In the following examples, can the segment joining the given points form a triangle ? If triangle is formed, state the type of the triangle considering sides of the triangle.
(1) L(6,4) , M(-5,-3) , N(-6,8)
8. In the following examples, can the segment joining the given points form a triangle ? If triangle is formed, state the type of the triangle considering sides of the triangle.
(1) L(6,4) , M(-5,-3) , N(-6,8)
L(6, 4), M(–5, –3), N(–6, 8)
By distance Formula,
By distance Formula,
LM = [sqrt{((x_2 – x_1)^2 + (y_2 – y_1)^2)}]
= [sqrt{((-5-6)^2 + (-3-4)^2)}]
= [sqrt{((-11)^2 + (-7)^2)}]
= [sqrt{(121 + 49)}]
= [sqrt{(170)}]
∴ LM = [sqrt{(170)} ……(i)]
By distance Formula,
MN = [sqrt{((x_2 – x_1)^2 + (y_2 – y_1)^2)}]
= [sqrt{((-6 – (-5))^2 + (8 – (- 3))^2)}]
= [sqrt{((-1)^2 + (11)^2)}]
= [sqrt{(1 + 121)}]
= [sqrt{(122)}]
∴ MN = [sqrt{(122)} ……(ii)]
By distance Formula,
LN = [sqrt{((x_2 – x_1)^2 + (y_2 – y_1)^2)}]
= [sqrt{((-6-6)^2 + (8 – 4)^2)}]
= [sqrt{((-12)^2 + (4)^2)}]
= [sqrt{(144 + 16)}]
= [sqrt{(160)}]
[∴ LN = sqrt{(160)} ……..(iii)]
(MN + LN) > LM
These points are not collinear.
∴ We can construct a triangle through 3 non-collinear points.
LM ≠ MN ≠ LN
∴ Triangle formed is a scalene triangle.
Explanation:-
Given points are L(6, 4), M(-5, -3), and N(-6, 8).
Using the distance formula, we can find the length of each side of the triangle.
The distance formula is given by:
d = √[(x2 – x1)^2 + (y2 – y1)^2]
Where d is the distance between two points (x1, y1) and (x2, y2).
Finding LM:
LM = √[(-5 – 6)^2 + (-3 – 4)^2]
= √[(-11)^2 + (-7)^2]
= √[121 + 49]
= √170
Finding MN:
MN = √[(-6 – (-5))^2 + (8 – (-3))^2]
= √[(-1)^2 + (11)^2]
= √[1 + 121]
= √122
Finding LN:
LN = &radic[(-6 – 6)^2 + (8 – 4)^2]
= √[(-12)^2 + (4)^2]
= √[144 + 16]
= √160
As (MN + LN) > LM, the given points are not collinear, and hence, we can construct a triangle through these points.
Also, since LM ≠ MN ≠ LN, the triangle formed is a scalene triangle.
Chapter 5. Co-ordinate Geometry – Problem set 5 (Page 122)