Question

Directions: 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 = 1/2
5. Passing through (1/2,1) and (4,2)
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Answer 1

Answer :- 1

slope, m = 3

y - intercept, c = 1

So, equation of line using slope intercept form is

y = mx + c

⇛ y = 3x + 1

⇛ 3x - y + 1 = 0

Answer :- 2

Passing through the point (0, 2) and slope = - 4.

Point, (a, b) = (0, 2)

slope, m = - 4

So, equation of line using slope point form is

y - b = m(x - a)

⇛ y - 2 = - 4(x - 0)

⇛ y - 2 = - 4x

⇛4x + y - 2= 0

Answer :- 3

Passing through (-1, 3) and (1, 1).

Equation of line using two point form is

$\bf \:y-y_1 = \dfrac{y_2-y_1}{x_2-x_1}(x-x_1)$

$\bf\implies \:x_2 = 1,x_1 = - 1,y_2 = 1,y_1 = 3$

on substituting the values, we get

$\bf\implies \:y -3 = \dfrac{1 - 3}{1 - ( - 1)} (x + 1)$

$\bf\implies \:y - 3 = - 1(x + 1)$

$\bf\implies \:y - 3 = - x - 1$

$\bf\implies \:x + y - 2 = 0$

Answer :- 4

Passing through the point (1, 3) and slope = 1/2.

Point, (a, b) = (1, 3)

slope, m = 1/2

So, equation of line using slope point form is

y - b = m(x - a)

⇛ y - 3 = 1/2 (x - 1)

⇛ 2y - 6 = x - 1

⇛x - 2y + 5 = 0

Answer :- 3

Passing through (1/2, 1) and (4, 2).

Equation of line using two point form is

$\bf \:y-y_1 = \dfrac{y_2-y_1}{x_2-x_1}(x-x_1)$

On substituting the values

$\bf\implies \:x_2 = \frac{1}{2} ,x_1 = 4,y_2 = 1,y_1 = 2$

we get,

$\bf\implies \:y - 2 = \dfrac{1 - 2}{ \frac{1}{2} - 4} (x - 4)$

$\bf\implies \:y - 2 = \frac{2}{7} (x - 4)$

$\bf\implies \:7y - 14 = 2x - 8$

$\bf\implies \:2x - 7y + 6 = 0$

Answer 2

Answer:

this is correct..........................