Question

read the instructions first and answer the questions​

Answer 1

Step-by-step explanation:

1)Yes,

2)Converting into linear equation by using slope and point and then write it as slope and intercept form then write the points and locate them on the Cartesian plane.

Answer 2

Concept Used :-

Let us consider a line having slope m and passes through the point (a, b), then equation of line is given by

$\bf \:  ⟼ y - b = m(x - a)$

$\bf \:\large \red{AηsωeR : 1.} ✍$

Equation of line having slope, m = 2 and passes through the point (3, 1) is

$\bf \:  ⟼ y - 1 = 2(x - 3)$

$\bf \:  ⟼ y - 1 = 2x - 6$

$\bf \:  ⟼ 2x - y - 5 = 0 \: \sf \: ⟼(1) \:$

$\bf \:\large \red{AηsωeR : 2.} ✍$

Equation of line having slope, m = 3/2 and passes through the point (0, 1) is

$\bf \:  ⟼ y - 1 = \dfrac{3}{2} (x - 0)$

$\bf \:  ⟼ 2y - 2 = 3x$

$\bf \:  ⟼ 3x - 2y + 2 = 0 \: ⟼(2)$

$\bf \:\large \red{AηsωeR : 3.} ✍$

Equation of line having slope, m = - 1 and passes through the point (- 1, 4) is

$\bf \:  ⟼ y - 4 = - 1(x + 1)$

$\bf \:  ⟼ y - 4 = - x - 1$

$\bf \:  ⟼ x + y - 3 = 0 \: ⟼ \: (3)$

$\bf \:\large \red{AηsωeR : 4.} ✍$

Equation of line having slope, m = - 3 and passes through the point (2, - 1) is

$\bf \:  ⟼ y + 1 = - 3(x - 2)$

$\bf \:  ⟼ y + 1 = - 3x + 6$

$\bf \:  ⟼ 3x + y - 5 = 0 \: ⟼ \: (4)$

Points on Line (1) :-

$\begin{gathered}\boxed{\begin{array}{c|c|c|c }\bf x&\sf 0&\sf 2.5&\s\\\frac{\qquad \quad \qquad}{}&\frac{\quad \qquad \qquad}{}&\frac{\qquad \quad\qquad}{}&\frac{\qquad \quad \qquad}{}&\\\bf y&\sf - 5&\sf 0&\end{array}}\end{gathered}$

Points on line (2)

$\begin{gathered}\boxed{\begin{array}{c|c|c|c }\bf x&\sf 0&\sf 2&\s\\\frac{\qquad \quad \qquad}{}&\frac{\quad \qquad \qquad}{}&\frac{\qquad \quad\qquad}{}&\frac{\qquad \quad \qquad}{}&\\\bf y&\sf 1&\sf 4&\end{array}}\end{gathered}$

Points on line (3)

$\begin{gathered}\boxed{\begin{array}{c|c|c|c }\bf x&\sf 0&\sf 3&\s\\\frac{\qquad \quad \qquad}{}&\frac{\quad \qquad \qquad}{}&\frac{\qquad \quad\qquad}{}&\frac{\qquad \quad \qquad}{}&\\\bf y&\sf 3&\sf 0&\end{array}}\end{gathered}$

Points on line (4)

$\begin{gathered}\boxed{\begin{array}{c|c|c|c }\bf x&\sf 0&\sf 2&\s\\\frac{\qquad \quad \qquad}{}&\frac{\quad \qquad \qquad}{}&\frac{\qquad \quad\qquad}{}&\frac{\qquad \quad \qquad}{}&\\\bf y&\sf -5&\sf - 1&\end{array}}\end{gathered}$

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1. The graph is plot on the Cartesian Plane.

2. Yes, we can graph a line by getting the equation of line using slope point form.