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

  • Last Updated : 16 May, 2021

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 ∆ABC = 1/2[x1(y2 – y3) + (y3 – y1)x2 + x3(y1 – y2)]

= 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 ∆ABC = 0

⇒1/2(ab – a – b) = 0

⇒ab – a – b = 0

⇒ ab = a + b

Now on dividing by ab, we get

\frac{ab}{ab}=\frac{a}{ab}+\frac{b}{ab}

1=\frac{1}{b}+\frac{1}{a}⇒\frac{1}{a}+\frac{1}{b}=1



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 ∆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 (\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n})   

Or (\frac{k*3+1*(-5)}{k+1},\frac{k(5)+1(1)}{k+1}) or (\frac{3k-5}{k+1},\frac{5k+1}{k+1})

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

So, Area of ∆ABC = 1/2[x1(y2 – y3) + (y3 – y1)x2 + x3(y1 – y2)]

2=\frac{1}{2}[\frac{3k-5}{k+1}(5+2)+1(-2-\frac{5k+1}{k+1}+7(\frac{5k+1}{k+1}-5)]

2=\frac{1}{2}[\frac{3k-5}{k+1}*7-2-\frac{5k+1}{k+1}+7(\frac{5k+1}{(k+1)-35})]\\ ⇒4=[\frac{21k-35}{k+1}-\frac{5k+1}{k+1}+\frac{35k+7}{k+1}-37]\\ ⇒4=[\frac{21k-35-5k+1+35k+7-37k-37}{k+1}]\\ ⇒|\frac{14k-66}{k+1}|=±4

(i) (14k – 66)(k + 1) = 4 ⇒ 14k – 66 = 4k + 4

⇒ 14k – 4k = 66 + 4



⇒10k = 70

⇒ k = 70/10 = 7

(ii) (14k – 66)(k + 1) = -4 ⇒ 14 – 66 = -4k – 4

⇒ 14k + 4k = -4 + 66 ⇒ 18k = 62 

⇒ 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 ∆ABC = 1/2[x1(y2 – y3) + (y3 – y1)x2 + x3(y1 – y2)]

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

⇒ 4x = 17 – 3

4x = 14

⇒ 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 \frac{△DBC}{△ABC}=\frac{1}{2}  , 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[x1(y2 – y3) + (y3 – y1)x2 + x3(y1 – y2)]

= 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 △DBC,

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

=\frac{(14x-7)*2}{49}

|\frac{2(14x-7)}{49}|=\frac{1}{2}⇒4|14x-7|=±49

If 56x – 28 = 49

⇒ 56x = 49 + 28 = 77

⇒ x = 77/56 = 11/8

If 4(14x – 7) = 49

⇒56x – 28 = -49

⇒56x = -49 + 28 = -21

⇒x = -21/56 = -3/8

x = 11/8 or -3/8  

Question 27. If three points (x1, y1), (x2, y2), (x3, y3)  lie on the same line, prove that

\frac{y_2-y_3}{x_2x_3}+\frac{y_3-y_1}{x_3x_1}+\frac{y_1-y_2}{x_1x_2}=0

Solution:

Let us assume that ABC is a triangle whose vertices are (x1, y1), (x2, y2), (x3, y3

Area of ∆ABC = 1/2[x1(y2 – y3) + (y3 – y1)x2 + x3(y1 – y2)]

The points be on the same line, so

[x1(y2 – y3) + (y3 – y1)x2 + x3(y1 – y2)] = 0



On dividing by x1 x2 x3, we get

\frac{x_1(y_2-y_1)}{x_1x_2x_3}+\frac{x_2(y_3-y_1)}{x_1x_2x_3}+\frac{x_3(y_1-y_2)}{x_1x_2x_3}=0\\ \frac{y_2-y_3}{x_2x_3}+\frac{y_3-y_1}{x_3x_1}+\frac{y_1-y_2}{x_1x_2}=0

Question 28. Find the area of a parallelogram ABCD if three of its vertices are A (2, 4), B (2 + √3, 5) and C (2, 6). 

Solution:

Given that ABCD is a ||gm whose vertices are A (2, 4), B (2 + √3, 5) and C (2, 6).

Now draw one diagonal AC of ||gm ABCD

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

Area of ∆ABC = 1/2[x1(y2 – y3) + (y3 – y1)x2 + x3(y1 – y2)]

= 1/2 [2(5 – 6) + (2 + √3)(6 – 4) + 2(4 – 5)]

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

= 1/2 [-2 + 4 + 2√3 – 2] = 1/2 (2√3) = √3sq.units

So, the area of ||gm ABCD = 2 * area(△ABC)

= 2 * (√3) = 2√3 sq.units

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

Solution:

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

and A, B, C are collinear

Area of ∆ABC = 1/2[x1(y2 – y3) + (y3 – y1)x2 + x3(y1 – y2)]

= 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 [6k2 – 15k – 2k + 5 – 2k2 + 5k – 5] 

= 1/2[4k2 – 12k] = 2k2 – 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

Area of ∆ABC = 1/2[x1(y2 – y3) + (y3 – y1)x2 + x3(y1 – y2)]

= 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

⇒3b – 6c = 21

3b – 6c = 21             ———–(i)

and 2b + c = 4        ———-(ii)

⇒c = 4 – 2b

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

⇒3b – 6(4 – 2b) = 21

⇒3b – 24 + 12b = 21

15b = 21 + 24 = 45

⇒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 – 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

Area of ∆ABC = 1/2[x1(y2 – y3) + (y3 – y1)x2 + x3(y1 – y2)]

= 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 ⇒ 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



(\frac{1+a}{2},(\frac{-2+2}{2})=(\frac{2-4}{2},\frac{3-3}{2})\\ ⇒\frac{1+a}{2}=\frac{2-4}{2}=\frac{-2}{2}=-1

= 1/2[x1(y2 – y3) + (y3 – y1)x2 + x3(y1 – y2)]

⇒1 + a = -2

⇒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 

(y-y_1)=\frac{y_2-y_1}{x_2-x_1}(x-x_1)

⇒(y + 2)=\frac{3+2}{2-1}(x-1)

⇒(y + 2) = 5(x – 1)    

⇒5x – y = 7             ————(i)

So, the slope of AB, say m_1=\frac{y_2-y_1}{x_2-x_1}=\frac{3+2}{2-1}=5

Let the slope of DP be m2.

Since, DP is perpendicular to AB.

By condition of perpendicularity

m1 * m2 = -1 ⇒ 5 * m2 = -1

⇒m2 = -1/5

Now, eq. of DP, having slope (-1/5)

and passing the point (-4,-3) is (y – y1) = m2(x – x1)



(y+3)=-\frac{1}{5}(x+4)

⇒5y + 15 = -x – 4

⇒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)]

⇒x + 25x – 35 = -19

⇒26x = 16

x = 8/13

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

y=5(\frac{8}{13})-7=\frac{40}{13}-7\\ ⇒y=\frac{40-91}{13}⇒y=\frac{-51}{13}

So, the coordinates of point P=(\frac{8}{13},\frac{-51}{13})

And, the length of the height of a parallelogram,

DP=\sqrt{(\frac{8}{13}+4)^2+(\frac{-51}{13}+3)^2}   

Now, by using distance formula, we get

d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\ ⇒DP=\sqrt{(\frac{60}{13})^2+(\frac{-12}{13})^2}\\ =\frac{1}{13}\sqrt{3600+144}\\ =\frac{1}{13}.\sqrt{3744}=\frac{12\sqrt{26}}{13}

Hence, the required length of height of a parallelogram is 12√26/13   

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 ∆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

(\frac{8+x}{2},\frac{2+y}{2})=(\frac{6+9}{2},\frac{1+4}{2})

(\frac{8+x}{2},\frac{2+y}{2})=(\frac{15}{2},\frac{5}{2})\\ \frac{8+x}{2}=\frac{15}{2}

⇒8+x=15⇒x=7

and \frac{2+y}{2}=\frac{5}{2}

⇒2 + y = 5 ⇒ y = 3

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

Now, mid-point of side DC=(\frac{7+9}{2},\frac{3+4}{2})

E = (8, 7/2)

Now we find the area of ∆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 ∆ADE is 3/4sq. units

Question 34. 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.

Solution:

Given that ABC is a triangle, so let us assume that the vertices of triangle ABC are A(x1, y1), B(x2, y2) and C(x3, y3

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, \frac{x_2+x_3}{2}=-\frac{1}{2}

and \frac{y_2+y_3}{2}=\frac{5}{2}

⇒x2 + x3 = -1          ————(i)

and y2 + y3 = 5

As we know that E(7, 3) is the mid-point of CA.

So, \frac{x_3+x_1}{2}=7 and \frac{y_3+y_1}{2}=3

⇒x3 + x1 = 14                      ———–(iii)



and y3 + y1 = 6                   ————(iv)

Also, F(7/2, 7/2) is the mid-point of AB \frac{x_1+x_2}{2}=\frac{7}{2}

 and \frac{y_1+y_2}{2}=\frac{7}{2}

⇒x1 + x2 = 7             ————–(v)

and y1 + y2 = 7        —————-(vi) 

Now on adding Eq.(i), (ii) and (v), we get

2(x1 + x2 + x3) = 20

x1 + x2 + x3 = 10                    —————(vii)

On subtracting Eq. (i) (iii) and (v) from Eq. (vii), we get

x1 = 11, x2 = -4, x3 = 3

On adding Eq.(ii), (iv) and (vi), we get

2(y1 + y2 + y3) = 18

⇒y1 + y2 + y3 = 9    ————-(viii)

On subtracting Eq. (ii), (iv) and (vi) from Eq. (viii), we get

y1 = 4, y2 = 3, y3 = 9

Hence, the vertices of ∆ABC are A(11, 4), B(-4, 3) and C(3, 2)

Area of ∆ABC = 1/2[x1(y2 – y3) + (y3 – y1)x2 + x3(y1 – y2)]

= 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 ∆ABC is 11 sq.unit

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