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3 years ago

Given J (x, -8) and K (-1,-5) and the graph of line l below, find the value of x so that JK ll L.

Mathematics

1 answer:

Bingel [31]3 years ago

5 0

Answer:

x = -2

Step-by-step explanation:

slope of line l = 3

we want the line containing points J and K to have a slope of 3 also

-8-(-5) = 3

x-(-1) 1

cross-multiply to solve for 'x':

3(x+1) = -3

3x + 3 = -3

3x = -6

x = -2

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$\bold{\huge{\blue{\underline{ Solution }}}}$

<h3>Given :-</h3>

The right angled below is formed by 3 squares A, B and C
The area of square B has an area of 144 inches ²
The area of square C has an of 169 inches ²

We have to find the area of square A?

<h3>Let's Begin :- </h3>

The right angled triangle is formed by 3 squares

We have, 

Area of square B is 144 inches²
Area of square C is 169 inches²

We know that,

$\bold{ Area \: of \: square = Side × Side }$

Let the side of square B be x

Subsitute the required values,

$\sf{ 144 = x × x }$

$\sf{ 144 = x² }$

$\sf{ x = √144}$

$\bold{\red{ x = 12\: inches }}$

Thus, The dimension of square B is 12 inches

Area of square C = 169 inches

Let the side of square C be y

Subsitute the required values,

$\sf{ 169 = y × y }$

$\sf{ 169 = y² }$

$\sf{ y = √169}$

$\bold{\green{ y = 13\: inches }}$

Thus, The dimension of square C is 13 inches.

It is mentioned in the question that, the right angled triangle is formed by 3 squares

The dimensions of square be is x and y

Let the dimensions of square A be z

<h3>Therefore, By using Pythagoras theorem, </h3>

The sum of squares of base and perpendicular height equal to the square of hypotenuse

That is,

$\bold{\pink{ (Perpendicular)² + (Base)² = (Hypotenuse)² }}$

Here, 

Base = x = 12 inches
Perpendicular = z
Hypotenuse = y = 13 inches

Subsitute the required values,

$\sf{ (z)² + (x)² = (y)² }$

$\sf{ (z)² + (12)² = (169)² }$

$\sf{ (z)² + 144 = 169}$

$\sf{ (z)² = 169 - 144 }$

$\sf{ (z)² = 25}$

$\bold{\blue{ z = 5 }}$

Thus, The dimensions of square A is 5 inches

<h3>Therefore,</h3>

Area of square

$\sf{ = Side × Side }$

$\sf{ = 5 × 5 }$

$\bold{\orange{ = 25\: inches }}$

Hence, The area of square A is 25 inches.

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3 years ago

Given J (x, -8) and K (-1,-5) and the graph of line l below, find the value of x so that JK ll L.​

Given J (x, -8) and K (-1,-5) and the graph of line l below, find the value of x so that JK ll L.