A non-right triangle with known x,y coordinates. Lets say A,B,C is the points of triangle. I want to find a new point D which is perpendicular line from point B and lies on the line AC. Any simpled questions for D(x,y)?

abbracciopj
2022-06-26
Answered

A non-right triangle with known x,y coordinates. Lets say A,B,C is the points of triangle. I want to find a new point D which is perpendicular line from point B and lies on the line AC. Any simpled questions for D(x,y)?

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lodosr

Answered 2022-06-27
Author has **24** answers

First, given coordinates of $A$ and $C$, you can have the formula of the line AC.

Secondly, we know that line $BD$ and $AC$are perpendicular, so we can get the slope of $BD$ (the product of slopes of perpendicular lines in ${\mathbb{R}}^{2}$ is $-1$). With coordinates of $B$, we can then get the formula of $BD$. And $D$ is just the point of intersection of $AC$ and $BD$.

Secondly, we know that line $BD$ and $AC$are perpendicular, so we can get the slope of $BD$ (the product of slopes of perpendicular lines in ${\mathbb{R}}^{2}$ is $-1$). With coordinates of $B$, we can then get the formula of $BD$. And $D$ is just the point of intersection of $AC$ and $BD$.

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but when I simplify that I don't get$2{\mathrm{sec}}^{2}x\mathrm{tan}x$ which is the correct answer?

I'm using the quotient rule and so I rewrite

Then set

I then get

but when I simplify that I don't get

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