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2 years ago

Given the points K(-2;5), L(1,3) and M(-5; 2p). Calculate the value

1 answer:

4 0

The product of the perpendicular lines is a negative one. Then the value of p will be negative 3.75.

<h3>What is coordinate geometry?</h3>

Coordinate geometry is the study of geometry using the points in space. Using this, it is possible to find the distance between the points, the dividing line is m:n ratio, finding the mid-point of the line, etc.

Given the points K(-2,5), L(1,3) and M(-5, 2p).

If KL is perpendicular to LM. Then we have

The slope of KL will be

KL = (1+2)/(3-5)

KL = -1.5

The slope of LM will be

LM = (1+6)/(3-2p)

LM = 7/(3-2p)

We know the product of the perpendicular lines is a negative one.

KL x ML = -1

-1.5 x 7/(3-2p) = -1

-10.5 = -3 + 2p

2p = -7.5

p = -3.75

More about the coordinate geometry link is given below.

brainly.com/question/1601567

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General Formulas and Concepts:

Calculus

Differentiation

Derivatives
Derivative Notation

Derivative Property [Multiplied Constant]: $\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

Basic Power Rule:

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Integrals

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U-Substitution

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx$

Step 2: Integrate Pt. 1

Identify variables for u-substitution.

Set u: $\displaystyle u = 4x^{-2}$
[u] Differentiate [Basic Power Rule, Derivative Properties]: $\displaystyle du = \frac{-8}{x^3} \ dx$
[Bounds] Switch: $\displaystyle \left \{ {{x = 9 ,\ u = 4(9)^{-2} = \frac{4}{81}} \atop {x = 5 ,\ u = 4(5)^{-2} = \frac{4}{25}}} \right.$

Step 3: Integrate Pt. 2

[Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}\int\limits^9_5 {\frac{-8}{x^3}e^\big{4x^{-2}}} \, dx$
[Integral] U-Substitution: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}\int\limits^{\frac{4}{81}}_{\frac{4}{25}} {e^\big{u}} \, du$
[Integral] Exponential Integration: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}(e^\big{u}) \bigg| \limits^{\frac{4}{81}}_{\frac{4}{25}}$
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Simplify: $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{1}{8} \bigg( e^\Big{\frac{4}{25}} - e^\Big{\frac{4}{81}} \bigg)$

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Integration

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3 years ago