We have provided you with Extra and Important Questions from Class 10 Maths Chapter 7 Coordinate Geometry. This Extra and Important Questions will help you to score 100% in your Board Exams. These extra questions will be helpful to revise the important topics and concepts.

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**Coordinate Geometry Class 10 Important Questions with Answers Maths Chapter 7**

**Extra Questions for Class 10 Maths Chapter 7 Very Short Answer Type**

**Question:**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:**

**Question:** 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:**

**Question:**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 = 1/3

**Question:**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)

**Question:**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

**Question:**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

**Question:**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

**Question:**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) .

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

**Solution:**

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

**Solution:**

**Extra Questions for Class 10 Maths Chapter 7 Short Answer Type**

**Question:**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:**

**Question:**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.

**Question:**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.

**Question:**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:**

**Question:**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.

**Question:**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:**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:**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:**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:**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:**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:**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).

**Extra Questions for Class 10 Maths Chapter 7 Long Answer Type**

**Question:**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.

**Answer:**

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

**Answer:**

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:**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).

**Answer:**

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:**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.

**Answer:**

The ratio in which p divides the line segment is 3/5, i.e., 3 : 5.

**Question:**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.

**Answer:**

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:**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.

**Answer:**

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

According to the question,

PA = QA

**Question:**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.

**Answer:**

∵ 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:**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.

**Answer:**

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