Answer:
( -1,-5 )
Step-by-step explanation:
We have the co-ordinates A( -3,3 ), B( -1,7 ) and C( 3,3 ).
We will find the orthocenter using below steps:
1. First, we find the equations of AB and BC.
The general form of a line is y=mx+b where m is the slope and b is the y-intercept.
Using the formula of slope given by [tex]m=\frac{y_{2}-y_{1}}{x_{2}-x_{1} }[/tex], we will find the slope of AB and BC.
Now, slope of AB is [tex]m=\frac{7-3}{-1+3}[/tex] i.e. [tex]m=\frac{4}{2}[/tex] i.e. [tex]m=2[/tex].
Putting this 'm' in the general form and using the point B( -1,1 ), we get the y-intercept as,
y = mx + b i.e. 1 = 2 × (-1) + b i.e. b = 3.
So, the equation of AB is y = 2x + 3.
Also, slope of BC is [tex]m=\frac{3-7}{3+1}[/tex] i.e. [tex]m=\frac{-4}{4}[/tex] i.e. [tex]m=-1[/tex].
Putting this 'm' in the general form and using the point B( -1,1 ), we get the y-intercept as,
y = mx + b i.e. 1 = (-1) × (-1) + b i.e. b = 0.
So, the equation of BC is y = -x.
2. We will find the slope of line perpendicular to AB and BC.
When two lines are perpendicular, then the product of their slopes is -1.
So, slope of line perpendicular to AB is [tex]m \times 2 = -1[/tex] i.e. [tex]m=\frac{-1}{2}[/tex]
So, slope of line perpendicular to BC is [tex]m \times (-1) = -1[/tex] i.e. m = 1.
3. We will now find the equations of line perpendicular to AB and BC.
Using the slope of line perpendicular to AB i.e. [tex]m=\frac{-1}{2}[/tex] and the point opposite to AB i.e. C( 3,3 ), we get,
y = mx+b i.e. [tex]3=\frac{-1}{2} \times 3 + b[/tex] i.e. [tex]b=\frac{9}{2}[/tex]
So, the equation of line perpendicular to AB is [tex]y=\frac{-x}{2} +\frac{9}{2}[/tex]
Again, using the slope of line perpendicular to BC i.e. m = 1 and the point opposite to BC i.e. A( -3,3 ), we get,
y = mx + b i.e. 3 = 1 × -3 + b i.e. b = 6.
So, the equation of line perpendicular to BC is y = x+6
4. Finally, we will solve the obtained equations to find the value of ( x,y ).
As, we have y = x+6 and [tex]y=\frac{-x}{2}+\frac{9}{2}[/tex]
This gives, [tex]y=\frac{-x}{2}+\frac{9}{2}[/tex] → [tex]x+6=\frac{-x}{2} +\frac{9}{2}[/tex] → 2x+12 = -x+9 → 3x = -3 → x = -1.
So, y = x+6 → y = -1+6 → y=5.
Hence, the orthocenter of the ΔABC is ( -1,5 ).
