Find the equations to the sides of the triangles the coordinates of whose angular points are respectively (0, 1), (2, 0) and (-1, -2)

Asked by Aaryan | 1 year ago |  35

##### Solution :-

(0, 1), (2, 0) and (-1, -2)

Given:

Points A (0, 1), B (2, 0) and C (-1, -2).

Let us assume,

m1, m2, and m3 be the slope of the sides AB, BC and CA, respectively.

So,

The equation of the line passing through the two points (x1, y1) and (x2, y2).

Then,

$$m_3=\dfrac{1+2}{1+0}$$

m1 = $$\dfrac{ -1}{2}$$, m2 = $$\dfrac{ -2}{3}$$ and m3= 3

So, the equation of the sides AB, BC and CA are

By using the formula,

y – y1= m (x – x1)

=> y – 1 = ($$\dfrac{ -1}{2}$$) (x – 0)

2y – 2 = -x

x + 2y = 2

=> y – 0 = ($$\dfrac{ -2}{3}$$) (x – 2)

3y = -2x + 4

2x – 3y = 4

= y + 2 = 3(x+1)

y + 2 = 3x + 3

y – 3x = 1

So, we get

x + 2y = 2, 2x – 3y =4 and y – 3x = 1

The equation of sides are x + 2y = 2, 2x – 3y =4 and y – 3x = 1

Answered by Aaryan | 1 year ago

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