Question

Find the rate of change of a line that passes through the points (20,-3) and (19,-20)

Answer 1

Question :- Find the rate of change of a line that passes through the points (20,-3) and (19,-20) ?

Formula used :-

• The rate of change of a line passes through (x₁, x₂) and (y₁, y₂) is Equal to slope of line = (y₂ - y₁) / (x₂ - x₁) .

Solution :-

we have given that :-

• x₁ = 20
• x₂ = 19
• y₁ = (-3)
• y₂ = (-20)

Putting all values Now, we get :-

→ Rate of change of line = Slope of line = (y₂ - y₁) / (x₂ - x₁) .

→ Slope = {(-20) - (-3)} / (19 - 20)

→ Slope = {(-20) + 3} / (-1)

→ Slope = (3 - 20) / (-1)

→ Slope = (-17) / (-1)

→ Slope = 17 .

Therefore,

Rate of change of line is 17 .

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Extra :-

✰) To find the distance between two points say P (x₁,y₁) and Q(x₂,y₂) is :

$\tt\sqrt{{(x_2-x_1)}^{2}+{(y_2-y_1)}^{2}}$

✰) To find the distance of a point say P (x,y) from origin is :

$\pink{\bf\sqrt {{x}^{2}+{y}^{2}}}$

✰) Coordinates of the point P (x,y) which device the line segment joining the points A (x₁,y₁) and B (x₂,y₂)

internally in the ratio m₁ : m₂ is :

$\red{\sf\dfrac {m_1 x_2 + m_2 x_1}{m_1+m_2},,,\dfrac {m_1 y_2 + m_2 y_1}{m_1+m_2}}$

✰) The midpoint of the line segment joining the points P (x₁,y₁) and Q(x₂,y₂) is : ---

$\purple{\tt\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}}$

✰) The area of the triangle formed by the points (x₁,y₁)(x₂,y₂) and (x_3,y_3) is the numerical value of the expression:----

$\green{\tt\dfrac{1}{2}[x_1(y_2-y_3)+x_2 (y_3-y_1)+x_3(y_1-y_2)]}$

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Answer 2

Question:-

Find the rate of change of a line that passes through the points (20,-3) and (19,-20) .

Solution:-

To find the rate of change of a

line we find slope of the line that passes through the points (20,-3) and (19,-20) .

Let, (x1 , y1) = (20 , - 3) & (x2 , y2)

= (19, - 20)

As we know slope is denoted

by " m "

Now to find slope we use formula

=> m = [y2 - y1] / [x2 - x1]

=> m = [- 20 - (- 3)] / [19 - 20]

=> m = [- 20 + 3] / [-1 ]

=> m = -17 / -1

=> m = 17

Hence, rate of change of line is 17 .

Extra info :-

${\tt{\red{1.{\purple{equation \: of \: the \: line \: having }}}}} \\ {\tt{\red{{\purple{slope \: m \: and \: y - intercept \: c \: is}}}}} \\ {\tt{\blue{\underline {\blue{\pink{y = mx + c}}}}}} \\ {\tt{\red{2.{\purple{if \: x \: and \: y \: intecepts\: of \: a \: line }}}}} \\ {\tt{\red{{\purple{are \: a \: and \: b \: respectively \: then }}}}} \\ {\tt{\red{{\purple{eq. \: of \:the \: line \: is \:{ \pink{ \frac{x}{a} + \frac{y}{b} = 1. }}}}}}}$

i hope it helps you.