Question

In the triangle ABC with vertices \[A\left( 2,3 \right),B\left( 4, - 1 \right)\] and \[C\left( 1,2 \right)\] . Find the equation and length of altitude from the vertex \[A\].

Answer 1

Hint: In the given question, we need to find the equation and length of the altitude from the vertex \[A\] in the triangle \[{ABC}\] . By using slope formula and distance formula we can find the solution of the question.
Formula used :
Slope ,
\[m\ = \dfrac{{change\ in\ y}}{{change\ in\ x}}\]

Complete answer:

Given, a triangle \[{ABC}\] with the \[A\left( 2,3 \right),B\left( 4, - 1 \right)\] and \[C\left( 1,2 \right)\].Let \[{AM}\] be the perpendicular from the vertex \[A\]
Now, we know that if two lines are perpendicular, then the product of their slopes = \[- 1\]
Slope of \[AM\]—Slope of \[{BC} = - 1\]
\[Slope\ of\ AM\ = - \dfrac{1}{{Slope\ of\ BC}}\] •••••(1)
First we need to find the slope of BC,
Slope of \[{BC}\] joining \[(4,\ - 1)\ \&\ \left( 1,\ 2 \right)\]
\[= \dfrac{2 - \left( - 1 \right)}{1 – 4}\]
On simplifying,
We get,
\[= \dfrac{2 + 1}{- 3}\]
\[= \dfrac{3}{- 3}\]
By simplifying,
We get,
\[= - 1\]
\[Slope\ of\ BC\ = \ - 1\]
From (1) ,
\[Slope\ of\ AM\ = \ - \dfrac{1}{- 1}\]
\[= 1\]
Equation of line having one coordinate and the slope is given by
\[(y-y{_{1}})= m(x-x{_{1}})\]
\[y – 3 = 1\left( x – 2 \right)\]
By moving variables to one side,
We get,
\[2 – 3 = x – y\]
On simplifying,
We get,
\[- 1 = x – y\]
\[- x + y – 1 = 0\]
The equation of \[{AM}\] is \[y – x = 1\]
Now finding the equation of \[{BC}\]
The equation of \[{BC}\] is
\[y - \left( - 1 \right) = - 1\left( x – 4 \right)\]
\[y + 1 = - x + 4\]
By moving variables to one side,
We get,
\[y + x = - 1 + 4\]
On further simplifying, we get,
\[y + x = 3\]
 \[x + y – 3 = 0\]
The equation of \[{BC}\] is
\[x + y – 3 = 0\]
The perpendicular distance (d) of a line \[Ax + By + c = 0\] from the point \[(x{_{1}}, x{{_2}} )\] is
\[d =\]$ |Ax_1 + By_1 + C |$/\[\sqrt{}(A^{2} + B^{2}\]
\[d = \dfrac{\left| 1\left( 2 \right) + 1\left( 3 \right) + \left( - 3 \right) \right|}{\sqrt{1^{2} + 1^{2}}}\]
\[= \dfrac{\left| 2 + 3 – 3 \right|}{\sqrt{2}}\]
\[= \dfrac{\left| 2 \right|}{\sqrt{2}}\]
On multiplying both numerator and denominator by \[\sqrt{2}\],
We get,
\[= \dfrac{2}{\sqrt{2}} \times \dfrac{\sqrt{2}}{\sqrt{2}}\]
\[= \dfrac{2\sqrt{2}}{2}\]
By dividing,
We get,
\[= \sqrt{2}\]
The length if altitude \[AM\ = \sqrt{2}\]
Final answer :
The equation of \[{AM}\] is \[y – x = 1\] and The length of altitude \[AM\ = \sqrt{2}\].

Note:
The slope of a line is defined as the measure of its Steepness. It is calculated by dividing the change in \[y\] coordinate by change in \[x\] co-ordinate. Mathematically, slope is denoted by the letter \[m\] . Slope is positive when m is greater than \[0\] and when m is less than \[0\] , slope is negative.If the slope is equal to \[0\] that means it is a constant function.