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Does any one know how to find the distance between the point (2,3) and the line 4x + 3y = 10?

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First, find the equation of the line perpendicular to the given line that passes thorugh (2, 3). Once you have the equation of the perpendicular, solve a system of equations of the two equations to find where the lines intersect. Then, find the distance between the point of intersection and point (2, 3).
Could x=2 @mathstudent55
could that work?

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Ok this is a big pain to solve, but here it is. First put the equation of the line into y = m*x+b format y = (4/3)x + 3.333 The perpendicular line will have slope -3/4 and passes through (2,3) so it's equation will be y = (-3/4)x+4.5 The two lines intersect at (.56,4.08) The triangle with tip (.56,4.08) and right point at (2,3) has base 1.44 and height of 1.08 \[\sqrt{1.44^{2}+1.08^{2}} = 1.8 \] 1.8 is the answer
when they intersect i get (-2,6)?
What did you get for the equation of the perpendicular?
i plugged in the one that you had calculated
@Matt.Mawson calculated it, but I think he made a mistake.
Let's start from the beginning. The given line is: 4x + 3y = 10
Solve it for y to get the slope: 3y = -4x + 10 y = (-4/3)x + 10/3
He made a mistake right in the beginning when he subtracted the 4x from both sides. It should be -4x on the right side, but he wrote 4x.
Yes i get that for my equation as well. and for my perpendicular i calculated y=(3/4)x +1.5
Yes, I have both of those.
for my point of intercept i have .88, 2.16
Yes, I got the same x and y as you.
so how will i find the distance?
The final step is the distance between (0.88, 2.16) and (2,3)
For that distance I get 1.4
yes i have that too... Okay. Thank You!

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