**Important Questions Class 10 Maths Chapter 7** Coordinate Geometry are given at BYJU’S with stepwise solutions. Students who are preparing for the **board exams of 2022-23** are advised to practice these important questions of Coordinate Geometry to score high marks in the Maths exam. These questions will provide good practice to students to solve any problem asked in the exam from this chapter. Solving these questions will also help in revision, and it will boost the confidence level of students.

This chapter has some important formulas to help students with their board exam preparation and competitive exams. Some of the important questions from the distance formula, section formula, and the triangle area are provided here. Students can also get the solutions for all the questions of the Class 10 **NCERT **textbook for Maths.

## Coordinate Geometry Formulas List for Classes 9, 10 and 11

Coordinate geometry is an integral topic in classes 9, 10 and 11. All the important coordinate geometry formulas for class 9, class 10 and class 11 are given below.

All Formulas of Coordinate Geometry | |
---|---|

General Form of a Line | Ax + By + C = 0 |

Slope Intercept Form of a Line | y = mx + c |

Point-Slope Form | y − y_{1}= m(x − x_{1}) |

The slope of a Line Using Coordinates | m = Δy/Δx = (y_{2 }− y_{1})/(x_{2 }− x_{1}) |

The slope of a Line Using General Equation | m = −(A/B) |

Intercept-Intercept Form | x/a + y/b = 1 |

Distance Formula | |P_{1}P_{2}| = √[(x_{2 }− x_{1})^{2 }+ (y_{2 }− y_{1})^{2}] |

For Parallel Lines, | m_{1 }= m_{2} |

For Perpendicular Lines, | m_{1}m_{2} = -1 |

Midpoint Formula | M (x, y) = [½(x_{1 }+ x_{2}), ½(y_{1 }+ y_{2})] |

Angle Formula | tan θ = [(m_{1 }– m_{2})/ 1 + m_{1}m_{2}] |

Area of a Triangle Formula | ½ |x_{1}(y_{2}−y_{3})+x_{2}(y_{3}–y_{1})+x_{3}(y_{1}–y_{2})| |

Distance from a Point to a Line | d = [|Ax_{0 }+ By_{0 }+ C| / √(A^{2 }+ B^{2})] |

Section Formula (Internal division) | P(x, y) = [(m_{1}x_{2} + m_{2}x_{1})/(m_{1} + m_{2}), (m_{1}y_{2} + m_{2}y_{1})/(m_{1} + m_{2})] |

Section Formula (External division) | P(x, y) = [(m_{1}x_{2} – m_{2}x_{1})/(m_{1} – m_{2}), (m_{1}y_{2} – m_{2}y_{1})/(m_{1} – m_{2})] |

## Solved Examples

**Example 1: **

What is the slope of the line joining the points P(-2, 3) and Q(2, 7)?

**Solution:**

Let the given points be:

P(-2, 3) = (x_{1}, y_{1})

Q(2, 7) = (x_{2}, y_{2})

Solpe of the line PQ = (y_{2} – y_{1})/(x_{2} – x_{1})

= (7 – 3)/ [2 – (-2)]

= 4/4

= 1

**Example 2:**

If the distance between the points A(2, –2) and B(–1, x) is 5, find the value of x.

**Solution:**

Let the given points be:

A(2, -2) = (x_{1}, y_{1})

B(-1, x) = (x_{2}, y_{2})

Distance between A and B = √[(x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2}]

√[(-1 – 2)^{2} + (x + 2)^{2}] = 5 [given distance is 5]

Squaring on both sides, we get;

(3)^{2} + (x + 2)^{2} = 25

(x + 2)^{2} = 25 – 9

(x + 2)^{2} = 16

x + 2 = ± 4

When x + 2 = 4, x = 2

When x + 2 = -4, x = -6

## Important Questions & Answers For Class 10 Maths Chapter 7 Coordinate Geometry

**Q. 1: Find the distance of the point P (2, 3) from the x-axis.**

**Solution:**

We know that,

(x, y) = (2, 3) is a point on the Cartesian plane in the first quadrant.

x = Perpendicular distance from y-axis

y = Perpendicular distance from x-axis

Therefore, the perpendicular distance from x-axis = y coordinate = 3

**Q. 2:** **Find a relation between x and y such that the point (x, y) is equidistant from the points (7, 1) and (3, 5).**

**Solution:**

Let P(x, y) be equidistant from the points A(7, 1) and B(3, 5).

Then, AP = BP

AP^{2} = BP^{2}

Using distance formula,

(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

x – y = 2

Hence, the relation between x and y is x – y = 2.

**Q. 3:** **Find the coordinates of the points of trisection (i.e., points dividing into three equal parts) of the line segment joining the points A(2, – 2) and B(– 7, 4).**

**Solution:**

Let P and Q be the points of trisection of AB, i.e., AP = PQ = QB.

Therefore, P divides AB internally in the ratio 1: 2.

Let (x_{1}, y_{1}) = (2, -2)

(x_{2}, y_{2}) = (-7, 4)

m_{1} : m_{2} = 1 : 2

Therefore, the coordinates of P, by applying the section formula,

**Q. 5: Find the value of k if the points A(2, 3), B(4, k) and C(6, –3) are collinear.**

**Solution:**

Given,

A(2, 3)= (x_{1}, y_{1})

B(4, k) = (x_{2}, y_{2})

C(6, -3) = (x_{3}, y_{3})

If the given points are collinear, the area of the triangle formed by them will be 0.

½ [x_{1}(y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2})] = 0

½ [2(k + 3) + 4(-3 -3) + 6(3 – k)] = 0

½ [2k + 6 – 24 + 18 – 6k] = 0

½ (-4k) = 0

4k = 0

k = 0

**Q. 6: Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, –1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle.**

**Solution:**

Let the vertices of the triangle be A(0, -1), B(2, 1) and C(0, 3).

Let D, E, F be the mid-points of the sides of this triangle.

Using the mid-point formula, coordinates of D, E, and F are:

D = [(0+2)/2, (-1+1)/2] = (1, 0)

E = [(0+0)/2, (-1+3)/2] = (0, 1)

F = [(0+2)/2, (3+1)/2] = (1, 2)

We know that,

Area of triangle = ½ [x_{1}(y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2})]

Area of triangle DEF = ½ {(1(2 – 1) + 1(1 – 0) + 0(0 – 2)}

= ½ (1 + 1)

= 1

Area of triangle DEF = 1 sq.unit

Area of triangle ABC = ½ {0(1 – 3) + 2(3 – (-1)) + 0(-1 – 1)}

= ½ (8)

= 4

Area of triangle ABC = 4 sq.units

Hence, the ratio of the area of triangle DEF and ABC = 1 : 4.

**Q. 7:** **Name the type of triangle formed by the points A (–5, 6), B (–4, –2) and C (7, 5).**

**Solution:**

The points are A (–5, 6), B (–4, –2) and C (7, 5).

Using distance formula,

d = √ ((x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2})

AB = √((-4+5)² + (-2-6)²)

= √(1+64)

=√65

BC=√((7+4)² + (5+2)²)

=√(121 + 49)

=√170

AC=√((7+5)² + (5-6)²)

=√144 + 1

=√145

Since all sides are of different lengths, ABC is a scalene triangle.

**Q.8: Find the area of triangle PQR formed by the points P(-5, 7), Q(-4, -5) and R(4, 5).**

**Solution:**

Given,

P(-5, 7), Q(-4, -5) and R(4, 5)

Let P(-5, 7) = (x_{1}, y_{1})

Q(-4, -5) = (x_{2}, y_{2})

R(4, 5) = (x_{3}, y_{3})

Area of the triangle PQR = (½)|x_{1}(y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2})|

= (½) |-5(-5 – 5) + (-4)(5 – 7) + 4(7 + 5)|

= (½) |-5(-10) -4(-2) + 4(12)|

= (½) |50 + 8 + 48|

= (½) × 106

= 53

Therefore, the area of triangle PQR is 53 sq. units.

**Q.9: 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.**

**Solution:**

Given,

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

Here,

A(2, 5) = (x_{1}, y_{1})

B(x, y) = (x_{2}, y_{2})

m : n = 3 : 4

Using section formula,

C(-1, 2) = [(mx_{2} + nx_{1})/(m + n), (my_{2} + ny_{1})/(m + n)]

= [(3x + 8)/(3 + 4), (3y + 20)/(3 + 7)]

By equating the corresponding coordinates,

(3x + 8)/7 = -1

3x + 8 = -7

3x = -7 – 8

3x = -15

x = -5

And

(3y + 20)/7 = 2

3y + 20 = 14

3y = 14 – 20

3y = -6

y = -2

Therefore, the coordinates of B(x, y) = (-5, -2).

**Q.10: Find the ratio in which the line x – 3y = 0 divides the line segment joining the points (-2, -5) and (6, 3). Find the coordinates of the point of intersection.**

**Solution:**

Let the given points be:

A(-2, -5) = (x_{1}, y_{1})

B(6, 3) = (x_{2}, y_{2})

The line x – 3y = 0 divides the line segment joining the points A and B in the ratio k : 1.

Using section formula,

Point of division P(x, y) = [(kx_{2} + x_{1})/(k + 1), (ky_{2} + y_{1})/(k + 1)]

x = (6k – 2)/(k + 1) and y = (3k – 5)/(k + 1)

Here, the point of division lies on the line x – 3y = 0.

Thus,

[(6k – 2)/(k + 1)] – 3[(3k – 5)/(k + 1)] = 0

6k – 2 – 3(3k – 5) = 0

6k – 2 – 9k + 15 = 0

-3k + 13 = 0

-3k = -13

k = 13/3

Thus, the ratio in which the line x – 3y = 0 divides the line segment AB is 13 : 3.

Therefore, x = [6(13/3) – 2]/ [(13/3) + 1]

= (78 – 6)/(13 + 3)

= 72/16

= 9/2

And

y = [3(13/3) – 5]/ [(13/3) + 1]

= (39 – 15)/(13 + 3)

= 24/16

= 3/2

Therefore, the coordinates of the point of intersection = (9/2, 3/2).

**Q.11: Write the coordinates of a point on the x-axis which is equidistant from points A(-2, 0) and B(6, 0).**

**Solution:**

Let P(x, 0) be a point on the x-axis.

Given that point, P is equidistant from points A(-2, 0) and B(6, 0).

AP = BP

Squaring on both sides,

(AP)² = (BP)²

Using distance formula,

(x + 2)² + (0 – 0)² = (x – 6)² + (0 – 0)²

x² + 4x + 4 = x² – 12x + 36

4x + 12x = 36 – 4

16x = 32

x = 2

Therefore, the coordinates of a point on the x-axis = (2, 0).

**Q.12: If A(-2, 1), B(a, 0), C(4, b) and D(1, 2) are the vertices of a parallelogram ABCD, find the values of a and b. Hence, find the lengths of its sides.**

**Solution:**

Given vertices of a parallelogram ABCD are:

A(-2, 1), B(a, 0), C(4, b) and D(1, 2)

We know that the diagonals of a parallelogram bisect each other.

So, midpoint of AC = midpoint of BD

[(-2 + 4)/2, (1 + b)/2] = [(a + 1)/2, (0 + 2)/2]

By equating the corresponding coordinates,

2/2 = (a + 1)/2 and (1 + b)/2 = 2/2

a + 1 = 2 and b + 1 = 2

a = 1 and b = 1

Therefore, a = 1 and b = 1.

Let us find the lengths of sides of a parallelogram, i.e. AB, BC, CD and DA

Using the distance formula,

AB = √[(1 + 2)² + (0 – 1)²] = √(9 + 1) = √10 units

BC = √[(4 – 1)² + (1 – 0)²] = √(9 + 1) = √10 units

And CD = √10 and DA = √10 {the opposite sides of a parallelogram are parallel and equal}

Hence, the length of each side of the parallelogram ABCD = √10 units.

**Q.13: If A(-5, 7), B(-4, -5), C(-1, -6) and D(4, 5) are the vertices of a quadrilateral, find the area of the quadrilateral ABCD.**

**Solution:**

Given vertices of a quadrilateral are:

A(-5, 7), B(-4, -5), C(-1, -6) and D(4, 5)

The quadrilateral ABCD can be divided into two triangles ABD and BCD.

Area of the triangle with vertices (x_{1}, y_{1}), (x_{2}, y_{2}), and (x_{3}, y_{3}) = (½) |x_{1}(y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2})|

Area of triangle ABD = (½) |-5(-5 – 5) + (-4)(5 – 7) + 4(7 + 5)|

= (½) |-5(-10) -4(-2) + 4(12)|

= (½) |50 + 8 + 48|

= (½) × 106

= 53

Area of triangle BCD = (½) |-4(-6 – 5) + (-1)(5 + 5) + 4(-5 + 6)|

= (½) |-4(-11) -1(10) + 4(1)|

= (½) |44 – 10 + 4|

= (½) × 38

= 19

Therefore, the area of quadrilateral ABCD = Area of triangle ABD + Area of triangle BCD

= 53 + 19

= 72 sq.units

**Q.14: Find the ratio in which P(4, m) divides the line segment joining the points A(2, 3) and B(6, -3). Hence, find m.**

**Solution:**

Let P(4, m) divides the line segment joining the points A(2, 3) and B(6, -3) in the ratio k : 1.

Here,

P(4, m) = (x, y)

A(2, 3) = (x_{1}, y_{1})

B(6, -3) = (x_{2}, y_{2})

Using section formula,

p(x, y) = [(kx_{2} + x_{1})/(k + 1), (ky_{2} + y_{1})/(k + 1)]

(4, m) = [(6k + 2)/(k + 1), (-3k + 3)/(k + 1)]

By equating the x-coodinate,

(6k + 2)/(k + 1) = 4

6k + 2 = 4k + 4

6k – 4k = 4 – 2

2k = 2

k = 1

Thus, the point P divides the line segment joining A and B in the ratio 1 : 1.

Now by equating the y-coodinate,

(-3k + 3)/(k + 1) = m

Substituting k = 1,

[-3(1) + 3]/(1 + 1) = m

m = (3 – 3)/2

m = 0

**Q.15: Find the distance of a point P(x, y) from the origin.**

**Solution:**

Given,

P(x, y)

Coordinates of origin = O(0, 0)

Let P(x, y) = (x_{1}, y_{1})

O(0, 0) = (x_{2}, y_{2})

Using distance formula,

OP = √[(x_{2} – x_{1})² + (y_{2} – y_{1})²]

= √[(x – 0)² + (y – 0)²]

= √(x² + y²)

Hence, the distance of the point P(x, y) from the origin is √(x² + y²) units.