Answer:
[tex]\sf \left(\dfrac{1}{11},\dfrac{3}{11}\right)[/tex]
Step-by-step explanation:
The orthocenter is the point of intersection of the altitudes of a triangle.
(The altitude is the line that passes through the vertex and is perpendicular to the side opposite the vertex).
**Refer to attached diagram**
DC, EA and FB are the perpendicular lines drawn from the three vertices D, E and F.
[tex]\sf Let\:(x_1,y_1)=D\implies(x_1,y_1)=(1,3)[/tex]
[tex]\sf Let\:(x_2,y_2)=E\implies (x_2,y_2)= (-1,0)[/tex]
[tex]\sf Let\:(x_3,y_3)= F\implies (x_3,y_3)=(2,-1)[/tex]
G (x, y) is the intersection point of the three altitudes of the triangle, i.e. the orthocenter.
To find the point of intersection (orthocenter), we need to create linear equations for the altitudes, then equate them to find the point of intersection.
We only need to work with 2 altitudes.
Calculate the slopes of two sides of the triangle.
[tex]\sf \implies Slope\:of\:EF=\dfrac{y_3-y_2}{x_3-x_2}=\dfrac{-1-0}{2-(-1)}=-\dfrac{1}{3}[/tex]
[tex]\sf \implies Slope\:of\:DF=\dfrac{y_3-y_1}{x_3-x_1}=\dfrac{-1-3}{2-1}=-4[/tex]
As the slopes of the altitudes are perpendicular to the slopes of the sides:
[tex]\textsf{Slope of DC}=\sf \dfrac{-1}{\textsf{slope of EF}}=\dfrac{-1}{-\frac{1}{3}}=3[/tex]
[tex]\textsf{Slope of EA}=\dfrac{-1}{\textsf{slope of DF}}=\sf \dfrac{-1}{-4}=\dfrac{1}{4}[/tex]
Using point D (1, 3) and the slope of DC to create a linear equation for DC using the point-slope form of a linear equation:
[tex]\sf \implies y-y_1=m(x-x_1)[/tex]
[tex]\sf \implies y-3=3(x-1)[/tex]
[tex]\implies \sf y=3x[/tex]
Using point E (-1, 0) and the slope of EA to create a linear equation for EA using the point-slope form of a linear equation:
[tex]\sf \implies y-y_1=m(x-x_1)[/tex]
[tex]\sf \implies y-0=\dfrac{1}{4}(x-(-1))[/tex]
[tex]\sf \implies y=\dfrac{1}{4}x+\dfrac{1}{4}[/tex]
To find the x-value of the point of intersection (the orthocenter), equate the equations and solve for x:
[tex]\implies \sf 3x=\dfrac{1}{4}x+\dfrac{1}{4}[/tex]
[tex]\implies \sf \dfrac{11}{4}x=\dfrac{1}{4}[/tex]
[tex]\implies \sf 11x=1[/tex]
[tex]\implies \sf x=\dfrac{1}{11}[/tex]
To find the y-value of the point of intersection (the orthocenter), substitute the found value of x into the equation for line DC:
[tex]\implies \sf y=3\left(\dfrac{1}{11}\right)[/tex]
[tex]\implies \sf y=\dfrac{3}{11}[/tex]
Therefore, the orthocenter is at point:
[tex]\sf \left(\dfrac{1}{11},\dfrac{3}{11}\right)[/tex]
