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## Extra Questions for Class 10 Maths Coordinate Geometry with Answers Solutions

**Extra Questions for Class 10 Maths Chapter 7 Coordinate Geometry with Solutions Answers**

### Coordinate Geometry Class 10 Extra Questions Very Short Answer Type

Answer the following questions in one word, one sentence or as per the exact requirement of the question.

**Coordinate Geometry Class 10 Extra Questions Question 1.**

What is the area of the triangle formed by the points 0 (0, 0), A (-3, 0) and B (5, 0)?

Solution:

Area of ∆OAB = \(\frac{1}{2}\) [0(0 – 1) – 3(0 – 0) + 5(0 – 0)] = 0

⇒ Given points are collinear

**Coordinate Geometry Extra Questions Class 10 Question 2.**

If the centroid of triangle formed by points P (a, b), Q (b, c) and R (c, a) is at the origin, what is the value of a + b + c?

Solution:

**Coordinate Geometry Class 10 Extra Questions With Solutions Pdf Question 3.**

AOBC is a rectangle whose three vertices are A (0, 3), 0 (0, 0) and B (5, 0). Find the length of its diagonal.

Solution:

**Class 10 Coordinate Geometry Extra Questions Question 4.**

Find the value of a, so that the point (3, a) lie on the line 2x – 3y = 5.

Solution:

Since (3, a) lies on the line 2x – 3y = 5

Then 2(3) – 3(a) = 5

– 3a = 5 – 6

– 3a = -1

⇒ a = \(\frac{1}{3}\)

**Coordinate Geometry Class 10 Important Questions Question 5.**

Find distance between the points (0, 5) and (-5, 0).

Solution:

Here x_{1} = 0, y_{1} = 5, x_{2} = -5 and y_{2} = 0)

**Extra Questions Of Coordinate Geometry Class 10 Question 6.**

Find the distance of the point (-6,8) from the origin.

Solution:

Here x_{1} = -6, y_{1} = 8

x_{2} = 0, y_{2} = 0

**Class 10 Maths Chapter 7 Extra Questions With Solutions Question 7.**

If the distance between the points (4, k) and (1, 0) is 5, then what can be the possible values of k?

Solution:

Using distance formula

**Coordinate Geometry Class 10 Extra Questions With Solutions Question 8.**

If the points A (1, 2), B (0, 0) and C (a, b) are collinear, then what is the relation between a and b?

Solution:

Points A, B and C are collinear

⇒ 1(0 – b) + 0 (b – 2) + a(2 – 0) = 0

⇒ -b + 2a = 0 or 2a = b

**Extra Questions On Coordinate Geometry Class 10 Question 9.**

Find the ratio in which the line segment joining the points (-3, 10) and (6, – 8) is divided by (-1, 6).

Solution:

In Fig. 6.6, let the point P(-1, 6) divides the line joining A(-3, 10) and B (6, -8) in the ratio k : 1

Hence, the point P divides AB in the ratio 2 : 7.

**Coordinate Geometry Class 10 Extra Questions Pdf Question 10.**

The coordinates of the points P and Q are respectively (4, -3) and (-1, 7). Find the abscissa of a point R on the line segment PQ such that \(\frac{P R}{P Q}\) = \(\frac{3}{5}\).

Solution:

### Coordinate Geometry Class 10 Extra Questions Short Answer Type 1

**Coordinate Geometry Class 10 Questions With Solutions Question 1.**

Write the coordinates of a point on x-axis which is equidistant from the points (-3, 4) and (2, 5).

Solution:

Let the required point be (x, 0).

Since, (x, 0) is equidistant from the points (-3, 4) and (2, 5) .

**Class 10 Maths Coordinate Geometry Extra Questions Question 2.**

Find the values of x for which the distance between the points P (2, -3) and Q (x, 5) is 10.

Solution:

**Coordinate Geometry Questions Class 10 Question 3.**

What is the distance between the points (10 cos 30°, 0) and (0, 10 cos 60°)?

Solution:

**Extra Questions On Distance Formula Class 10 Question 4.**

In Fig. 6.8, if A(-1, 3), B(1, -1) and C (5, 1) are the vertices of a triangle ABC, what is the length of the median through vertex A?

Solution:”

**Ch 7 Maths Class 10 Extra Questions Question 5.**

Find the ratio in which the line segment joining the points P (3, -6) and Q (5,3) is divided by the x-axis.

Solution:

Let the required ratio be λ : 1

Given that this point lies on the x-axis

Thus, the required ratio is 2 : 1.

**Class 10 Maths Ch 7 Extra Questions Question 6.**

Point P (5, -3) is one of the two points of trisection of the line segment joining the points A (7, -2) and B (1, -5). State true or false and justify your answer.

Solution:

Points of trisection of line segment AB are given by

∴ Given statement is true.

**Coordinate Geometry Extra Questions Question 7.**

Show that ∆ABC, where A(-2, 0), B(2, 0), C(0, 2) and APQR where P(-4, 0), Q(4, 0), R(0,4) are similar triangles.

OR

Show that ∆ABC with vertices A(-2, 0), B(0, 2) and C(2, 0) is similar to ∆DEF with vertices D(-4, 0), F(4,0) and E(0, 4).

[∆PQR is replaced by ∆DEF]

Solution:

**Chapter 7 Maths Class 10 Extra Questions Question 8.**

Point P (0, 2) is the point of intersection of y-axis and perpendicular bisector of line segment joining the points, A (-1, 1) and B (3, 3). State true or false and justify your answer.

Solution:

The point P (0, 2) lies on y-axis

AP ≠ BP

∴ P(0, 2) does not lie on the perpendicular bisector of AB. So, given statement is false.

**Coordinate Geometry Class 10 Questions Question 9.**

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

Solution:

Let A (5,-2), B (6, 4) and C (7, -2) be the vertices of a triangle

Here, AB = BC

∴ ∆ABC is an isosceles triangle.

**Class 10 Coordinate Geometry Important Questions Question 10.**

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

Solution:

Let A(1, 2), B(4, y), C(x, 6) and D(3, 5) be the vertices of a parallelogram ABCD.

Since, the diagonals of a parallelogram bisect each other.

Hence, x = 6 and y = 3.

Question 11.

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

Solution:

Let the point on y-axis be P(0, y) and AP : PB = k : 1

Question 12.

Let P and Q be the points of trisection of the line segment joining the points A(2, -2) and B(-7, 4) such that P is nearer to A. Find the coordinates of P and Q.

Solution:

∵ P divides AB in the ratio 1 : 2.

Question 13.

Find the ratio in which the point (-3, k) divides the line-segment joining the points (-5, 4) and (-2, 3). Also find the value of k.

Solution:

Let Q divide AB in the ratio of p : 1

Question 14.

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

Solution:

Let the point P be (2y, y)

Hence, coordinates of point P are (16, 8).

Question 15.

If two adjacent vertices of a parallelogram are (3, 2) and (-1, 0) and the diagonals intersect at (2, -5), then find the coordinates of the other two vertices.

Solution:

Let other two coordinates are (x, y) and (x’, y’)

O is mid point of AC and BD

Hence, co-ordinates are (1, -12) and (5, -10).

### Coordinate Geometry Class 10 Extra Questions Short Answer Type 2

Question 1.

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

Solution:

Let A (1, 5), B (2, 3) and C (-2, -11) be the given points. Then we have

Clearly, AB + BC ≠ AC

∴ A, B, C are not collinear.

Question 2.

Find the distance between the following pairs of points:

(i) (-5, 7), (-1, 3)

(ii) (a, b), (-a, -b)

Solution:

(i) Let two given points be A (-5, 7) and B (-1, 3).

Thus, we have x_{1} = -5 and x_{2} = -1

y_{1} = 7 and y_{2} = 3

(ii) Let two given points be A (a,b) and B(-a, -b)

Here, x_{1} = a and x_{2} = -a; y_{1} = b and y_{2} = -b

Question 3.

Name the type of quadrilateral formed, if any, by the following points, and give reasons for

your answer: (i) (-1, -2), (1, 0), (- 1, 2), (-3, 0) (ii) (4, 5), (7, 6), (4, 3), (1, 2)

Solution:

(i) Let A(-1, -2), B(1, 0), C(-1, 2) and D(-3, 0) be the four given points.

Then, using distance formula, we have,

Hence, four sides of quadrilateral are equal and diagonals AC and BD are also equal.

∴ Quadrilateral ABCD is a square.

(ii) Let A (4, 5), B (7, 6), C (4, 3) and D (1, 2) be the given points. Then,

∴ ABCD is a parallelogram.

Question 4.

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

OR

A line segment is of length 10 units. If the coordinates of its one end are (2, -3) and the abscissa of the other end is 10, find its ordinate.

Solution:

We have, PQ = 10

Squaring both sides, we have

⇒ (8)^{2} + (y + 3)^{2} = 100

⇒ (y + 3)^{2} = 100 – 64

⇒ (y + 3)^{2} = 36 or y + 3 ± 16

⇒ y + 3 = 6, y + 3 = -6 or y = 3, y = -9

Hence, values of y are – 9 and 3.

Question 5.

If Q(0, 1) is equidistant from P(5,-3) and R(x, 6) find the value of x. Also, find the distances of QR and PR.

Solution:

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

Therefore, QP = QR

Squaring both sides, we have, Qp^{2} = QR^{2}

⇒ (5 – 0)^{2} + (-3 -1)^{2} = (x – 0)^{2} + (6 – 1)^{2}

25 + 16 = x^{2} + 25

x^{2} = 16

∴ x = ±4

Thus, R is (4, 6) or (-4, 6).

Question 6.

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

OR

Find the coordinates of a point on the x-axis which is equidistant from the points A(2, -5) and B(-2, 9).

Solution:

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

Now, P(x, 0) is equidistant from point A(2,-5) and B(-2, 9)

∴ AP = BP

Squaring both sides, we have

(x – 2)^{2} + 25 = (x + 2)^{2} + 81

⇒ x^{2} + 4 – 4x + 25 = x^{2} + 4 + 4x + 81

⇒ -8x = 56

∴ x = \(\frac{56}{-8}\)

∴ The point on the x-axis equidistant from given points is (-7,0).

Question 7.

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

Solution:

Given points are A(x, y), B(1, 2) and C(7, 0)

These points will be collinear if the area of the triangle formed by them is zero.

Now, ar(∆ABC) = \(\frac{1}{2}\) [x_{1}(y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2})]

⇒ 0 = \(\frac{1}{2}\) [x(2 – 0) + 1(0 – y) + 7(y – 2)]

⇒ 0 = \(\frac{1}{2}\) (2x – y + 7y – 14)

⇒ 2x + by – 14 = 0

⇒ x + 3y = 7, which is the required relation between x and y.

Question 8.

Find a relation between x and y such that the point (x, y) is equidistant from the point (3,6) and (-3, 4).

Solution:

Let P(x, y) be equidistant from the points A(3, 6) and B(-3, 4)

i.e., PA = PB

Squaring both sides, we get

AP^{2} = BP^{2}

⇒ (x – 3)^{2} + (y – 6)^{2} = (x + 3)^{2} + (1 – 4)^{2}

⇒ x^{2} – 6x + 9 + y^{2} – 12y + 36 = x^{2} + 6x + 9 +y^{2} – 8y + 16

⇒ -12x – 4y + 20 = 0

⇒ 3x + y – 5 = 0, which is the required relation.

Question 9.

Find the coordinates of the point which divides the line joining of (- 1, 7) and (4, – 3) in the ratio 2 : 3.

Solution:

Let P(x, y) be the required point. Thus, we have

So, the coordinates of P are (1, 3).

Question 10.

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

Solution:

Let the given points be A(4, -1) and B(-2,-3) and points of trisection be P and Q.

Lęt AP = PQ = QB = k

PB = PQ + QB = k + k = 2k

AP : PB = k : 2k = 1 : 2.

Therefore, coordinates of P are

Question 11.

Find the ratio in which the line segment joining A(1, -5) and B(-4, 5) is divided by the x-axis. Also find the coordinates of the point of division.

Solution:

Let the required ratio bek: 1. Then, the coordinates of the point of division is

Question 12.

The points A(4, -2), B(7, 2), C(0, 9) and D(-3, 5) form a parallelogram. Find the length of the altitude of the parallelogram on the base AB.

Solution:

Let, DE = h be the height of parallelogram ABCD w.r.t. base AB.

Question 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).

Solution:

Let the coordinates of A be (x, y)

Now, C is the centre of circle therefore, the coordinates of

Hence, coordinates of A are (3, -10).

Question 14.

If A and B are (-2, -2) and (2, -4), respectively, find the coordinates of P such that AB = 5 AB and P lies on the line segment AB.

Solution:

Question 15.

Find the coordinates of the points which divide the line segment joining A (-2, 2) and B (2, 8) into four equal parts (Fig. 6.21).

Solution:

Let P, Q, R be the points that divide the line segment joining A(-2, 2) and B(2, 8) into four equal parts.

Since, Q divides the line segment AB into two equal parts, i.e., Q is the mid-point of AB.

Question 16.

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

Solution:

Let A(3, 0), B(4, 5), C(-1, 4) and D(-2, -1) be the vertices of a rhombus.

Therefore, its diagonals

Question 17.

Find the area of the triangle whose vertices are: (-5, -1), (3, -5), (5, 2)

Solution:

Let A(x_{1}, y_{1}) = (-5, -1), B(x_{2}, y_{2}) = (3 – 5), C(x_{3}, y_{3}) = (5, 2)

∴ area of ∆ABC = \(\frac{1}{2}\) [x_{1} (y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2})]

= \(\frac{1}{2}\)(-5(-5 – 2) + 3 (2 + 1) + 5(-1 + 5)]

= \(\frac{1}{2}\)(35 + 9 + 20) = \(\frac{1}{2}\) × 64 = 32 sq units.

Question 18.

If the point A (0, 2) is equidistant from the points B(3, p) and C(p, 5), find p. Also find the length of AB.

Solution:

Given that A (0, 2) is equidistant from B (3, p) and C (p, 5)

∴ AB = AC

or

AB^{2} = AC^{2}

(3 – 0)^{2} + (0 – 2)^{2} = (p – 0)^{2} + (5 – 2)^{2}

3^{2} + p^{2} + 4 – 4p = p^{2} + 9

= 4 – 4p = 0

⇒ 4p = 4

⇒ p = 1

Question 19.

If the points A (-2, 1), B (a, b) and C (4, -1) are collinear and a – b = 1, find the values of a and b.

Solution:

Since the given points are collinear, then area of ∆ABC = 0

⇒ \(\frac{3}{5}\)[x1 ( y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2})] = 0

Given, x_{1} = -2, y_{1} = 1, x_{2} = 0, y_{2} = b, x_{3} = 4, y_{3} = -1

Putting the values,

\(\frac{1}{2}\)[-2(b + 1) + a(-1 – 1) + 4(1 – b)] = 0

⇒ -26 – 2 – 2a + 4- 4b = 0

⇒ 2a + 65 = 2

a + 3b = 1 ….. (i)

Given, a – b = 1 … (ii)

Subtracting (i) from (ii), we have

– 4b = 0

⇒ b = 0

Substituting the value of b in (ii), we have a = 1.

Question 20.

If the point P (k-1, 2) is equidistant from the points A (3, k) and B (k, 5), find the values of k.

Solution:

Let the given line segment be divided by point Q. Since P is equidistant from A and B,

AP = BP or Ap^{2} = BP^{2}

[3 – (k – 1)]^{2} + (k – 2)^{2} = [k – (k – 1)]^{2} + (5 – 2)^{2}

(3 – k + 1)^{2} + (k – 2)^{2} = (k – k + 1)^{2} + (3)^{2}

(4 – k)^{2} + (k – 2)^{2} = (1)^{2} + (3)^{2}

⇒ 16 + k^{2} – 8k + k^{2} + 4 – 4k = 1 + 9

⇒ 2k^{2} – 12k + 20 = 10

⇒ k^{2} – 6k + 10 = 5

⇒ ķ^{2} – 6k + 5 = 0

⇒ k^{2} – 5k – k + 5 = 0

⇒ k (k – 5) -1(k – 5) = 0

⇒ k = 1 or k = 5

Question 21.

Find the ratio in which the line segment joining the points A(3, -3) and B(-2, 7) is divided by x-axis. Also find the coordinates of the point of division.

Solution:

Here, point Q is on x-axis so its ordinate is O.

Let ratio be k : 1 and coordinate of point Q be (x, 0)

We are given that A(3, -3) and B(-2, 7)

Question 22.

Find the values of k if the points Ask + 1, 2k), B(3k, 2k + 3) and C(5k – 1, 5k) are collinear.

Solution:

Points Ask + 1, 2k), B (3k, 2k + 3) and C (5k – 1, 5k) are collinear

∴ Area of ∆ABC = 0

Question 23.

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.

Solution:

Given, PA = PB or (PA)^{2} = (PB)^{2}

(a + b – x)^{2} + (b – a – y)^{2} = (a – b – x)^{2} + (a + b – y)^{2}

⇒ (a + b)^{2} + x^{2} – 2ax – 25x + (b – a)^{2} + y^{2} – 2by + 2ay

⇒ (a – b)^{2} + x^{2} – 2ax + 2bx + (a + b)^{2} + y^{2} – 2ay – 2by

⇒ 4ay = 4bx or bx = ay

Hence proved.

Question 24.

If the point C(-1, 2) divides internally the line segment joining the points A(2, 5) and B(x, y) in the ratio of 3 : 4, find the value of x^{2} + y^{2}.

Solution:

Question 25.

In the given figure, ∆ABC is an equilateral triangle of side 3 units. Find the coordinates of the other two vertices.

Solution:

Since AB = BC = AC = 3 units

∴ Co-ordinates of B are (5, 0)

Let co-ordinates of C be (x, y)

AC^{2} = BC^{2}

[∵ ∆ABC is an equilateral triangle]

Using distance formula

⇒ (x – 2)^{2} + (y – 0)^{2} = (x – 5)^{2} + (y – 0)^{2}

⇒ x^{2}+ 4 – 4x + y = x^{2}+ 25 – 10x + y^{2 }6x = 21

⇒ x = \(\frac{21}{6}\) = \(\frac{7}{2}\)

Again (x – 2)^{2} + (y – 0)^{2} = 9

### Coordinate Geometry Class 10 Extra Questions Long Answer Type

Question 1.

Find the value of ‘k”, for which the points are collinear: (7, -2), (5, 1), (3, k).

Solution:

Let the given points be

A (x_{1}, y_{1}) = (7, -2), B (x_{2}, Y_{2}) = (5, 1) and C (x_{3}, y_{3}) = (3, k)

Since these points are collinear therefore area (∆ABC) = 0

⇒ \(\frac{1}{2}\) [x_{1}(y_{2} – Y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2})] = 0

⇒ x_{1}(y_{2} – y_{3}) + x_{2}(y_{3} – y_{1}) + x_{3}(y_{1} – y_{2}) = 0

⇒ 7(1 – k) + 5(k + 2) + 3(-2 -1) = 0

⇒ 7 – 7k + 5k + 10 – 9 = 0

⇒ -2k + 8 = 0

⇒ 2k = 8

⇒ k = 4

Hence, given points are collinear for k = 4.

Question 2.

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 A (x_{1}, y_{1}) = (0, -1), B (x_{2}, y_{2}) = (2, 1), C (x_{3}, y_{3}) = (0, 3) be the vertices of ∆ABC.

Now, let P, Q, R be the mid-points of BC, CA and AB, respectively.

So, coordinates of P, Q, R are

Ratio of ar (∆PQR) to the ar (∆ABC) = 1 : 4.

Question 3.

Find the area of the quadrilateral whose vertices, taken in order, are (-4, -2), (-3, -5), (3, -2) and (2, 3).

Solution:

Let A(4, -2), B(-3, -5), C(3, -2) and D(2, 3) be the vertices of the quadrilateral ABCD.

Now, area of quadrilateral ABCD

= area of ∆ABC + area of ∆ADC

Question 4.

A median of a triangle divides it into two triangles of equal areas. Verify this result for ∆ABC whose vertices are A (4,-6), B (3, -2) and C (5, 2).

Solution:

Since AD is the median of ∆ABC, therefore, D is the mid-point of BC.

Hence, the median divides it into two triangles of equal areas.

Question 5.

Find the ratio in which the point P (x, 2), divides the line segment joining the points A (12, 5) and B (4, -3). Also find the value of x.

Solution:

The ratio in which p divides the line segment is \(\frac{3}{5}\), i.e., 3 : 5.

Question 6.

If A (4, 2), B (7, 6) and C (1, 4) are the vertices of a ∆ABC and AD is its median, prove that the median AD divides into two triangles of equal areas.

Solution:

Given: AD is the median on BC.

⇒ BD = DC

The coordinates of midpoint D are given by.

Hence, AD divides ∆ABC into two equal areas.

Question 7.

If the point A (2, -4) is equidistant from P (3, 8) and Q (-10, y), find the values of y. Also find distance PQ.

Solution:

Given points are A(2, 4), P(3, 8) and Q(-10, y)

According to the question,

PA = QA

Question 8.

The base BC of an equilateral triangle ABC lies on y-axis. The coordinates of point Care (0, -3). The origin is the mid-point of the base. Find the coordinates of the points A and B. Also find the coordinates of another point D such that BACD is a rhombus.

Solution:

∵ O is the mid-point of the base BC.

∴ Coordinates of point B are (0, 3). So,

BC = 6 units Let the coordinates of point A be (x, 0).

Using distance formula,

∴ Coordinates of point A = (x, 0) = (3√3, 0)

Since BACD is a rhombus.

∴ AB = AC = CD = DB

∴ Coordinates of point D = (-3√3, 0).

Question 9.

Prove that the area of a triangle with vertices (t, t-2), (t + 2, t + 2) and (t + 3, t) is independent of t.

Solution:

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

Area of the triangle = \(\frac{1}{2}\)[t + 2 – t) + (t + 2) (t – t + 2) + (t + 3) (t – 2 – t – 2)]

= \(\frac{1}{2}\) [2t + 2t + 4 – 4t – 12 ]

= 4 sq. units

which is independent of t.

Hence proved.

Question 10.

The area of a triangle is 5 sq units. Two of its vertices are (2, 1) and (3, -2). If the third vertex is (\(\frac{7}{2}\), y), find the value of y.

Solution:

Given: ar(∆ABC) = 5 sq. units

Question 11.

The coordinates of the points A, B and Care (6, 3), (-3,5) and (4,-2) respectively. P(x, y) is any point in the plane. Show that \(\frac { ar(∆PBC) }{ ar(∆ABC) } \) = \(\frac{x+y-2}{7}\)

Solution:

P(x, y), B(-3, 5), C(4, -2), A(6, 3)

Question 12.

In Fig. 6.32, the vertices of ∆ABC are A(4, 6), B(1, 5) and C(7, 2). A line-segment DE is drawn to intersect the sides AB and AC at D and E respectively such that \(\frac{A D}{A B}=\frac{A E}{A C}=\frac{1}{3}\) Calculate the area of ∆ADE and compare it with area of ∆ABC.

Solution:

\

Question 13.

If a = b = 0, prove that the points (a, a^{2}), (b, b^{2}) (0, 0) will not be collinear.

Solution:

∵ We know that three points are collinear if area of triangle = 0

∴ Area of triangle with vertices (a, a^{2}), (b, b^{2}) and (0, 0)

∵ Area of ∆ ≠ 0

∴ Given points are not collinear.

### Coordinate Geometry Class 10 Extra Questions HOTS

Question 1.

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

Solution:

As line segment AB is trisected by the points P and Q.

Therefore,

Case I: When AP : PB = 1 : 2.

⇒ P (3, -2)

Since the point P (3,-2) lies on the line

2x – y + k = 0 =

⇒ 2 × 3-(-2) + k = 0

⇒ k = -8

Case II: When AP : PB = 2 : 1.

Coordinates of point P are

Since the point P(4, -5) lies on the line

2x – y + k = 0

∴ 2 × 4-(-5) + k = 0

∴ k = -13

Question 2.

Prove that the diagonals of a rectangle bisect each other and are equal.

Solution:

Let OACB be a rectangle such that OA is along x-axis and OB is along y-axis. Let OA = a and OB = b.

Then, the coordinates of A and B are (a,0) and (0, b) respectively.

Since, OACB is a rectangle. Therefore,

AC = OB

⇒ AC = b

Also, OA = a

⇒ BC = a

So, the coordinates of Care (a, b).

∴ OC = AB.

Question 3.

In what ratio does the y-axis divide the line segment joining the point P (4, 5) and Q (3, -7)?

Also, find the coordinates of the point of intersection.

Solution:

Suppose y-axis divides PQ in the ratio k : 1. Then, the coordinates of the point of division are|

Question 4.

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

Solution:

Let O(x, y) be the centre of circle. Given points are A(6, -6), B(3, -7) and C(3, 3).

Question 5.

If the coordinates of the mid-points of the sides of a triangle are (1, 1), (2, – 3) and (3, 4). Find its centroid.

Solution:

Let P(1, 1), Q(2, -3), R(3, 4) be the mid-points of sides AB, BC and CA respectively, of triangle ABC. Let A(x_{1}, y_{1}), B(x_{2}, y_{2}) and C(x_{3}, y_{3}) be the vertices of triangle ABC. Then, P is the mid-point of AB.

Question 6.

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).

Solution:

Let P(x_{1}, y_{1}) be common point of both lines and divide the line segment joining A(2, -2) and B(3, 7) in ratio k : 1.

Question 7.

Show that ∆ABC with vertices A (-2, 0), B (2, 0) and C (0, 2) is similar to ∆DEF with vertices

D(4, 0) E (4, 0) and F (0, 4).

Solution:

Given vertices of ∆ABC and ∆DEF are

A(-2, 0), B(2, 0), C(0, 2), D(-4, 0), E(4, 0) and F(0, 4)

Here, we see that sides of ∆DEF are twice the sides of a ∆ABC.

Hence, both triangles are similar.