## RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5

These Solutions are part of RD Sharma Class 10 Solutions. Here we have given RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5

**Other Exercises**

- RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.1
- RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.2
- RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.3
- RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.4
- RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5
- RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry VSAQS
- RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry MCQS

**Question 1.**

Find the area of a triangle whose vertices are :

(i) (6, 3), (-3, 5) and (4, -2)

(ii) (\({ at }_{ 1 }^{ 2 }\), 2at_{1}), (\({ at }_{ 2 }^{ 2 }\), 2at_{2}) and (\({ at }_{ 3 }^{ 2 }\), 2at_{3})

(iii) (a, c + a), (a, c) and (-a, c – a)

**Solution:**

(i) Co-ordinates of ∆ABC are A (6, 3), B (-3, 5) and C (4, -2)

**Question 2.**

Find the area of the quadrilaterals, the coordinates of whose vertices are

(i) (-3, 2), (5, 4), (7, -6) and (-5, -4)

(ii) (1, 2), (6, 2), (5, 3) and (3, 4)

(iii) (-4, -2), (-3, -5), (3, -2), (2, 3) **(C.B.S.E. 2009)**

**Solution:**

(i) Let vertices of quadrilateral ABCD be A (-3, 2), B (5, 4), C (7, -6) and D (-5, -4)

Join AC

**Question 3.**

The four vertices of a quadrilaterals are (1, 2), (-5, 6), (7, -4) and (k, -2) taken in order. If the area of the quadrilateral is zero, find the value of k ?

**Solution:**

Let the vertices of quadrilateral ABCD be

A (1, 2), B (-5, 6), C (7, -4) and D (k, -2)

Join AC

**Question 4.**

The vertices of ∆ABC are (-2, 1), (5, 4) and (2, -3) respectively. Find the area of the triangle and the length of the altitude through A.

**Solution:**

Vertices of ∆ABC are A (-2, 1), B (5, 4) and C (2, -3) and AD ⊥ BC, let AD = h

Now area of ∆ABC

**Question 5.**

Show that the following sets of points are collinear

(a) (2, 5), (4, 6) and (8, 8)

(b) (1, -1), (2, 1) and (4, 5)

**Solution:**

We know that points are collinear if the area of the triangle formed by them is zero

(a) Vertices of ∆ABC are (2, 5), (4, 6) and (8, 8)

**Question 6.**

Find the area of a quadrilateral ABCD, the coordinates of whose varities are A (-3, 2), B (5, 4), C (7, -6) and D (-5, -4). **[CBSE 2016]**

**Solution:**

Area of quadrilateral ABCD

= area of ∆ABC + area of ∆ACD

**Question 7.**

In ∆ABC, the coordinates of vertex A are (0, -1) and D (1, 0) and E (0, 1) respectively the mid-points of the sides AB and AC. If F is the mid-point of side C, find the area of ∆DEF. **[CBSE 2016]**

**Solution:**

Let B (p, q), C (r, s) and F (x, y)

Mid-point of AB = Coordinates of D

**Question 8.**

Find the area of the triangle PQR with Q (3, 2) and the mid-points of the sides through Q being (2, -1) and (1, 2). **[CBSE 2015]**

**Solution:**

In ∆PQR, L and N are mid points of QR and QP respectively coordinates of Q are (3, 2) of L are (2, -1) and of N are (1, 2)

**Question 9.**

If P (-5, -3), Q (-4, -6), R (2, -3) and S (1, 2) are the vertices of a quadrilateral PQRS, find its area. **[CBSE 2015]**

**Solution:**

P (-5, -3), Q (-4, -6), R (2, -3) and S (1,2) are the vertices of a quadrilateral PQRS

Join PR which forms two triangles PQR and PSR

**Question 10.**

If A (-3, 5), B (-2, -7), C (1, -8) and D (6, 3) are the vertices of a quadrilateral ABCD, find its area. **[CBSE 2014]**

**Solution:**

A (-3, 5), B (-2, -7), C (1,-8) and D (6, 3) are the vertices of a quadrilateral ABCD

Join AC

**Question 11.**

For what value of ‘a’ the points (a, 1), (1, -1) and (11, 4) are collinear?

**Solution:**

Let the vertices of ∆ABC are A (a, 1), B (1, -1) and C (11, 4)

**Question 12.**

Prove that the points (a, b), (a_{1}, b_{1}) and (a – a_{1}, b – b_{1}) are collinear if ab_{1} = a_{1}b.

**Solution:**

**Question 13.**

If the vertices of a triangle are (1, -3), (4, p) and (-9, 7) and its area is 15 sq. units, find the value(s) of p. **[CBSE 2012]**

**Solution:**

The vertices of a triangle are (1, -3), (4, p) and (-9, 7) and area of triangle = 15 sq. units

**Question 14.**

If (x, y) be on the line joining the two points (1, -3) and (-4, 2), prove that x + y + 2 = 0.

**Solution:**

Point (x, y) be on the line joining the two points (1, -3) and (-4, 2)

Points (x, y), (1, -3) and (-4, 2) are collinear

Let the points (x, y) (1, -3) and (-4, 2) are the vertices of a triangle, then

**Question 15.**

Find the value of k if points (k, 3), (6, -2) and (-3, 4) are collinear. **[CBSE 2008]**

**Solution:**

Let the points (k, 3), (6, -2) and (-3, 4) be the vertices of a triangle, then

**Question 16.**

Find the value of k, if the points A (7, -2), B (5, 1) and C (3, 2k) are collinear. **[CBSE 2010]**

**Solution:**

Points A (7, -2), B (5, 1) and C (3, 2k) are collinear

area of ∆ABC = 0

**Question 17.**

If the point P (m, 3) lies on the line segment joining the points A (\(\frac { -2 }{ 5 }\) , 6) and B (2, 8), find the value of m.

**Solution:**

**Question 18.**

If R (x, y) is a point on the line segment joining the points P (a, b) and Q (b, a), then prove that x + y = a + b. **[CBSE 2010]**

**Solution:**

Point R (x, y) lies on the line segment joining the points P (a, b) and Q (b, a)

Area of ∆PRQ = 0

**Question 19.**

Find the value of k, if the points A (8, 1), B (3, -4) and C (2, k) are collinear. **[CBSE 2010]**

**Solution:**

The points A (8, 1), B (3, -4) and C (2, k) are collinear

Area of ∆ABC = 0

**Question 20.**

Find the value of a for which the area of the triangle formed by the points A (a, 2a), B (-2, 6) and C (3, 1) is 10 square units.

**Solution:**

**Question 21.**

If a ≠ b ≠ 0, prove that the points (a, a²), (b, b²), (0, 0) are never collinear. **[CBSE 2017]**

**Solution:**

**Question 22.**

The area of a triangle is 5 sq. units. Two of its vertices are at (2, 1) and (3, -2). If the third vertex is (\(\frac { 7 }{ 2 }\) , y), find y. **[CBSE 2017]**

**Solution:**

**Question 23.**

Prove that the points (a, 0), (0, b) and (1, 1) are collinear if, \(\frac { 1 }{ a }\) + \(\frac { 1 }{ b }\) = 1.

**Solution:**

Let the points are A (a, 0), B (0, b) and C (1, 1) which form a triangle

**Question 24.**

The point A divides the join of P (-5, 1) and Q (3, 5) in the ratio k : 1. Find the two values of k for which the area of ∆ABC where B is (1, 5) and C (7, -2) is equal to 2 units.

**Solution:**

Let the coordinates of A be (x, y) which divides the join of P (-5, 1) and Q (3, 5) in the ratio. Then

**Question 25.**

The area of a triangle is 5. Two of its vertices are (2, 1) and (3, -2). The third vertex lies on y = x + 3. Find the third vertex.

**Solution:**

Let the coordinates of third vertex of the triangle be (x, y) and other two vertices are (2, 1) and (3, 2)

**Question 26.**

If a ≠ b ≠ c, prove that the points (a, a²), (b, b²), (c, c²) can never be collinear.

**Solution:**

**Question 27.**

Four points A (6, 3), B (-3, 5), C (4, -2) and D (x, 3x) are given in such a way that \(\frac { \triangle DBC }{ \triangle ABC } =\frac { 1 }{ 2 }\) , find x?

**Solution:**

Let A (6, 3), B (-3, 5), C (4, -2) and D (x, 3x) are the vertices of quadrilateral ABCD

AC and BD are joined

**Question 28.**

If three points (x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3}) lie on the same line, prove that

**Solution:**

Let the points (x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3}) are the vertices of a triangle

**Question 29.**

Find the area of a parallelogram ABCD if three of its vertices are A (2, 4), B (2 + √3, 5) and C (2, 6). **[CBSE 2013]**

**Solution:**

Three vertices of a ||gm ABCD are A (2, 4), B (2 + √3 , 5) and C (2, 6).

Draw one diagonal AC of ||gm ABCD

**Question 30.**

Find the value (s) of k for which the points (3k – 1, k – 2), (k, k – 7) and (k – 1, -k – 2) are collinear.** [CBSE 2014]**

**Solution:**

**Question 31.**

If the points A (-1, -4), B (b, c) and C (5, -1) are collinear and 2b + c = 4, find the values of b and c. **[CBSE 2014]**

**Solution:**

**Question 32.**

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 6. **[CBSE 2014]**

**Solution:**

Points A (-2, 1), B (a, b) and C (4, -1) are

collinear if area ∆ABC = 0

Now area of ∆ABC

**Question 33.**

If the points A (1, -2), B (2, 3), C (a, 2) and D (-4, -3) form a parallelogram, find the value of a and height of the parallelogram taking AB as base. **[NCERT Exemplar]**

**Solution:**

In parallelogram, we know that, diagonals bisects each other

i.e., mid-point of AC = mid-point of BD

**Question 34.**

A (6, 1), B (8, 2) and C (9, 4) are three vertices of a parallelogram ABCD. If E is the mid-point of DC, find the area of ∆ADE. **[NCERT Exemplar]**

**Solution:**

Given that, A (6,1), B (8,2) and C (9,4) are three vertices of a parallelogram ABCD.

Let the fourth vertex of parallelogram be (x, y).

We know that, the diagonal of a parallelogram bisect each other.

**Question 35.**

If D (\(\frac { -1 }{ 2 }\), \(\frac { 5 }{ 2 }\)) E (7, 3) and F (\(\frac { 7 }{ 2 }\), \(\frac { 7 }{ 2 }\)) are the mid-points of sides of ∆ABC, find the area of ∆ABC. **[NCERT Exemplar]**

**Solution:**

Hope given RD Sharma Class 10 Solutions Chapter 6 Co-ordinate Geometry Ex 6.5 are helpful to complete your math homework.

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