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4 years ago

(-2,3)(0,8) what is en equation for these points

Mathematics

1 answer:

jenyasd209 [6]4 years ago

4 0

Answer:

The line equation that passes through the given points is 5x – 2y + 16 = 0

Explanation:

Given:

Two points are A(-2, 3) and B(0, 8).

To find:

The line equation that passes through the given two points.

Solution:

We know that, general equation of a line passing through two points (x1, y1), (x2, y2) is given by

$\frac{(y- y1)}{(x-x_1)}= \frac{((y_2- y_1)}{(x_2- x_1 )}$

${(y- y1)= \frac{((y_2- y_1)}{(x_2- x_1 )}\times(x-x_1)$ .............(1)

here, in our problem x1 = 0, y1 = 8, x2 = -2 and y2 = 3.

Now substitute the values in (1)

$y-8 = \frac{(3-8)}{(- 2 - 0)}\times(x- 0)$

$y-8 = \frac{(- 5)}{(-2)}\times(x)$

$y-8 =\frac{5}{2}x$

2y – 16 = 5x

5x – 2y + 16 = 0

Hence, the line equation that passes through the given points is 5x – 2y + 16 = 0.

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3 years ago

(9)(8/9) in find the product?

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3 years ago

26. Define a relation ∼ ∼ on R 2 R2 by stating that ( a , b ) ∼ ( c , d ) (a,b)∼(c,d) if and only if a 2 + b 2 ≤ c 2 + d 2 . a2+

Tresset [83]

Answer:

~ is reflexive.

~ is asymmetric.

~ is transitive.

Step-by-step explanation:

~ is reflexive:

i.e., to prove $\forall (a, b) \in \mathbb{R}^2$ , $(a, b) R(a, b)$ .

That is, every element in the domain is related to itself.

The given relation is $\sim: (a,b) \sim (c, d) \iff a^2 + b^2 \leq c^2 + d^2$

Reflexive:

$(a, b) \sim (a, b)$ since $a^2 + b^2 = a^2 + b^2$

This is true for any pair of numbers in $\mathbb{R}^2$ . So, $\sim$ is reflexive.

Symmetry:

$\sim$ is symmetry iff whenever $(a, b) \sim (c, d)$ then $(c, d) \sim (a, b)$ .

Consider the following counter - example.

Let (a, b) = (2, 3) and (c, d) = (6, 3)

$a^2 + b^2 = 2^2 + 3^2 = 4 + 9 = 13$

$c^2 + d^2 = 6^2 + 3^2 = 36 + 9 = 42$

Hence, $(a, b) \sim (c, d)$ since $a^2 + b^2 \leq c^2 + d^2$

Note that $c^2 + d^2 \nleq a^2 + b^2$

Hence, the given relation is not symmetric.

Transitive:

$\sim$ is transitive iff whenever $(a, b) \sim (c, d) \hspace{2mm} \& \hspace{2mm} (c, d) \sim (e, f)$ then $(a, b) \sim (e, f)$

To prove transitivity let us assume $(a, b) \sim (c, d)$ and $(c, d) \sim (e, f)$ .

We have to show $(a, b) \sim (e, f)$

Since $(a, b) \sim (c, d)$ we have: $a^2 + b^2 \leq c^2 + d^2$

Since $(c, d) \sim (e, f)$ we have: $c^2 + d^2 \leq e^2 + f^2$

Combining both the inequalities we get:

$a^2 + b^2 \leq c^2 + d^2 \leq e^2 + f^2$

Therefore, we get: $a^2 + b^2 \leq e^2 + f^2$

Therefore, $\sim$ is transitive.

Hence, proved.

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3 years ago

Rewrite the following linear equations in general form

Romashka-Z-Leto [24]

Answer:

5y = x +15

Step-by-step explanation:

here you have to multiply both sides with 5 to make it a general form ,

5(y) = 5(1/5x) + 5(3)

5y = x + 15

Got a little help from app called Gauthmath. It got concept videos and tutors. You can check it on Play Store and App Store. Really good app in my opinion.

Hope this helps. :)

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3 years ago

What did you get for the answer and how to solve -7(k-3)

Alinara [238K]

Answer:

Step-by-step explanation:

-7k+21

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4 years ago

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