Name: What Is Transformation of Axes?
Uploaded: 2019-07-19
Duration: 4 min 23 s
Description: Let's learn how the coordinates of a point change when we shift the coordinated axis by some distance. Watch this video and explore this concept of coordinat...

Question

Prove analytically:

The altitudes of a triangle are concurrent.

Accepted Answer

Let $A (x_{1}, y_{1}), B (x_{2}, y_{2}) and C (x_{3}, y_{3})$ be the vertices of the triangle $A B C$

If $m_{1}$ is the slope of $A B$ , then we use the two points formula to find the slope of the line $m_{1} = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} . . . (i)$

Since the altitude $C D$ is perpendicular to the side $A B$ , its slope $m$ is given by using a condition of the perpendicular slope $m = - \frac{1}{m_{1}} = - \frac{x_{2} - x_{1}}{y_{2} - y_{1}} . . . (i i)$

The equation of altitude $C D$ is passing through $C (x_{3}, y_{3})$ with slope $m$ is $(y - y_{3}) = m (x - x_{3})$

$\Rightarrow (y - y_{3}) = - \frac{x_{2} - x_{1}}{y_{2} - y_{1}} (x - x_{3})$

$\Rightarrow (y_{2} - y_{1}) (y - y_{3}) = - (x_{2} - x_{1}) (x - x_{3})$

$\Rightarrow (x_{2} - x_{1}) (x - x_{3}) + (y_{2} - y_{1}) (y - y_{3}) = 0$

$\Rightarrow (x_{2} - x_{1}) x - (x_{2} - x_{1}) x_{3} + (y_{2} - y_{1}) y - (y_{2} - y_{1}) y_{3} = 0$

$\Rightarrow (x_{2} - x_{1}) x + (y_{2} - y_{1}) y - (x_{2} - x_{1}) x_{3} - (y_{2} - y_{1}) y_{3} = 0 . . . . (i i i)$

For equation of altitude $A E$ , we just replace $x_{1}$ with $x_{2}$ , $x_{2}$ with $x_{3}$ , $x_{3}$ with $x_{1}$ and $y_{1}$ with $y_{2}$ , $y_{2}$ with $y_{3}$ , $y_{3}$ with $y_{1}$ in $(i i i)$

$\Rightarrow (x_{3} - x_{2}) x + (y_{3} - y_{2}) y - (x_{3} - x_{2}) x_{1} - (y_{3} - y_{2}) y_{1} = 0 . . . . (i v)$

For equation of altitude $B F$ , we just replace $x_{1}$ with $x_{2}$ , $x_{2}$ with $x_{3}$ , $x_{3}$ with $x_{1}$ and $y_{1}$ with $y_{2}$ , $y_{2}$ with $y_{3}$ , $y_{3}$ with $y_{1}$ in $(i v)$

$\Rightarrow (x_{1} - x_{3}) x + (y_{1} - y_{3}) y - (x_{1} - x_{3}) x_{2} - (y_{1} - y_{3}) y_{2} = 0 . . . . (v)$

To see whether altitudes $(i i i), (i v) and (v)$ are concurrent, consider the determinant

$= |\begin{array}{l} x_{2} - x_{1} & y_{2} - y_{1} & - (x_{2} - x_{1}) x_{3} - (y_{2} - y_{1}) y_{3} \\ x_{3} - x_{2} & y_{3} - y_{2} & - (x_{3} - x_{2}) x_{1} - (y_{3} - y_{2}) y_{1} \\ x_{1} - x_{3} & y_{1} - y_{3} & - (x_{1} - x_{3}) x_{2} - (y_{1} - y_{3}) y_{2} \end{array}|$

$= |\begin{matrix} x_{2} - x_{1} & y_{2} - y_{1} & - (x_{2} - x_{1}) x_{3} - (y_{2} - y_{1}) y_{3} \\ x_{3} - x_{2} & y_{3} - y_{2} & - (x_{3} - x_{2}) x_{1} - (y_{3} - y_{2}) y_{1} \\ 0 & 0 & 0 \end{matrix}| = 0$ $(R_{3} \to R_{3} + (R_{1} + R_{2}))$

Hence, the altitudes of a triangle are concurrent.

Prove analytically:

The altitudes of a triangle are concurrent.

Important Questions on Straight Lines

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