CLASS 10

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Maths Ch7 Coordinate Geometry Important Questions 2026

1. △ABC is a triangle such that AB:BC = 1:2. Point A lies on the y-axis and the coordinates of B and C are known.

Which of the following formula can DEFINITELY be used to find the coordinates of A?

i) Section formula

ii) Distance formula

a. only i)

b. only ii)

c. both i) and ii)

d. neither i) or ii)

Soln. (c) both i) and ii)

Explanation: 

Section Formula: The section formula can be used to find the coordinates of a point that divides a line segment internally in a given ratio.

• Here, the ratio AB:BC=1:2 can help us determine A's position relative to B and C, especially if we treat A, B, and C as collinear points.
• Given that A lies on the y-axis, we can use the section formula to confirm its y-coordinate while fixing the x-coordinate as 0.

Thus, the section formula can be used to find A's coordinates.

Distance Formula: The distance formula can be used to calculate the distance between two points.

• Using the known coordinates of B and C, as well as the condition AB:BC=1:2, we can compute the distances AB and BC.
• By ensuring the point A lies on the y-axis, we can solve for the y-coordinate of A using the distance condition.

Thus, the distance formula can also be used to find A's coordinates.

2. Preeti and Arun are both driving to their respective offices from the same home. Preeti drives towards the east at an average speed of 30 km per hour for 12 minutes and then towards the south at an average speed of 60 km per hour for 3 minutes. Arun drives towards the west at an average speed of 30 km per hour for 4 minutes and then towards the north at an average speed of 45 km per hour for 4 minutes.

What is the straight-line distance between Preeti's office and Arun's office? Show your steps and represent the given scenario on the coordinate plane.

Soln.  Preeti's final position

Distance traveled east for 12 minutes: = Speed × Time 

= 30×1/5

=6 km.

(6,0)

Distance traveled  south for 3 minutes: = Speed × Time 

= 60×1/20

=3 km.

(3,0)

Since she is driving south, this means she moves 3 km along the negative y-axis. So, her final position is: (6,−3)

Arun's final position

Distance traveled west for 4 minutes: = Speed × Time 

= 30×1/15

=2 km.

(-2,0)

Distance traveled north for 4 minutes: = Speed × Time 

= 45×1/20

=2 km.

(2,0)

Since he is driving north, this means he moves 3 km along the positive y-axis. So, his final position is: (−2,3)

d= √(x₂​−x₁​)²+(y_2​−y₁​)²

d= √(-2​-6)​)²+(3- (-3))²

d= √(-8​)²+(6)²

d= √64+36

d= √100

d= 10km

The straight-line distance between Preeti's office and Arun's office is 10 km.

3. A circle with centre O(2,-5) has a chord with end-points A(1, 2) and B. M(5, -2) is the point where the perpendicular to the chord from the centre touches AB.

Find the coordinates of point B. Show your steps with valid reasons.

Soln. The center of the circle is O(2,−5)

A(1,2) and B(x,y) are the endpoints of the chord ABABAB.

M(5,−2) is the midpoint of AB where the perpendicular from the center O touches the chord.

M= x1​+x2 / 2, y1​+y2 / 2

(5,-2)= 1+x / 2, 2+y / 2

1+x / 2 = 5

x=9

2+y / 2 = -2

y=−6

B(9,−6)

4. Raaji and Gagan are finding a treasure that is exactly on the straight line joining them. Raaji's location is at (-6, -5) and Gagan's location is at (10, 11). The distance from the treasure to Raaji's location is three times that of the distance to Gagan's location.

Find the coordinates of the location of the treasure. Show your steps.

Soln. RT=3⋅TG.

T(x,y)= ((mx₂​+nx₁)​/m+n, (my₂​+ny₁)m+n)

x= 3(10)+1(−6)​/3+1

x= 30−6/4

x= 24/4

x= 6

y= 3(11)+1(−5)​/3+1

y= 33−5/4

y= 28/4

y= 7

The coordinates of the treasure are:

T(6,7)

5. On a golf course, three holes A(-6, -1), B and C(9, -4) lie on a straight line in that order.

The distance between B and C is two times that between B and A.Rahul strikes the ball, which is at point P(2, 3), such that it goes in the hole B.

i) Find the coordinates of hole B.

ii) Find the shortest distance covered by the ball.Show your steps.

Soln. i) Find the coordinates of hole B.

BC=2×BA

B divides AC in the ratio 1:2 (from A to C)

B(x,y)= ((mx₂​+nx₁)​/m+n, (my₂​+ny₁)m+n)

B(x,y)= (1(9)+2(−6)​/1+2, 1(−4)+2(−1)/1+2)

B(−1,−2)

ii) Find the shortest distance covered by the ball.

d=√(x₂​−x₁​)²+(y₂​−y₁​)²

d=√(-1​−2​)²+((-2)−3)²

d=√(-3​)²+(-5)²

d=√9+25

d=√34

6. The three vertices of a rhombus PQRS are P(2, -3), Q(6, 5) and R(-2, 1).

a) Find the coordinates of the point where both the diagonals PR and QS intersect.

b) Find the coordinates of the fourth vertex S.Show your steps and give valid reasons.

Soln. The vertices of the rhombus are given as:

• P(2,−3),
• Q(6,5)
• R(−2,1)

Key Properties of a Rhombus:

1. The diagonals of a rhombus bisect each other at right angles.
2. This means the point of intersection of the diagonals is the midpoint of both diagonals.

a) Find the coordinates of the point where both the diagonals PR and QS intersect. 

M= (x₁​+x₂​)/2 , (y₁​+y₂​)/2 , 

M= (2​+(-2)​)/2 , (-3+1)/2 , 

M= 0/2 , (-2)/2 , 

M= 0 , -1 

Thus, the diagonals intersect at the point (0,−1)

b) Find the coordinates of the fourth vertex S.

M(0,−1) is the midpoint,

Q(6,5) and S(x,y) are the endpoints.

(6+x)/2, (5+y)/2=(0,−1)

For x:

(6+x)/2 = 0

6+x=0

x=−6

For y

(5+y)/2= -1

5+y=−2

y=−7

Thus, the coordinates of S are (−6,−7)

7. The line x + 2y =2 forms a triangle OPQ, with the coordinate axes.

(i) What are the coordinates of points P and Q?

(ii) What is the area of the triangle formed? Show your steps.

Soln.  

(i) What are the coordinates of points P and Q?

The given line is: 

x+2y=2

For the x-axis 

Substitute y=0 into x+2y=2

x+2(0)=2⟹x=2

So, Q has coordinates (2,0)

For the y-axis 

Substitute x=0 nto x+2y=2

0+2y=2⟹y=1

So, P has coordinates (0,1)

Thus, the coordinates of P and Q are:

P(0,1)and Q(2,0)

(ii) What is the area of the triangle formed? Show your steps.

O(0,0)

P(0,1),

Q(2,0)

area of a △ = ½ ∣x_1​(y₂​−y₃​)+x₂​(y₃​−y₁​)+x₃​(y₁​−y₂​)∣

area of a △ = ½ ∣0_​(1-0​)+0​(​0−0​)+02(0​−1​)∣

area of a △ = ½ ∣0_​+0​+2​(​−1​)∣

area of a △ = ½ ∣0_​+0​-2​∣

area of a △ = ½ ∣​-2​∣

area of a △ = ½ 2​

area of a △ = 1

8. A(5, 1), B(1, 4) and C(8, 5) are the coordinates of the vertices of a triangle.

Which of the following types of triangle will △ABC be?

a. Equilateral triangle

b. Isosceles right-angled triangle

c. Scalene right-angled triangle

d. Isosceles acute-angled triangle

Soln.  (b) Isosceles right-angled triangle

Explanation:

AB=√(x_2​−x₁​)²+(y₂​−y₁​)²

AB= √(1_​−5​)²+(4-1​)²

AB= √(4)²+(3)²

AB= √16+9

AB= √25

AB= 5

BC =√(x_2​−x₁​)²+(y₂​−y₁​)²

BC = √(8_​−1​)²+(5-4​)²

BC = √(7)²+(1)²

BC = √49+1

BC = √50

BC = 5√2

CA =√(x_2​−x₁​)²+(y₂​−y₁​)²

CA = √(8_​−5)²+(5-1)²

CA = √(3)²+(4)²

CA = √9+16

CA = √25

CA = 5

Two sides are equal: AB=CA=5

This makes the triangle isosceles.

Checking for a right angle:

BC²=AB²+CA²

(5√2​)²=5²+5²

(50​)=25 + 25

(50​)=50

The condition holds true, so the triangle is right-angled.

The triangle △ABC is an isosceles right-angled triangle.

9. The radius of a circle with centre at the origin is ½ units.

Find all the points on the circle which are of the form (-y, y ). Show your steps.

Soln.  

The equation of a circle with center at the origin (0,0) and radius r is:

x²+y²=r²

x²+y²=(½)²

x²+y²=(1/4)

x=−y

y=y

(−y)²+y²=¼

(y)²+y²=¼

2y²=¼

y²=¼ / 2

y²= 8

y= ± √2/4

(−y,y)= (−√2/4, √2/4)

(−y,y)= -(−(√2/4), -√2/4)

(−y,y)= (√2/4, -√2/4)

points on the circle of radius 1/2 that are of the form (−y,y) are:(−√2/4, √2/4) or (√2/4, -√2/4)

Some More Important Question Answers of Class 10 Maths

Question 1. Find the ratio in which the y-axis divides the line segment joining A(5, −6) and B(−1, −4). Also find the point of division.

Solution.

1. Let P be the point where the y-axis meets AB. On the y-axis x = 0, so P = (0, y).
2. Suppose P divides AB internally in the ratio k : 1 (i.e. AP : PB = k : 1). Use the section formula Px = (k·xB + 1·xA) / (k + 1). Here xA = 5, xB = −1.
3. Set Px = 0 and solve:
   0 = (k·(−1) + 1·5) / (k + 1) → −k + 5 = 0 → k = 5.
   So the ratio is 5 : 1 (AP : PB = 5 : 1).
4. Now find Py using the section formula: Py = (k·yB + 1·yA) / (k + 1) = (5·(−4) + 1·(−6)) / 6 = (−20 − 6)/6 = −26/6 = −13/3.
5. Answer: ratio = 5 : 1, point = (0, −13/3).

Question 2. The x-coordinate of point P is twice its y-coordinate. If P is equidistant from Q(2, −5) and R(−3, 6), find P.

Solution.

1. Let P = (x, y) and give x = 2y. So P = (2y, y).
2. Condition: distances PQ and PR are equal. Use squared distances to avoid square roots:
   PQ² = (2y − 2)² + (y + 5)².
   PR² = (2y + 3)² + (y − 6)².
3. Set PQ² = PR²:
   (2y − 2)² + (y + 5)² = (2y + 3)² + (y − 6)².
4. Expand both sides:
   Left: (4y² − 8y + 4) + (y² + 10y + 25) = 5y² + 2y + 29.
   Right: (4y² + 12y + 9) + (y² − 12y + 36) = 5y² + 0y + 45.
5. Equate and simplify:
   5y² + 2y + 29 = 5y² + 45 → 2y + 29 = 45 → 2y = 16 → y = 8.
6. Then x = 2y = 16. So P = (16, 8).

Question 3. Find points of trisection P and Q of segment AB where A(2, −2) and B(−7, 4).

Solution.

1. Trisection means dividing AB into three equal parts: AP = PQ = QB.
2. For P (nearest A) ratio AP : PB = 1 : 2 → use section formula with m1 = 1, m2 = 2:
   Px = (1·xB + 2·xA) / 3 = (1·(−7) + 2·2)/3 = (−7 + 4)/3 = −3/3 = −1.
   Py = (1·yB + 2·yA) / 3 = (1·4 + 2·(−2))/3 = (4 − 4)/3 = 0.
   So P = (−1, 0).
3. For Q (nearer B) ratio AP : PB = 2 : 1 (so m1 = 2, m2 = 1):
   Qx = (2·(−7) + 1·2)/3 = (−14 + 2)/3 = −12/3 = −4.
   Qy = (2·4 + 1·(−2))/3 = (8 − 2)/3 = 6/3 = 2.
   So Q = (−4, 2).

Question 4. Show that the points (3,0), (6,4) and (−1,3) form a right-angled isosceles triangle. Identify the right angle.

Solution.

1. Label A(3,0), B(6,4), C(−1,3)
   AB = √[(6−3)² + (4−0)²] = √[3² + 4²] = √[9 + 16] = √25 = 5.
   BC = √[(−1−6)² + (3−4)²] = √[(−7)² + (−1)²] = √[49 + 1] = √50.
   CA = √[(−1−3)² + (3−0)²] = √[(−4)² + 3²] = √[16 + 9] = √25 = 5.
2. Two sides AB and CA are equal (AB = CA = 5) ⇒ triangle is isosceles with vertex A between those equal sides.
3. Check right angle using Pythagoras at A: AB² + AC² = 5² + 5² = 25 + 25 = 50 = BC². So AB² + AC² = BC² → angle at A is 90°.
4. Conclusion: triangle is right-angled at A and isosceles (legs AB and AC).

Question 5. Point (−3, k) divides the segment joining (−5, −4) and (−2, 3) internally. Find the ratio and k.

Solution.

1. Let A = (−5, −4), B = (−2, 3), and P = (−3, k). Suppose AP : PB = m : 1 (m unknown).
2. Use section formula for x-coordinate:
   Px = (m·xB + 1·xA)/(m+1) = (m(−2) + (−5))/(m+1) = (−2m − 5)/(m+1).
   Set equal to Px = −3:
   (−2m − 5)/(m + 1) = −3 → −2m − 5 = −3(m + 1) = −3m − 3.
   Rearranging: −2m − 5 = −3m − 3 → add 3m both sides → m − 5 = −3 → m = 2.
3. So ratio AP : PB = 2 : 1 (or 2:1). Now find y:
   Py = (m·yB + 1·yA)/(m+1) = (2·3 + (−4))/3 = (6 − 4)/3 = 2/3.
4. Hence k = 2/3 and P = (−3, 2/3).

Question 6. Show that (2, −2), (−2, 1), (5, 2) form a right triangle. Also find its area.

Solution.

1. Label A(2,−2), B(−2,1), C(5,2). Compute side lengths:
   AB = √[(−2 − 2)² + (1 + 2)²] = √[(−4)² + 3²] = √[16 + 9] = √25 = 5.
   BC = √[(5 + 2)² + (2 − 1)²] = √[7² + 1²] = √[49 + 1] = √50.
   CA = √[(5 − 2)² + (2 + 2)²] = √[3² + 4²] = √[9 + 16] = √25 = 5.
2. Two sides AB and CA equal ⇒ isosceles. Check right angle:
   AB² + AC² = 25 + 25 = 50 = BC². So the angle at A is 90°.
3. Area = (1/2) × (leg1) × (leg2) = (1/2) × AB × AC = (1/2) × 5 × 5 = 12.5 square units.

Question 7. In ΔABC, A(0, −1). D and E are midpoints of AB and AC with D(1,0), E(0,1). If F is the midpoint of BC, find areas of ΔABC and ΔDEF.

Solution.

1. Use midpoint data to find B and C. Let B = (xB, yB). Midpoint D = ((0 + xB)/2, (−1 + yB)/2) = (1, 0).
   So (0 + xB)/2 = 1 → xB = 2. And (−1 + yB)/2 = 0 → −1 + yB = 0 → yB = 1. So B = (2, 1).
2. Let C = (xC, yC). Midpoint E = ((0 + xC)/2, (−1 + yC)/2) = (0, 1).
   So (0 + xC)/2 = 0 → xC = 0. And (−1 + yC)/2 = 1 → −1 + yC = 2 → yC = 3. So C = (0, 3).
3. Now we have A(0, −1), B(2, 1), C(0, 3). Compute area ABC by shoelace formula:
   Sum1 = xA·yB + xB·yC + xC·yA = 0·1 + 2·3 + 0·(−1) = 6.
   Sum2 = yA·xB + yB·xC + yC·xA = (−1)·2 + 1·0 + 3·0 = −2.
   Area = (1/2)·|Sum1 − Sum2| = (1/2)·|6 − (−2)| = (1/2)·8 = 4.
   So area(ΔABC) = 4 square units.
4. Midpoint F of BC: F = ((2 + 0)/2, (1 + 3)/2) = (1, 2).
   Now the triangle DEF has D(1,0), E(0,1), F(1,2). Use shoelace:
   Sum1 = 1·1 + 0·2 + 1·0 = 1.
   Sum2 = 0·0 + 1·1 + 2·1 = 0 + 1 + 2 = 3.
   Area = (1/2)·|1 − 3| = (1/2)·2 = 1.
5. Answer: area(ΔABC) = 4, area(ΔDEF) = 1. (ΔDEF is one-fourth of ΔABC.)

Question 8. If P(x,y) is equidistant from A(a + b, b − a) and B(a − b, a + b), prove that bx = ay.

Solution.

1. Equidistant means PA = PB → PA² = PB². Write coordinates:
   PA² = (x − (a + b))² + (y − (b − a))².
   PB² = (x − (a − b))² + (y − (a + b))².
2. Expand both sides and cancel common x² + y² terms:
   Left expansion:
   −2x(a + b) + (a + b)² − 2y(b − a) + (b − a)².
   Right expansion:
   −2x(a− b) + (a − b)² − 2y(a + b) + (a + b)².
3. Move all to left side and cancel identical squared terms (note (a+b)² cancels):
   [−2x(a + b) + 2x(a − b)] + [−2y(b − a) + 2y(a + b)] + [(b − a)² − (a − b)²] = 0.
4. Simplify each bracket:
   −2x(a + b) + 2x(a − b) = −2x( (a+b) − (a−b) ) = −2x(2b) = −4bx.
   −2y(b − a) + 2y(a + b) = −2y(b − a − a − b)?? Better recompute directly:
   −2y(b − a) + 2y(a + b) = (−2yb + 2ya) + (2ya + 2yb) = 4ay.
   And (b − a)² − (a − b)² = 0 (they are equal).
5. So equation becomes −4bx + 4ay = 0 → divide by 4 → −bx + ay = 0 → ay = bx → bx = ay. QED.

Question 9. Point C(−1, 2) divides A(2, 5) and B(x, y) in ratio 3 : 4 (internally). Find x² + y².

Solution.

1. Given AC : CB = 3 : 4 → use section formula with m1 = 3, m2 = 4, A = (2,5), B = (x,y), and C = (−1, 2).
   Coordinates of C = ( (3·x + 4·2)/7 , (3·y + 4·5)/7 ).
2. Set equal to C's given coordinates:
   (3x + 8)/7 = −1 → 3x + 8 = −7 → 3x = −15 → x = −5.
   (3y + 20)/7 = 2 → 3y + 20 = 14 → 3y = −6 → y = −2.
3. So (x, y) = (−5, −2). Then x² + y² = (−5)² + (−2)² = 25 + 4 = 29.

Question 10. Prove that the area of a triangle with vertices (t, t − 2), (t + 2, t + 2), (t + 3, t) is independent of t.

Solution.

1. Let P1 = (t, t − 2), P2 = (t + 2, t + 2), P3 = (t + 3, t). Use area determinant formula:
   Area = (1/2) · | x1(y2 − y3) + x2(y3 − y1) + x3(y1 − y2) |.
2. Compute differences:
   y2 − y3 = (t + 2) − t = 2 → x1(y2 − y3) = t·2 = 2t.
   y3 − y1 = t − (t − 2) = 2 → x2(y3 − y1) = (t + 2)·2 = 2t + 4.
   y1 − y2 = (t − 2) − (t + 2) = −4 → x3(y1 − y2) = (t + 3)·(−4) = −4t − 12.
3. Sum = 2t + (2t + 4) + (−4t − 12) = (2t + 2t − 4t) + (4 − 12) = 0 − 8 = −8.
4. Area = (1/2)·|−8| = 4 (constant, independent of t).
5. Hence the area is 4 square units for any t.

Question 11. A(4,6), B(1,5), C(7,2). Points D on AB and E on AC are such that AD/AB = AE/AC = 1/3 (i.e. D and E are at 1/3 from A). Find area(ΔADE) and compare with area(ΔABC).

Solution.

1. Coordinates: A(4,6), B(1,5), C(7,2). Since AD/AB = 1/3, AD : DB = 1 : 2. So use section formula with m1 = 1, m2 = 2 to find D:
   D = ( (1·xB + 2·xA)/3, (1·yB + 2·yA)/3 ) = ( (1·1 + 2·4)/3, (1·5 + 2·6)/3 ) = (9/3, 17/3 ) = (3, 17/3).
2. Similarly E on AC with AE : EC = 1 : 2:
   E = ( (1·xC + 2·xA)/3, (1·yC + 2·yA)/3 ) = ( (7 + 8)/3, (2 + 12)/3 ) = (15/3, 14/3 ) = (5, 14/3).
3. Area ΔABC by shoelace:
   Sum1 = 4·5 + 1·2 + 7·6 = 20 + 2 + 42 = 64.
   Sum2 = 6·1 + 5·7 + 2·4 = 6 + 35 + 8 = 49.
   Area ABC = (1/2)·|64 − 49| = (1/2)·15 = 15/2 = 7.5.
4. Area ΔADE using A(4,6), D(3,17/3), E(5,14/3):
   Sum1 = 4·(17/3) + 3·(14/3) + 5·6 = 68/3 + 42/3 + 30 = (110/3) + 30 = (110 + 90)/3 = 200/3.
   Sum2 = 6·3 + (17/3)·5 + (14/3)·4 = 18 + 85/3 + 56/3 = 18 + 141/3 = 18 + 47 = 65.
   Area ADE = (1/2)·|200/3 − 65| = (1/2)·|(200 − 195)/3| = (1/2)·(5/3) = 5/6 ≈ 0.8333.
5. Compare: area ADE = 5/6 and area ABC = 15/2. Ratio = (5/6) ÷ (15/2) = (5/6)*(2/15) = 1/9. So area ADE = (1/9) area ABC.

Question 12. Find area of quadrilateral ABCD with A(1,2), B(6,2), C(5,3), D(3,4).

Solution.

1. Use shoelace formula for vertices in order A → B → C → D → A.
   Sum1 = xA·yB + xB·yC + xC·yD + xD·yA
   = 1·2 + 6·3 + 5·4 + 3·2 = 2 + 18 + 20 + 6 = 46.
   Sum2 = yA·xB + yB·xC + yC·xD + yD·xA
   = 2·6 + 2·5 + 3·3 + 4·1 = 12 + 10 + 9 + 4 = 35.
   Area = (1/2)·|46 − 35| = (1/2)·11 = 5.5 square units.
2. Answer: Area = 5.5 sq. units.

Question 13. Find the area of quadrilateral ABCD with A(−3,2), B(5,4), C(7,−6), D(−5,−4).

Solution.

1. Shoelace with order A → B → C → D → A:
   Sum1 = (−3)·4 + 5·(−6) + 7·(−4) + (−5)·2 = (−12) + (−30) + (−28) + (−10) = −80.
   Sum2 = 2·5 + 4·7 + (−6)·(−5) + (−4)·(−3) = 10 + 28 + 30 + 12 = 80.
   Area = (1/2)·|Sum1 − Sum2| = (1/2)·|−80 − 80| = (1/2)·160 = 80.
2. Answer: Area = 80 square units.

Question 14. If A(5,2), B(2,−2), C(−2, t) and ∠B = 90°, find t.

Solution.

1. Right angle at B ⇒ BA · BC = 0 (dot product zero). Compute vectors with B as origin:
   BA = A − B = (5 − 2, 2 − (−2)) = (3, 4).
   BC = C − B = (−2 − 2, t − (−2)) = (−4, t + 2).
2. Dot product: BA · BC = 3·(−4) + 4·(t + 2) = −12 + 4t + 8 = 4t − 4.
3. Set = 0: 4t − 4 = 0 → t = 1.
4. Answer: t = 1.

Question 15. Find the ratio in which P(3/4, 5/12) divides A(1/2, 3/2) and B(2, −5).

Solution.

1. Let AP : PB = k : 1. Use section formula:
   Px = (k·xB + 1·xA) / (k + 1) = (k·2 + 1·(1/2)) / (k + 1).
   Set equal to 3/4:
   (2k + 1/2)/(k + 1) = 3/4.
2. Multiply both sides by 4(k + 1):
   4(2k + 1/2) = 3(k + 1) → 8k + 2 = 3k + 3 → 5k = 1 → k = 1/5.
3. Check y coordinate consistency:
   Py = (k·(−5) + 1·(3/2))/(k + 1) = ( (−5/5) + 3/2 )/(1 + 1/5) = (−1 + 1.5)/(6/5) = 0.5/(6/5) = 0.5 * (5/6) = 5/12. Works.
4. Answer: ratio AP:PB = 1 : 5.

Question 16.
Find p such that A(4,7), B(p,3), C(7,3) are vertices of a right triangle with right angle at B.

Solution.

1. Right angle at B ⇒ BA ⟂ BC. Compute vectors:
   BA = A − B = (4 − p, 7 − 3) = (4 − p, 4).
   BC = C − B = (7 − p, 3 − 3) = (7 − p, 0).
2. Dot product BA·BC = (4 − p)(7 − p) + 4·0 = 0 ⇒ (4 − p)(7 − p) = 0.
3. Solve: either p = 4 or p = 7. If p = 7 then B = C (degenerate). So valid p = 4.
4. Answer: p = 4 (then BA = (0,4) vertical, BC = (3,0) horizontal → perpendicular).

Question 17. Find relation between x and y if points A(x,y), B(−5,7), C(−4,5) are collinear.

Solution.

1. Compute slope BC = (5 − 7)/(−4 + 5) = (−2)/1 = −2.
2. For A to be collinear, slope AB must equal −2:
   (y − 7)/(x + 5) = −2 → y − 7 = −2(x + 5) = −2x − 10.
3. So y = −2x − 3. Rearranged: 2x + y + 3 = 0.

Question 18. Show that points (a,a), (−a,−a), (−√3·a, √3·a) are vertices of an equilateral triangle.

Solution.

1. Let A = (a, a), B = (−a, −a), C = (−√3·a, √3·a). Compute squared distances:
   AB² = (−a − a)² + (−a − a)² = (−2a)² + (−2a)² = 4a² + 4a² = 8a².
   AC² = (−√3 a − a)² + (√3 a − a)²
   = a²( (−√3 − 1)² + (√3 − 1)² )
   = a²( (3 + 2√3 + 1) + (3 − 2√3 + 1) ) = a²(4 + 4) = 8a².
   BC² = (−√3 a + a)² + (√3 a + a)²
   = a²( (−√3 + 1)² + (√3 + 1)² )
   = a²( (3 − 2√3 + 1) + (3 + 2√3 + 1) ) = a²(4 + 4) = 8a².
2. All three squared distances are equal → AB = AC = BC → triangle is equilateral.

Question 19. For which values of k are the points (8,1), (3, −2k), (k, −5) collinear?

Solution.

1. Let points be P1(8,1), P2(3, −2k), P3(k, −5). Collinearity means slope P1P2 = slope P2P3.
2. Slope P1P2 = (−2k − 1)/(3 − 8) = (−2k − 1)/(−5) = (2k + 1)/5.
   Slope P2P3 = (−5 + 2k)/(k − 3) = (2k − 5)/(k − 3).
3. Equate: (2k + 1)/5 = (2k − 5)/(k − 3). Cross-multiply:
   (2k + 1)(k − 3) = 5(2k − 5).
   Expand left: 2k² − 6k + k − 3 = 2k² − 5k − 3. Right: 10k − 25.
   Move all to the left: 2k² − 5k − 3 − 10k + 25 = 0 → 2k² − 15k + 22 = 0.
4. Solve quadratic: Discriminant Δ = (−15)² − 4·2·22 = 225 − 176 = 49. So √Δ = 7.
   k = [15 ± 7]/(2·2) = (15 ± 7)/4.
   So k = (15 + 7)/4 = 22/4 = 11/2 or k = (15 − 7)/4 = 8/4 = 2.
5. Answer: k = 2 or k = 11/2.

Question 20. Show that diagonals of rectangle ABCD with A(2, −1), B(5, −1), C(5, 6), D(2, 6) bisect each other.

Solution.

1. For any parallelogram/rectangle diagonals bisect each other. Show midpoints equal:
   Midpoint of AC = ((2 + 5)/2, (−1 + 6)/2) = (7/2, 5/2) = (3.5, 2.5).
   Midpoint of BD = ((5 + 2)/2, (−1 + 6)/2) = (7/2, 5/2) = (3.5, 2.5).
2. Since midpoints coincide, diagonals AC and BD bisect each other.

Extra Coordinate Geometry Practice Questions to ace your boards!

Q1. Distance Between Two Points: Find the value of xxx for which the distance between the points P(2,−3)P(2, -3)P(2,−3) and Q(x,5)Q(x, 5)Q(x,5) is 10 units.

Q2. Equidistant Points: If the points (3,2)(3, 2)(3,2) and (2,−3)(2, -3)(2,−3) are equidistant from a point (x,y)(x, y)(x,y), show that x+5y=0x + 5y = 0x+5y=0.

Q3. Collinearity Check: For what value of ppp, are the points (2,1)(2, 1)(2,1), (p,−1)(p, -1)(p,−1), and (−1,3)(-1, 3)(−1,3) collinear?

Q4. Midpoint of a Line Segment: Find the coordinates of the midpoint of the line segment joining the points A(1,−2)A(1, -2)A(1,−2) and B(3,4)B(3, 4)B(3,4).

Q5. Distance Between Two Points: Calculate the distance between the points A(3,5)A(3, 5)A(3,5) and B(8,−7)B(8, -7)B(8,−7). 

Q6. Area of a Triangle: Find the area of the triangle with vertices at A(0,0)A(0, 0)A(0,0), B(4,0)B(4, 0)B(4,0), and C(0,3)C(0, 3)C(0,3).

Q7. Point Dividing a Line Segment: Find the coordinates of the point that divides the line segment joining A(2,3)A(2, 3)A(2,3) and B(4,7)B(4, 7)B(4,7) in the ratio 1:2.

Q8. Equation of a Line: Find the equation of the line passing through the points (1,2)(1, 2)(1,2) and (3,4)(3, 4)(3,4).

Q9. Distance from a Point to a Line: Find the perpendicular distance from the point P(3,4)P(3, 4)P(3,4) to the line 2x−y+1=02x - y + 1 = 02x−y+1=0.

Q10. Centroid of a Triangle: Find the coordinates of the centroid of the triangle with vertices at A(1,2)A(1, 2)A(1,2), B(4,5)B(4, 5)B(4,5), and C(7,8)C(7, 8)C(7,8).

How to Ace These Class 10 Coordinate Geometry Important Questions

This chapter involves clear concepts and direct application of formulas. Follow this step-by-step approach for efficient preparation:

Step 1: Revise Basic Coordinate Concepts

Understand how to locate points in the Cartesian plane using ordered pairs (x, y). Learn how the x-axis, y-axis, and four quadrants divide the plane. Remember that the signs of coordinates vary in different quadrants.

Step 2: Learn the Distance Formula

The distance between two points (x₁, y₁) and (x₂, y₂) is given by
√[(x₂ – x₁)² + (y₂ – y₁)²]
Practise this formula for questions based on finding distance between two points or verifying geometric shapes like triangles or rectangles.

Step 3: Practise the Section Formula

Learn the formula for finding coordinates of a point dividing a line segment joining (x₁, y₁) and (x₂, y₂) internally in the ratio m:n.
It is given by:
[(mx₂ + nx₁) / (m + n), (my₂ + ny₁) / (m + n)]
Understand its application for midpoint calculation by taking m = n = 1.

Step 4: Revise the Midpoint Formula

The midpoint (M) of a line segment joining (x₁, y₁) and (x₂, y₂) is given by
M = [(x₁ + x₂)/2, (y₁ + y₂)/2]
This is a direct application of the section formula and is often asked in short questions.

Step 5: Learn the Area of a Triangle Formula

For points A(x₁, y₁), B(x₂, y₂), and C(x₃, y₃), the area of triangle ABC is given by:
½ |x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)|
Practise this formula for both direct numerical and reasoning-based questions.

Step 6: Understand Verification-Based Questions

Some questions ask you to verify whether points are collinear or form a specific shape. Use the distance or area formulas logically to verify the given condition.

Step 7: Practise Mixed Problems

Combine formulas to solve complex questions involving both distance and area or midpoint and ratio together. This will help you handle multi-step exam questions easily.

How to Use Class 10 Important Questions with Solutions for Maximum Benefits?

Effectively utilizing coordinate geometry in class 10 important questions is a crucial aspect of exam preparation. This detailed guide will help you maximize these resources:

• Use the extra questions and solutions to identify where you're struggling. If you find that you're consistently making errors in a particular type of question, dedicate additional time to that topic. You can revise the theory, practice more questions, and then review the solution PDFs to make sure you're on the right track.
• Once you have enough practice, it's a good idea to solve the important class 10 coordinate geometry questions under timed conditions. Set a timer for the same amount of time as an exam and solve the questions without solutions or notes. This will help you build exam discipline and improve time management.
• Regularly review the solutions to the questions you’ve solved. Sometimes, we think we’ve mastered a concept, only to realize later that we’ve forgotten the details. Every week, dedicate a day to review the solutions from the Coordinate Geometry Class 10 Extra Questions with Solutions PDF that you have practiced. This way, the concepts stay fresh in your mind.
• Coordinate geometry is a visual subject. Make use of graphs and diagrams while solving questions. Draw the coordinate plane, plot the points, and see the relationships between the geometric shapes and the coordinate system. This not only enhances understanding but also helps in retaining information longer.

How to Use Class 10 Ch7 Maths Important Questions Effectively?

To truly excel in the coordinate geometry chapter of Class 10, here are some unique tips:

• It’s tempting to simply memorize the formulas for distance, section, and area. However, understanding the derivation of these formulas can significantly improve your ability to apply them correctly in unfamiliar situations.
• When practicing, use graph paper. Plotting points and lines on graph paper makes the geometric relationships clearer and helps understand the nuances of questions better.
• Many questions in exams are word problems. Ensure you understand how to translate these word problems into coordinate geometry equations or diagrams.
• Take periodic self-tests using extra questions to check your progress. This will help you stay focused on your preparation and determine the additional work required in specific areas.
• Refer to past year's papers and sample papers to understand the latest trends in the types of questions asked. Occasionally, similar patterns or repetitions of questions occur.

FAQs

Q1. How many marks are usually allotted to this chapter in Class 10 board exams?

Ans. This chapter usually carries 6 to 8 marks, mostly through numerical and application-based questions.

Q2. Which topics are most important for exams?

Ans. Distance formula, section formula, and area of a triangle are the most important and frequently asked topics.

Q3. How can I avoid making sign errors in formulas?

Ans. Always write down the coordinates clearly, check the quadrant they belong to, and substitute carefully with brackets around negative numbers.

Q4. Are derivations of formulas required in exams?

Ans. No, derivations are not required. Only direct application of formulas is expected in CBSE exams.

Q5. How can I prepare this chapter quickly before exams?

Ans. Revise all formulas and solve 3 to 4 mixed problems daily from sample papers and previous years’ question papers.