If A (1, -1),B (0, 4),C (-5, 3) are vertices of a triangle then find the slope of each side
If A (1, -1),B (0, 4),C (-5, 3) are vertices of a triangle then find the slope of each side
A (1, –1), B (0, 4), C (–5, 3) form a triangle.
Slope of AB = \[\frac{4 - \left( - 1 \right)}{0 - 1} = \frac{5}{- 1} = - 5\]
Slope of BC = \[\frac{3 - 4}{- 5 - 0} = \frac{- 1}{- 5} = \frac{1}{5}\]
Slope of AC = \[\frac{3 - \left( - 1 \right)}{- 5 - 1} = \frac{4}{- 6} = \frac{- 2}{3}\]
Explanation:-
To determine whether the given triangle is a right-angled triangle or not, we can use the Pythagorean theorem, which states that in a right triangle, the square of the hypotenuse (the longest side) is equal to the sum of the squares of the other two sides.
Let’s label the sides of the triangle as follows:
- AB is the side opposite to point C
- BC is the side opposite to point A
- AC is the side opposite to point B
Using the distance formula, we can calculate the lengths of these sides:
AB = sqrt((0 – 1)^2 + (4 – (-1))^2) = sqrt(1 + 25) = sqrt(26) BC = sqrt((-5 – 0)^2 + (3 – 4)^2) = sqrt(25 + 1) = sqrt(26) AC = sqrt((-5 – 1)^2 + (3 – (-1))^2) = sqrt(36 + 16) = sqrt(52) = 2sqrt(13)
Now, we can check whether the Pythagorean theorem holds:
- If AB^2 + BC^2 = AC^2, then the triangle is right-angled.
- If AB^2 + AC^2 = BC^2 or BC^2 + AC^2 = AB^2, then the triangle is obtuse-angled.
- If AB^2 + BC^2 < AC^2 or AB^2 + AC^2 < BC^2 or BC^2 + AC^2 < AB^2, then the triangle is acute-angled.
Plugging in the values, we get:
AB^2 + BC^2 = 26 + 26 = 52 AC^2 = (2sqrt(13))^2 = 52
Since AB^2 + BC^2 = AC^2, we can conclude that the given triangle is a right-angled triangle.
Chapter 5. Co-ordinate Geometry – Practice Set 5.3 (Page 122)