In the triangle ABC with vertices A(2, 3),B(4 | vedclass

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Question

(A) Let $AD$ be the altitude of triangle $ABC$ from vertex $A$. Accordingly,$AD \perp BC$. The slope of $BC$ is $m_{BC} = \frac{2 - (-1)}{1 - 4} = \fr

Vedclass · Accepted Answer

$y-x=1$ and $\sqrt{2}$

Vedclass · Answer

(A) Let $AD$ be the altitude of triangle $ABC$ from vertex $A$. Accordingly,$AD \perp BC$. The slope of $BC$ is $m_{BC} = \frac{2 - (-1)}{1 - 4} = \frac{3}{-3} = -1$. Since $AD \perp BC$,the slope of $AD$ is $m_{AD} = -\frac{1}{m_{BC}} = -\frac{1}{-1} = 1$. The equation of the line $AD$ passing through point $A(2, 3)$ with slope $1$ is: $(y - 3) = 1(x - 2)$ $\Rightarrow y - 3 = x - 2$ $\Rightarrow y - x = 1$. Length of $AD$ is the perpendicular distance from $A(2, 3)$ to the line $BC$. The equation of $BC$ is $(y - (-1)) = -1(x - 4)$ $\Rightarrow y + 1 = -x + 4$ $\Rightarrow x + y - 3 = 0$. The perpendicular distance $d$ from $(x_1, y_1)$ to $Ax + By + C = 0$ is $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$. For $A(2, 3)$ and $x + y - 3 = 0$: $d = \frac{|1(2) + 1(3) - 3|}{\sqrt{1^2 + 1^2}} = \frac{|2 + 3 - 3|}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$. Thus,the equation is $y - x = 1$ and the length is $\sqrt{2}$ units.

In the triangle $ABC$ with vertices $A(2, 3)$,$B(4, -1)$ and $C(1, 2)$,find the equation and length of the altitude from the vertex $A$.

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