please please please help me!!!

The point where the altitudes of a triangle meet called Ortho Centre

We have given a triangle ABC whose vertices are(0, 6),(4, 6), (1, 3)

*In Step 1 we find slopes Of AB, BC,CA Slope formulae y _{2-}y_{1⁄} x2-X1*

*slope AB= 6-6/4-0 = 0/4 =0
*

* .... BC= 3-6/ 1-4 = -3/-3 =1
*

*....... CA=6-3/ 0-1 =3/-1 =-3
*

*In Step 2
*

*But we know Orthocentre is the point where perpendeculars drawn from vertex to opposite side meet. So
*

*Let's think a triangle ABC and AD, BE, CF are perpendiculars drawn to the vertex.
*

*Slope AD = -1/slope BC = -1/1 =-1
*

*.......BE = -1/slope CA = -1/-3 = 1/3
*

*.....CF = -1/slope AB = -1/0 undefined
*

*Step 3 we have now vertices and slopes of AD,BE,CF we find equations of lines AD,BE and CF*

*we have A(0,6) and m =-1 we substitute in the equation y*

_{-}y_{1 =}m(x-X_{1)}*y-6=-1(x-0)*

*y+x=6 - eq 1*

*B(4,6) and slope BE (1/3)*

*y-6=1/3(x-4)*

*3Y-18=x-4*

*3y-x=14 -eq 2*

*C(1,3) and whose slope CF undefined*

*So line is vertical and x=1 is the eq*

*Now solving any of equations 1&2 we get values for( x,y) orhto centre*

*(x,y) =(*

*solving eq 1 and eq 2 we get x=1, y=5 (*

*I hope this helps you*

* *

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