For line B to AC: y - 6 = (1/3)(x - 4); y - 6 = (x/3) - (4/3); 3y - 18 = x - 4, so 3y - x = 14
For line A to BC: y - 6 = (-1)(x - 0); y - 6 = -x, so y + x = 6
Since these lines intersect at one point (the orthocenter), we can use simultaneous equations to solve for x and/or y:
(3y - x = 14) + (y + x = 6) => 4y = 20, y = +5; Substitute this into y + x = 6: 5 + x = 6, x = +1
<span>So the orthocenter is at coordinates (1,5), and the slopes of all three orthocenter lines are above.</span>
-2x + 5 = 1/3x - 22
-7/3x = -27
-7x = -81
x = 81/7 or 11.57
They will intersect when x = 81/7 or 11.57
Answer:
x+3y-6=0
Step-by-step explanation:
given eqn is y=3x-2 which is 3x-y-2=0
the eqn of line perpendicular to given eqn is -x+3y+k=0
it passes through (6,4)
-6+3*4+k=0
or,. -6+12+k=0
or, k= -6
therefore, the eqn of line perpendicular to given eqn is x+3y-6=0
Answer:
(2,0)
Step-by-step explanation:
Answer:
The equation of a line is typically written as y=mx+b where m is the slope and b is the y-intercept.
