Question

Find the equation of each line in standard form with the given properties:

1. SLOPE = 3, Y INTERCEPT − = 1
2. Passing through (0,2), SLOPE= −4
3. passing through (−1,3) and (1,1)
4. passing through (1,3), SLOPE = ½
5. passing through (1/2, 1) and (4,2)
MAY BRAINLIEST TAMANG SAGOT

Answer 1

Step-by-step explanation:

Solution

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We know that the equation of line passing through point (x

0

,y

0

) whose slope is m is

(y−y

0

)=m(x−x

0

)

Thus the equation of line passing through the point (−4,3) whose slope is

2

1

is,

(y−3)=

2

1

(x+4)

⇒2(y−3)=(x+4)

⇒x−2y+10=0

Answer 2

Given: Properties of different lines are given

1. Slope = 3, y- intercept = 1

2. Passing through (0,2), slope= −4

3. passing through (−1,3) and (1,1)

4. passing through (1,3), slope = ½

5. passing through (1/2, 1) and (4,2)

To find: Equation of the lines

Solution: 1- Slope=3 and y intercept= 1

The equation can be written in slope intercept form y = mx + c where m is the slope and c is the y intercept.

Using the form,

The equation is y = 3x+1.

2- Passing through (0,2), slope= −4

The equation can be written in slope point form y-y1 = m(x-x1) where m is the slope and (x1,y1) is the point.

Using the form:

y-2 = -4 (x-0)

=> y -2 = -4x

=> 4x+y = 2

The equation of the line is 4x+y=2.

3-passing through (−1,3) and (1,1)

The equation can be written in two point form

$y - y1 = \frac{y2 - y1}{x2 - x1} (x - x1)$

where (x1,y1)= (-1,3) and (x2,y2)= (1,1)

Using the form,

$y - 3 = \frac{1 - 3}{1 - ( - 1)} (x - ( - 1))$

$= > y - 3 = - 1(x + 1)$

=> y-3 = -x -1

=> x+y = 2

The equation of the line is x+y=2.

4-passing through (1,3), Slope = ½

The equation can be written in slope point form y-y1 = m(x-x1) where m is the slope and (x1,y1) is the point.

Using the form,

y-3 = (1/2) ( x-1)

=> 2(y-3) = x-1

=> 2y-6 = x-1

=> x-2y = -5

The equation of the line is x-2y= -5.

5-passing through (1/2, 1) and (4,2)

The equation can be written in two point form

$y - y1 = \frac{y2 - y1}{x2 - x1} (x - x1)$

where (x1,y1) = (4,2) and (x2,y2)= (1/2,1)

Using the form,

$y - 2 = \frac{1 - 2}{ \frac{1}{2} - 4} (x -4)$

$= > y - 2 = \frac{ - 1}{ - \frac{7}{2} } (x - 4)$

$= > y - 2 = \frac{2}{7} (x - 4)$

=> 7(y-2) = 2(x-4)

=> 7y - 14 = 2x -8

=> 2x - 7y +6 = 0

The equation of the line is 2x-7y+6=0.