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ML Aggarwal Solutions Class 10 Mathematics Solutions for Section Formula Exercise 11 in Chapter 11 - Section Formula

Question 49 Section Formula Exercise 11

A (10, 5), B (6, – 3) and C (2, 1) are the vertices of triangle ABC. L is the mid point of AB, M is the mid-point of AC.

Write down the co-ordinates of L and M. Show that LM = \frac{1}{2} BC.

Answer:

Solution:

The midpoint is determined by taking the total of the x- and y-coordinates of the two points and dividing it in half.

Given points are A(10,5), B(6,-3) and C(2,1).

Let L(x,y) be the midpoint of AB.

Here x1= 10, y1 = 5

x2 = 6, y2 = -3

\text{By midpoint formula, }x = \frac{(x_1+x_2)}{2}\\ x = \frac{(10+6)}{2} = \frac{16}{2} = 8\\ \text{By midpoint formula, }y = \frac{(y_1+y_2)}{2}\\ y = \frac{(5-3)}{2} = \frac{2}{2} = 1

So co-ordinates of L are (8,1).

Let M(x,y) be the midpoint of AC.

Here x1= 10, y1 = 5

x2 = 2, y2 = 1

\text{By midpoint formula, }x = \frac{(x_1+x_2)}{2}\\ x = \frac{(10+2)}{2} = \frac{12}{2} = 6\\ \text{By midpoint formula, }y = \frac{(y_1+y_2)}{2}\\ y = \frac{(5+1)}{2} = \frac{6}{2} = 3\\ \text{So, co-ordinates of }M\text{ are }(6,3).\\ \text{By distance formula, }d(LM) = \sqrt{[(x_2-x_1)^2+(y_2-y_1)^2]}\\ \text{The points are }L(8,1)\text{ and }M(6,3).\\ \text{So, }x_1= 8, y_1= 1\\ x_2= 6, y_2 = 3\\ d(LM) = \sqrt{[(x_2-x_1)^2+(y_2-y_1)^2]}\\ d(LM) = \sqrt{[(6-8)^2+(3-1)^2]}\\ d(LM) = \sqrt{[(-2)^2+(2)^2]}\\ d(LM) = \sqrt{(4+4)}\\ d(LM) = \sqrt{8} = 2\sqrt{2} …(i)\\ \text{By distance formula, }d(BC) = \sqrt{[(x_2-x_1)^2+(y_2-y_1)^2]}\\ \text{The points are }B(6,-3)\text{ and }C(2,1).\\ \text{So, }x_1= 6, y_1 = -3\\ x_2 = 2, y_2 = 1\\ d(BC) = \sqrt{[(x_2-x_1)^2+(y_2-y_1)^2]}\\ d(BC) = \sqrt{[(2-6)^2+(1-(-3))^2]}\\ d(BC) = \sqrt{[(-4)^2+(4)^2]}\\ d(BC) = \sqrt{(16+16)}\\ d(BC) = \sqrt{32} = 4\sqrt{2} …(ii)\\ \text{From (i) and (ii), }LM = \frac{1}{2} BC.

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