Question

Please help me
Consider triangle ABC with vertices A(4,8), B(0,0) and C(6,0). AQ and BP are two altitudes of triangle ABC. AQ and BP meet at H.
(a) Find the equations of AQ and BP.
(b) Find the coordinates of H.
(c) If CR is the third altitude of ABC, show that CR also passes through H.

Answer 1

I have sketched a diagram, i found that AQ is perpendicular to BC, since BC is a horizontal line, so AQ is a vertical line, which the equation of AQ is x=4 (i.e. x-4-0), is it correct??

Answer 2

From the answer page:
(a) AQ: x-4=0
BP: x-4y = 0
(b) H (4, 1)

Answer 3

sorry dude i made silly mistake

Answer 4

the answer given is right do u wanna me explain?

Answer 5

Yes, thank you. The answer provided is correct but without calculation steps

Answer 6

first find out the slope of AC
m=\[\frac{ y2-y1 }{ x2-x1}\]
m=\[\frac{ 8-0 }{ 4-6 }=-4\]
bp is perpendicular to ac
so the slope is reciprocal to other
so the slope of BP IS reciprocal to slope OF AC
slope of BP=1/4

Answer 7

using the slope and and pt formula arive the eqn
\[y-y1=m(x-x1)\]

Answer 8

let me see..

Answer 9

have u understand what i wrote \

Answer 10

m ( difference of x ) = difference of y
1/4 ( x - 0 ) = y - 0
1/4 x = y
1/4 x - y = 0
x-4y = 0

I got it! Thank you!

Answer 11

medals pls

Answer 12

I finished the question, thank you very much!