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Citrus2011 [14]

3 years ago

9

What is an equation of the line that passes through the points (3,1) and (6,6)?

1 answer:

koban [17]3 years ago

5 0

<h3>The equation of the line that passes through the points (3,1) and (6,6) is:</h3>

$y = \frac{5}{3}x -4$

<em><u>Solution:</u></em>

Given that,

We have to find the equation of the line that passes through the points (3,1) and (6,6)

<em><u>Find the slope of line</u></em>

$m = \frac{y_2-y_1}{x_2-x_1}$

From given,

$(x_1, y_1) = (3, 1)\\\\(x_2, y_2) = (6, 6)$

Substituting the values we get,

$m = \frac{6-1}{6-3}\\\\m = \frac{5}{3}$

<em><u>The slope intercept form of line is given as:</u></em>

y = mx + c ------ eqn 1

Where,

m is the slope

c is the y intercept

Substitute m = 5/3 and (x, y) = (3, 1) in eqn 1

$1 = \frac{5}{3} \times 3 + c\\\\1 = 5 + c\\\\c = -4$

Substitute m = 5/3 and c = -4 in eqn 1

$y = \frac{5}{3}x -4$

Thus the equation of line in slope intercept form is found

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1 year ago

1. A box contains 10 blue Stickers, 15 red stickers , and 25 orange stickers . A sticker is selected and its color noted. Then i

denis-greek [22]

Answer:

1/25 ; 3/20 ; 3/50

Step-by-step explanation:

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P(First blue) = 10/50 = 1/5

P(second blue) = 10/50 = 1/5

1/5 * 1/5 = 1 / 25

b. Selecting one red sticker and then one orange sticker

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2 years ago

If y varies directly as x, and y is 20 when x is 4, what is the constant of variation for this relation?

arlik [135]

Answer:

5

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2 years ago

Read 2 more answers

ΔCAR has coordinates C (2, 4), A (1, 1), and R (3, 0). A translation maps point C to C' (3, 2). Find the coordinates of A' and R

shutvik [7]

Answer:

$A'= (2,-1)$ and $R'=(4,-2)$ under this translation.

Step-by-step explanation:

A translation in $R^{2}$ is a mapping T from $R^{2}$ to $R^{2}$ defined by $T(x,y) = (x + v_1,y+v_2)$ , where $v=(v_1,v_2)$ is a fixed vector in $R^{2}$ .

From the problem we know that $T(2,4)=(3,2)$ , so we need to find the values $v_1$ and $v_2$ such that $T(2,4) = (2 + v_1,4+v_2)=(3,2)$ , so $3=2 + v_1$ and $4+v_2=2$ , thus $v_1=1$ and $v_2=2$ .

Then $T(x,y) = (x + 1,y-2)$ and

$T(1,1)=(1+1,1-2)=(2,-1)=A'$

$T(3,0)=(3+1,0-2)=(4,-2)=R'$

Therefore $A'= (2,-1)$ and $R'=(4,-2)$ . The triangles CAR and C'Q'R' are shown in the figure below.

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3 years ago

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Citrus2011 [14]

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