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Class 10 RD Sharma Solutions – Chapter 14 Coordinate Geometry – Exercise 14.4

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  • Last Updated : 30 Nov, 2022
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Question 1. Find the centroid of the triangle whose vertices are:

(i) (1, 4), (-1, -1), and (3, -2)

Solution:

Given, vertices of triangle are (1, 4), (-1, -1), and (3, -2)

As we know that the coordinates of the centroid of a triangle whose vertices are (x1, y1), (x2, y2), (x3, y3)

(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})

So, the coordinates of the centroid of a triangle are(\frac{1-1+3}{3},\frac{4-1-2}{3})

= (1, 1/3)        

Hence, the centroid of the triangle whose vertices are (1, 4), (-1, -1), and (3, -2) is (1, 1/3).       

(i) (-2, 3), (2, -1), and (4, 0)

Solution:

Given, vertices of triangle are (-2, 3), (2, -1), and (4, 0)

As we know that the coordinates of the centroid of a triangle whose vertices are (x1, y1), (x2, y2), (x3, y3)

(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})

So, the coordinates of the centroid of a triangle are(\frac{-2+2+4}{3},\frac{3-1+0}{3})

= (4/3, 2/3)        

Hence, the centroid of the triangle whose vertices are (-2, 3), (2, -1), and (4, 0) is (4/3, 2/3).  

Question 2. Two vertices of a triangle are(1, 2), (3, 5), and its centroid is at the origin. Find the coordinates of the third vertex.

Solution:

According to question 

Two vertices of a triangle are(1, 2), (3, 5), and its centroid is at the origin

Find: the coordinates of the third vertex

So, let us assume that the coordinates of the third vertex is (x, y), 

So, the coordinates of the centroid of a triangle are (\frac{x+1+3}{3},\frac{y+2+5}{3})

 As we know that the centroid is at origin. So, 

(x + 1 + 3)/3 = 0

x + 4 = 0

x = -4

(y + 2 + 5)/3 = 0

y + 7 = 0

y = -7

Hence, the third coordinate is (-4, -7)

Question 3. Find the third vertex of a triangle, If two of Its vertices are at (−3, 1) and (0, −2) and the centroid is at the origin.

Solution:

According to the question

ABC is a triangle in which the coordinates of A are (−3, 1), B are (0, −2), and C are (a, b)

and the centroid of the triangle ABC is (0, 0)

so, 

(-3 + 0 + a)/3 = 0

a = 3

(1 – 2 + b)/3 = 0

b = 1

Hence, the coordinates of third vertex is (3, 1)

Question 4. A (3, 2) and B (−2, 1) are two vertices of a triangle ABC whose centroid G has the coordinates (5/3, -1/3). Find the coordinates of the third vertex C of the triangle.

Solution:

According to the question

ABC is a triangle in which the coordinates of A are (3, 2), B are (−2, 1), and C are (a, b)

and the centroid of the triangle ABC is (5/3, -1/3)

so, 

(3 – 2 + a)/3 = 5/3

a = 4

(1 + 2 + b)/3 = -1/3

b = -4

Hence, the coordinates of third vertex is (4, -4)

Question 5. If (−2, 3), (4, −3), and (4, 5) are the mid-points of the sides of a triangle, Find the coordinates of Its centroid.

Solution:

Let us assume ABC is a triangle 

Now in this triangle P, Q, R be the mid point of side, AC, AB, BC. 

The coordinates of P(−2, 3), Q (4, −3), and R(4, 5) 

Let us assume that the coordinates of A, B, and C are (x1, y1), (x2, y2), and (x3, y3)

As we know that P is the mid point of side AC, 

So, 

(x1 + x3) /2 = -2 

= x1 + x3 = -4 …..(1)

(y1 + y3) /2 = 3

= y1 + y3 = 6 …..(2)

As we know that Q is the mid point of side AB, 

So, 

(x1 + x2) /2 = 4 

= x1 + x2 = 8 …..(3)

(y1 + y2) /2 = -3

= y1 + y2 = -6 …..(4)

As we know that R is the mid point of side BC, 

So, 

(x2 + x3) /2 = 4 

= x2 + x3 = 8 …..(5)

(y2 + y3) /2 = 5

= y2 + y3 = 10 …..(6)

Now add c

2(x1 + x2 + x3) = -4 + 8 + 8 = 12  …..( 7)

x1 + x2 + x3 = 6

Now subtract eq(1), (3), (5) from eq(7), we get

x1 = 10

x2 = -2

x3 = -2

Similarly, add (2), (4), and (6), we get

2(y1 + y2 + y3) = -6 + 6 + 10 = 10 

y1 + y2 + y3 = 5   ……(8)

Now subtract eq(2), (4), (6) from eq(8), we get

y1 = -1

y2 = 11

y3 = -5

so, the coordinate of  ABC triangle are (10, -1), (-2, 11), and (-2, -5)

Hence, the centroid of the triangle is ((10 – 2 – 2)/3, (-1 + 11 – 5)/3)  = (2, 5/3)

Question 6. Prove analytically that the middle points of two sides of a triangle is equal to half to the third side.

Solution:

In triangle ABC,

D and E are the mid-points of the sides AB and AC 

The DE = 1/2 BC 

Let the coordinates of the vertices of a ∆ABC be (x1, y2), B(x2, y2), and C(x3, y3)

Then coordinates of D will be

(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})

and coordinates of E will be

(\frac{x_1+x_3}{2},\frac{y_1+y_3}{2})

Length of BC=\sqrt{(x_2-x_3)^2+(y_2-y_3)^2} ……(1)

and the length of DE

=\sqrt{(\frac{x_1+x_2}{2}-\frac{x_1+x_3}{2})^2+(\frac{y_1+y_2}{2}-\frac{y_1+y_3}{2})^2}\\ =\sqrt{(\frac{x_1+x_2-x_1-x_3}{2})^2+(\frac{y_1+y_2-y_1-y_3}{2})^2}\\ =\sqrt{\frac{(x_2-x_3}{2})^2+(\frac{y_2-y_3}{2})^2}\\ =\frac{1}{2}\sqrt{(x_2-x_3)^2+(y_2-y_3)^2}

= 1/2 BC 

Hence Proved

Question 7. Prove that the lines joining the middle points of the opposite sides of a quadrilateral and the join of the points of its diagonals meet in a point and bisect another.

Solution:

Let us assume the vertices of ABCD quadrilateral are A (x1, y1), B (x2, y2), C (x3, y3) and D (x4, y4) 

In this quadrilateral, E and F are the mid-points of side BC and AD 

and EF is joined G and H are the mid-points of diagonal AC and BD.

GH are joined

Now co-ordinates of E will be \frac{x_2+x_3}{2},\frac{y_2+y_3}{2}

and co-ordinates of F will be \frac{x_1+x_4}{2},\frac{y_1+y_4}{2}

Co-ordinates of G will be (\frac{x_1+x_3}{2},\frac{y_1+y_3}{2})

Co-ordinates of H will be(\frac{x_2+x_4}{2},\frac{y_2+y_4}{2})

Here, EF and GH intersect each other at M.

So, let M is the midpoint of EF, then its coordinates will be

(\frac{x_1+x_2+x_3+x_4}{4},\frac{y_1+y_2+y_3+y_4}{4})

Let M is the midpoint of GH, then its coordinates of M will be

(\frac{x_1+x_2+x_3+x_4}{4},\frac{y_1+y_2+y_3+y_4}{4})

Here, we conclude that the coordinates of in both cases are same

Therefore, EF and GH bisect each other at M

Question 8. If G be the centroid of a triangle ABC and P be any other point in the plane, prove that PA2 + PB2 + PC2 = GA2 + GB2 + GC2 + 3GP.

Solution:

According to the question 

ABC is a triangle, G is the centroid of it let P(h, x) is any point in the plane.

We have to prove that PA2 + PB2 + PC2 = GA2 + GB2 + GC2 + 3GP

Proof: 

In triangle ABC, 

Let us assume the coordinates of are(x1, y1) of B are(x2, y2), and of C are(x3, y3)

Therefore, co-ordinates of centroid G will be (u, v)

Where u = (x1 + x2 + 3)/3 and v = (y1 + y2 + 3)/3 

Now, we shall find L.H.S and R.H.S. separately

PA2 + PB2 + PC2 = (h – x1)2 + (k – y1)2 + (h – x2)2 + (k – y2)2 + (h – x3)2 + (k – y3)2

=3(h^2+k^2)+(x^2_1+x^2_2+x_3^2)+(y_1^2+y^2_2+y_3^2)-2h(x_1+x_2+x_3)-2k(y_1+y_2+y_3)\\ =3(h^2+k^2)+(x_1^2+x_2^2+x_3^2)+(y_1^2+y_2^2+y_3^2)-2h(3u)-2k(3v)\\ =3(h^2+k^2)-6hu-6kv+(x_1^2+x_2^2+x_3^2)+(x_1^2+x_2^2+x_3^2)GA^2+GB^2+GC^2+3GD^2\\ =(u-x_1)^2+(v-y_1)^2+(u-x_2)^2+(v-y_2)^2+(u-x_3)^2+(v-y_3)^2+3[(u-h)^2+(v-k)^2]\\

=3(u^2+v^2)+(x_1^2+y_1^2+x_2^2+y_2^2+x_3^2+y_3^2)-2u(x_1+x_2+x_3)-2v(y_1+y_2+y_3)+3(u^2+h^2-2uh+v^2+k^2-2vk)\\ =6(u^2+v^2)+(x_1^2+y_1^2+x_2^2+y_2^2+x_3^2+y_3^2)-2u(3u)-2v(3v)+3(h^2+k^2)-6uh-6vk\\ =(x_1^2+x^2_2+x_3^2)+(y_1^2+y^2_2+y^2_3)+3(h^2+k^2)-6uh-6vk\\

Hence proved

Question 9. If G be the centroid of triangle ABC, prove that AB2 + BC2 + CA2 = 3(GA2 + GB2 + GC2).

Solution:

According to the question 

ABC is a triangle and G be the centroid of this triangle

We have to prove that AB2 + BC2 + CA2 = 3(GA2 + GB2 + GC2)

Proof:

Let us assume that the coordinates of the vertices of ∆ABC be 

A(x1, y1), B(x2, y2), and C(x3, y3) and let G be the centroid of the triangle

Therefore, the coordinates of G will be

(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})

Now L.H.S.= AB2 + BC2 + CA2

= (x1 – x2)2 + (y1 – y2)2 + (x2 – x3)2 + (y2 – y3)2 + (x3 – x1)2 + (y3 – y1)2

=x_1^2+x^2_2-2x_1x_2+x_2^2+x_3^2-2x_2x_3+x_3^2+x_1^2-2x_3x_1+y_1^2+y_2^2-2y_1y_2+y^2_2+y^2_3-2y_2y_3+y_3^2+y_1^2-2y_3y\\ =2(x_1^2+x^2_2+x_3^2)-2(x_1x_2+x_2x_3+x_3x_1)+(y_1^2+y^2_2+y^2_3)-2(y_1y_2+y_2y_3+y_3y_2)\\

R.H.S = 3[GA2 + GB2 + GC2]

=3[(x_1-\frac{x_1+x_2+x_3}{3})^2+(y_1-\frac{y_1+y_2+y_3}{3})^2+(x_2-\frac{x_1+x_2+x_3}{3})^2+(y2-\frac{y_1+y_2+y_3}{3})^2+(x_3-\frac{x_1+x_2+x_3}{3})^3+(y_3-\frac{y_1+y_2+y_3}{3})]

=3[\frac{(2x_1-x_2-x_3}{3})^2+(\frac{2y_1+y_2-y_3}{3})^2+(\frac{2x_2-x_1-x_3}{3})^2+(\frac{2y_2-y_1-y_3}{3})^2+(\frac{2y_3-x_1-x_2}{3})^2+(\frac{2y_3-y_2-y_3}{3})^2]

=\frac{1}{3}[6x_1^2+6x_2^2-6x_1^2-6x_1x_2-6x_2x_3-6x_3x_1+6y_1^2+6y_2^2+6y_3^2-6y_1y_2-6y_2y_3-6y_3y_1]\\ =2[(x_1^2+x_2^2+x_3^2)-2(x_1x_2+x_2x_3+x_3x_1)+2(y_1^2+y_2^2+y_3^2+y_3^2+y_3^2)-2(y_1y_2+y_2y_3+y_3y_1)]

= L.H.S

Hence proved

Question 10. In the figure, a right triangle BOA is given. C is the mid-point of the hypotenuse AB. Show that it is equidistant from the vertices O, A, and B.

Solution:

In right triangle OAB, 

Co-ordinates of O are (0, 0), A are(2a, 0), and of B are(0, 2b)

C is the midpoint of AB. So, the co-ordinates of C will be

(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})

or 

(\frac{2a+0}{2},\frac{0+2b}{2})

Now CO = \sqrt{(a+0)^2+(b+0)^2}=\sqrt{a^2+b^2}

CA = \sqrt{(2a-a)^2+(0-b)^2}=\sqrt{(a)^2+(b)^2}\\ =\sqrt{a^2+b^2}

CB = \sqrt{(0-a)^2+(2b-b)^2}\\ =\sqrt{(-a)^2+(b)^2}=\sqrt{a^2+b^2}

So, we can conclude that, CO = CA = CB

Hence, C is equidistant to form the vertices O, A and B.

Hence proved


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