# Class 10 RD Sharma Solutions – Chapter 14 Coordinate Geometry – Exercise 14.5 | Set 3

### Question 23. Prove that the points (a, 0), (0, b) and (1, 1) are collinear if, 1/a + 1/b = 1.

**Solution:**

Let us assume that the points are A (a, 0), B (0, b) and C (1, 1) form a triangle ABC

Area of âˆ†ABC = 1/2[x

_{1}(y_{2 }– y_{3}) + (y_{3 }– y_{1})x_{2 }+ x_{3}(y_{1 }– y_{2})]= 1/2[a(b – 1) + 0(1 – 0) + 1(0 – b)]

= 1/2[ab – a + 0 + 1(-b)]

= 1/2[ab – a – b]

If the points are collinear, then area of âˆ†ABC = 0

â‡’1/2(ab – a – b) = 0

â‡’ab – a – b = 0

â‡’ ab = a + b

Now on dividing by ab, we get

â‡’

### Question 24. 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 âˆ†ABC where B is (1, 5) and C (7, -2) is equal to 2 units.

**Solution:**

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

So, the coordinates of A will be

Or or

It is given that area of âˆ†ABC is 2 units and vertices are B(1, 5), C(7, -2) and A are the vertices

So, Area of âˆ†ABC = 1/2[x

_{1}(y_{2 }– y_{3}) + (y_{3 }– y_{1})x_{2 }+ x_{3}(y_{1 }– y_{2})]â‡’

â‡’

(i) (14k – 66)(k + 1) = 4 â‡’ 14k – 66 = 4k + 4

â‡’ 14k – 4k = 66 + 4

â‡’10k = 70

â‡’ k = 70/10 = 7

(ii) (14k – 66)(k + 1) = -4 â‡’ 14 – 66 = -4k – 4

â‡’ 14k + 4k = -4 + 66 â‡’ 18k = 62

â‡’ k = 62/18 = 31/9

Hence, the value of k is 7, 31/9

### 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 us considered ABC is a triangle whose vertices are (2, 1), (3, -2) and (x, y)

Also the area of triangle ABC is 5

So,

Area of âˆ†ABC = 1/2[x

_{1}(y_{2 }– y_{3}) + (y_{3 }– y_{1})x_{2 }+ x_{3}(y_{1 }– y_{2})]5 = 1/2[x(1 + 2) + 2(-2 – y) + 3(y – 1)]

10 = [3x-4-2y+3y-3]

10 = 3x + y – 7

3x + y = 10 + 7

3x + y = 17

But it is given that the point (x, y) lies on y = x + 3

So, 3x + x + 3 = 17

â‡’ 4x = 17 – 3

4x = 14

â‡’ x = 7/2

and y = x + 3 = 7/2 +3 = 13/2

Hence, the third vertex is (7/2, 13/2)

### Question 26. Four points A (6, 3), B (-3, 5), C (4, -2) and D (x, 3x) are given in such a way that , find x?

**Solution:**

Let us assume that ABCD is a quadrilateral whose vertices are A (6, 3), B (-3, 5), C (4, -2) and D (x, 3x)

Now, AC and BD are joined

Area of triangle = 1/2[x

_{1}(y_{2 }– y_{3}) + (y_{3 }– y_{1})x_{2 }+ x_{3}(y_{1 }– y_{2})]= 1/2 [6(5 + 2) + (-3)(-2 – 3) + 4(3 – 5)]

= 1/2 [6 * 7 + (-3)(-5) + 4(-2)]

= 1/2 [42 + 15 – 8] = 49/2

and area of â–³DBC,

= 1/2[x(5 + 2) + (-3)(-2 – 3x) + 4(3x – 5)]

= 1/2[7x + 6 + 9x + 12x – 20]

= 1/2[28x – 14] = 14x – 7

Now,[Tex]=\frac{14x-7}{\frac{49}{2}} [/Tex]

If 56x – 28 = 49

â‡’ 56x = 49 + 28 = 77

â‡’ x = 77/56 = 11/8

If 4(14x – 7) = 49

â‡’56x – 28 = -49

â‡’56x = -49 + 28 = -21

â‡’x = -21/56 = -3/8

x = 11/8 or -3/8

### Question 27. If three points (x_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3}) lie on the same line, prove that

**Solution:**

Let us assume that ABC is a triangle whose vertices are (x

_{1}, y_{1}), (x_{2}, y_{2}), (x_{3}, y_{3})Area of âˆ†ABC = 1/2[x

_{1}(y_{2 }– y_{3}) + (y_{3 }– y_{1})x_{2 }+ x_{3}(y_{1 }– y_{2})]The points be on the same line, so

[x

_{1}(y_{2 }– y_{3}) + (y_{3 }– y_{1})x_{2 }+ x_{3}(y_{1 }– y_{2})] = 0On dividing by x

_{1 }x_{2 }x_{3}, we get

### Question 28. Find the area of a parallelogram ABCD if three of its vertices are A (2, 4), B (2 + âˆš3, 5) and C (2, 6).

**Solution:**

Given that ABCD is a ||gm whose vertices are A (2, 4), B (2 + âˆš3, 5) and C (2, 6).

Now draw one diagonal AC of ||gm ABCD

Diagonal bisects the ||gm into two triangle equal in area

_{1}(y_{2 }– y_{3}) + (y_{3 }– y_{1})x_{2 }+ x_{3}(y_{1 }– y_{2})]= 1/2 [2(5 – 6) + (2 + âˆš3)(6 – 4) + 2(4 – 5)]

= 1/2 [2 * (-1) + (2 + âˆš3) * 2 + 2 * (-1)]

= 1/2 [-2 + 4 + 2âˆš3 – 2] = 1/2 (2âˆš3) = âˆš3sq.units

So, the area of ||gm ABCD = 2 * area(â–³ABC)

= 2 * (âˆš3) = 2âˆš3 sq.units

### Question 29. Find the value (s) of k for which the points (3k â€“ 1, k â€“ 2), (k, k â€“ 7), and (k â€“ 1, -k â€“ 2) are collinear.

**Solution:**

Let us assume that the ABC is a triangle whose vertices are A(3k â€“ 1, k â€“ 2), B(k, k â€“ 7) and C(k â€“ 1, -k â€“ 2)

and A, B, C are collinear

_{1}(y_{2 }– y_{3}) + (y_{3 }– y_{1})x_{2 }+ x_{3}(y_{1 }– y_{2})]= 1/2[(3k – 1)(k – 7 + k + 2) + k(-k – 2 – k + 2) + (k – 1)(k – 2 – k + 7)]

= 1/2 [(3k – 1)(2k – 5) + k(-2k + (k – 1) * 5)]

= 1/2 [6k

^{2 }– 15k – 2k + 5 – 2k^{2 }+ 5k – 5]= 1/2[4k

^{2 }– 12k] = 2k^{2 }– 6k= 2k(k – 3)

As we know that the points are collinear

so, Area of ABC=0

2k(k – 3) = 0

Either 2k = 0 or k – 3 = 0, then k = 0 or k = 3

Therefore, k = 0, 3

### Question 30. 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.

**Solution:**

Let us assume that the ABC is a triangle whose vertices are A (-1, -4), B (b, c) and C (5, -1)

and A, B, C are collinear

_{1}(y_{2 }– y_{3}) + (y_{3 }– y_{1})x_{2 }+ x_{3}(y_{1 }– y_{2})]= 1/2 [-1(c + 1) + b(-1 + 4) + 5(-4 – c)]

= 1/2 [-c – 1 – b + 4b – 20 – 5c]

= 1/2 [3b – 6c – 21]

As we know that the points are collinear

so, Area of triangle ABC = 0

1/2(2b – 6c – 21) = 0

â‡’3b – 6c = 21

3b – 6c = 21 ———–(i)

and 2b + c = 4 ———-(ii)

â‡’c = 4 – 2b

On substituting the value of c in eq (i), we get

â‡’3b – 6(4 – 2b) = 21

â‡’3b – 24 + 12b = 21

15b = 21 + 24 = 45

â‡’b = 45/15 = 3

c = 4 – 2b = 4 – 2 * 3 = 4 – 6 = -2

b = 3, c = -2

### Question 31. If the points A (-2, 1), B (a, b), and C (4, -1) are collinear and a â€“ b = 1, find the values of a and 6.

**Solution:**

Let us assume that the ABC is a triangle whose vertices are A (-2, 1), B (a, b), and C (4, -1)

and A, B, C are collinear

_{1}(y_{2 }– y_{3}) + (y_{3 }– y_{1})x_{2 }+ x_{3}(y_{1 }– y_{2})]= 1/2 [-2(b + 1) + a(-1 – 1) + 4(1 – b)]

= 1/2[-2b – 2 – 2a + 4 – 4b]

= 1/2 [-2b – 2 – 2a + 4 – 4b]

= 1/2 [-2a – 6b + 2] = -a – 3b + 1

As we know that the points are collinear

so, Area of triangle ABC = 0

-a – 3b + 1 = 0

a + 3b = 1 ————(i)

a = 1 – 3b

and given that a – b = 1 ————(ii)

On solving eq(i) and (ii), we get

4b = 0 â‡’ b = 0

Therefore, a = 1 – 3b = 1 – 3 * 0 = 1 – 0 = 1

a = 1, b = 0

### Question 32. 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 a base.

**Solution:**

Given that ABCD is a parallelogram whose vertices are A (1, -2), B (2, 3), C (a, 2), and D (-4, -3)

As we know that, diagonals bisects each other

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

â‡’

= 1/2[x

_{1}(y_{2 }– y_{3}) + (y_{3 }– y_{1})x_{2 }+ x_{3}(y_{1 }– y_{2})]â‡’1 + a = -2

â‡’a = -3

So, the required value of a is -3

Given that, AB as base of a ||gm and drawn a perpendicular from A to AB

which meet AB at P. So, DP is a height of a ||gm.

Now, equation of base AB, passing through the points (1, -2) and (2, 3) is

â‡’

â‡’(y + 2)=

â‡’(y + 2) = 5(x – 1)

â‡’5x – y = 7 ————(i)

So, the slope of AB, say

Let the slope of DP be m

_{2}.Since, DP is perpendicular to AB.

By condition of perpendicularity

m

_{1 }* m_{2 }= -1 â‡’ 5 * m_{2 }= -1â‡’m

_{2 }= -1/5Now, eq. of DP, having slope (-1/5)

and passing the point (-4,-3) is (y – y

_{1}) = m_{2}(x – x_{1})â‡’

â‡’5y + 15 = -x – 4

â‡’x + 5y = -19 ————-(ii)

On adding eq (i) and (ii), then we get the intersection point P.

Now put the value of y from eq. (i) in eq. (ii), we get

x + 5(5x – 7) = -19 [using Eq. (i)]

â‡’x + 25x – 35 = -19

â‡’26x = 16

x = 8/13

Now put the value of x in Eq.(i), we get

So, the coordinates of point P

And, the length of the height of a parallelogram,

Now, by using distance formula, we get

Hence, the required length of height of a parallelogram is 12âˆš26/13

### Question 33. 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 âˆ†ADE.

**Solution:**

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

Let us assume that the fourth vertex of parallelogram be D(x, y).

Now as we know that, the diagonal of a parallelogram bisect each other.

so, Mid point of BD = Mid point of AC

â‡’

â‡’

â‡’8+x=15â‡’x=7

and

â‡’2 + y = 5 â‡’ y = 3

So, fourth vertex of ||gm is D(7, 3)

Now, mid-point of side DC

E = (8, 7/2)

Now we find the area of âˆ†ADE

= 1/2[6(3 – 7/2) + 7(7/2 – 1) + 8(1 – 3)]

= 1/2 [6 * (-1/2 + 7(5/2 + 8(-2)]

= 1/2 (-3 + 35/2 – 16)

= 1/2 ( 35/2 – 19)

= 1/2 (-3/2)

= -3/4 [As we know that area cannot be negative]

Hence, the required area âˆ†ADE is 3/4sq. units

### Question 34. If DE (7, 3) and F are the mid-points of sides of âˆ†ABC, find the area of âˆ†ABC.

**Solution:**

Given that ABC is a triangle, so let us assume that the vertices of triangle ABC are A(x

_{1}, y_{1}), B(x_{2}, y_{2}) and C(x_{3}, y_{3})And the mid points of side BC, CA, and AB are D(-1/2, 5/2), E(7, 3), and F(7/2, 7/2)

Since, D(-1/2, 5/2) is the mid-point of BC.

So,

and

â‡’x

_{2 }+ x_{3 }= -1 ————(i)and y

_{2 }+ y_{3 }= 5As we know that E(7, 3) is the mid-point of CA.

So, and

â‡’x

_{3 }+ x_{1 }= 14 ———–(iii)and y

_{3 }+ y_{1 }= 6 ————(iv)Also, F(7/2, 7/2) is the mid-point of AB

and

â‡’x

_{1 }+ x_{2 }= 7 ————–(v)and y

_{1 }+ y_{2 }= 7 —————-(vi)Now on adding Eq.(i), (ii) and (v), we get

2(x

_{1 }+ x_{2 }+ x_{3}) = 20x

_{1 }+ x_{2 }+ x_{3 }= 10 —————(vii)On subtracting Eq. (i) (iii) and (v) from Eq. (vii), we get

x

_{1 }= 11, x_{2 }= -4, x_{3 }= 3On adding Eq.(ii), (iv) and (vi), we get

2(y

_{1 }+ y_{2 }+ y_{3}) = 18â‡’y

_{1 }+ y_{2 }+ y_{3}= 9 ————-(viii)On subtracting Eq. (ii), (iv) and (vi) from Eq. (viii), we get

y

_{1 }= 4, y_{2 }= 3, y_{3 }= 9Hence, the vertices of âˆ†ABC are A(11, 4), B(-4, 3) and C(3, 2)

_{1}(y_{2 }– y_{3}) + (y_{3 }– y_{1})x_{2 }+ x_{3}(y_{1 }– y_{2})]= 1/2[11(3 – 2) + (-4)(2 – 4) + 3(4 – 3)]

= 1/2[11 * 1 + (-4)(-2) + 3(1)]

= 1/2(11 + 8 + 3)

= 22/2 = 11

Hence, the required area of âˆ†ABC is 11 sq.unit

## Please

Loginto comment...