Question

mathdude can you help me activity2 and activity 3​

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 (-1, 3) and slope = - 4.

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

slope, m = - 4

So, equation of line using slope point form is

y - b = m(x - a)

⇛ y - 3 = - 4(x + 1)

⇛ y - 3 = - 4x - 4

⇛4x + y + 1 = 0

Answer :- 3

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

Equation of line using two point form is

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

On substituting the values,

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

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 :- 5

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,

$\bf\implies \:x_2 = 4,x_1 = \dfrac{1}{2} ,y_2 = 2,y_1 = 3$

we get

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

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

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

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