How do i find the shorted distance between the point and the line

Mathematics

1 answer:

m_a_m_a [10]3 years ago

3 0

Answer:

5 units

Step-by-step explanation:

3x + 4y = 8

4y = -3x+8

y = -3/4+2

The shortest distance between a point and a line is the perpendicular line.

Slope of the perpendicular line: 4/3 and point (-3,-2)

b = -2-(4/3)(-3) = 2

Equation of the perpendicular line: y=4/3x+2

y is equal y

4/3x+2= -3/4x+2

4/3x +3/4x = 2-2

x = 0

Plug x=0 into one of the equations to find y

y = 4/3(0) + 2

y = 2

(0,2) and (-3,-2)

Distance = sqrt [(-3-0)^2 + (-2-2)^2]

Sqrt (-3)^2+ (-4)^2

Sqrt 25 = 5

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3 years ago

Lim t^4 - 6 / 2t^2 - 3t + 7

Harman [31]

I think you meant to say

$\displaystyle \lim_{t\to2}\frac{t^4-6}{2t^2-3t+7}$

(as opposed to x approaching 2)

Since both the numerator and denominator are continuous at t = 2, the limit of the ratio is equal to a ratio of limits. In other words, the limit operator distributes over the quotient:

$\displaystyle \lim_{t\to2} \frac{t^4 - 6}{2t^2 - 3t + 7} = \frac{\displaystyle \lim_{t\to2}(t^4-6)}{\displaystyle \lim_{t\to2}(2t^2-3t+7)}$

Because these expressions are continuous at t = 2, we can compute the limits by evaluating the limands directly at 2:

$\displaystyle \lim_{t\to2} \frac{t^4 - 6}{2t^2 - 3t + 7} = \frac{\displaystyle \lim_{t\to2}(t^4-6)}{\displaystyle \lim_{t\to2}(2t^2-3t+7)} = \frac{2^4-6}{2\cdot2^2-3\cdot2+7} = \boxed{\frac{10}9}$