Question

if line segment bc is considered the base of triangle abc, what is the corresponding height of the triangle? 1.25 units 1.6 units 0.625 units 0.8 units

Explanation:

Step1: Recall height - base concept

The height of a triangle with base $BC$ is the perpendicular distance from the opposite vertex $A$ to the line containing $BC$.

Step2: Use formula for distance from a point to a line

First, find the equation of the line passing through $B(3,2)$ and $C(- 1,-1)$. The slope $m=\frac{y_2 - y_1}{x_2 - x_1}=\frac{2+1}{3 + 1}=\frac{3}{4}$. Using the point - slope form $y - y_1=m(x - x_1)$ with point $C(-1,-1)$, we have $y+1=\frac{3}{4}(x + 1)$ or $3x-4y-1 = 0$.
The distance $d$ from the point $A(-1,1)$ to the line $Ax+By + C = 0$ (here $A = 3$, $B=-4$, $C=-1$) is given by the formula $d=\frac{\vert Ax_0+By_0 + C\vert}{\sqrt{A^{2}+B^{2}}}$. Substituting $x_0=-1$ and $y_0 = 1$ into the formula:
\[

$$\begin{align*} d&=\frac{\vert3\times(-1)-4\times1-1\vert}{\sqrt{3^{2}+(-4)^{2}}}\\ &=\frac{\vert-3 - 4-1\vert}{\sqrt{9 + 16}}\\ &=\frac{\vert-8\vert}{5}\\ &=1.6 \end{align*}$$

\]

Answer:

1.6 units