Suppose the vertices of a triangle are given | vedclass

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(D) The vertices of the triangle are $A(0,3)$,$B(-2,0)$,and $C(6,1)$.First,find the equations of the lines forming the sides of the triangle:Line $AB$

Vedclass · Accepted Answer

$\left(\frac{-6}{7}, \frac{3}{2}\right)$

Vedclass · Answer

(D) The vertices of the triangle are $A(0,3)$,$B(-2,0)$,and $C(6,1)$.First,find the equations of the lines forming the sides of the triangle:Line $AB$: Using the two-point form,$y - 0 = \frac{3-0}{0-(-2)}(x - (-2))$ $\Rightarrow y = \frac{3}{2}(x+2)$ $\Rightarrow 3x - 2y + 6 = 0$.Line $BC$: $y - 0 = \frac{1-0}{6-(-2)}(x - (-2))$ $\Rightarrow y = \frac{1}{8}(x+2)$ $\Rightarrow x - 8y + 2 = 0$.Line $AC$: $y - 3 = \frac{1-3}{6-0}(x - 0)$ $\Rightarrow y - 3 = \frac{-2}{6}x$ $\Rightarrow y = -\frac{1}{3}x + 3$ $\Rightarrow x + 3y - 9 = 0$.Let the point $P$ be $(\alpha, \alpha+1)$. For $P$ to lie inside the triangle,it must satisfy the inequalities defined by the lines such that it lies on the same side of each line as the third vertex.Alternatively,consider the line $L: y = x+1$. We find the intersection of this line with the sides of the triangle.Intersection with $AC$ $(x+3y-9=0)$: $x + 3(\alpha+1) - 9 = 0$ $\Rightarrow \alpha + 3\alpha + 3 - 9 = 0$ $\Rightarrow 4\alpha = 6$ $\Rightarrow \alpha = \frac{3}{2}$.Intersection with $BC$ $(x-8y+2=0)$: $\alpha - 8(\alpha+1) + 2 = 0$ $\Rightarrow \alpha - 8\alpha - 8 + 2 = 0$ $\Rightarrow -7\alpha = 6$ $\Rightarrow \alpha = -\frac{6}{7}$.Thus,for the point to lie inside the triangle,$\alpha$ must be in the interval $\left(\frac{-6}{7}, \frac{3}{2}\right)$.

Suppose the vertices of a triangle are given by $A(0,3)$,$B(-2,0)$,and $C(6,1)$. For the point $(\alpha, \alpha+1)$ to lie inside the triangle,$\alpha$ must lie in the interval:

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