Finding the x-Intercept of a Perpendicular Line Passing Through a Given Point

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To find the x-intercept of the line perpendicular to the line joining points (1, 1) and (3, 4) that passes through (1, 2), the slope of the original line is calculated as 3/2. The slope of the perpendicular line is determined to be -2/3, leading to the equation y = (-2/3)x + 8/3. Setting this equation to zero to find the x-intercept results in the solution (4, 0). The final answer is confirmed as (4, 0).
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Homework Statement



Find the x-intercept of the line that passes through the point (1; 2) and perpendicular to the
line joining (1; 1) and (3; 4):



The Attempt at a Solution



I want to confirm my steps if it's right ,,,,

M=(4-1)/(3-1)= 3/2

y-1=3/2(x-1) ====> y=(3/2)x-1/2

perpendicular = m1xm2=-1 ====> m2=-2/3

y-2=-2/3(x-1) ====> y=(-2/3)x+4/5


making it one equation

(3/2)x-(1/2)=(2/3)x+(4/3)
x=11/13
 
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MrNeWBiE said:
making it one equation

(3/2)x-(1/2)=(2/3)x+(4/3)
x=11/13

I am not sure what you did here. I am assuming you were trying to combine the equation of the first line with the second line. If so, you have calculated the point of intersection of the two lines, which is not what the question has asked for. The question asked you for the x-intercept of the line.
 
aha ,,,
so I should not combine them ,,,

(-2/3)x+4/5 = 0 ?

and the answer will be (10/13,0)
 
Looks good to me!
 
ops ,,,
I had a mistake in y=(-2/3)x+4/5 ,,, should be y=(-2/3)x+8/3

the answer is (4,0) ,,,

btw the for your helping
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
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