Question

the slope of a line -3. The line passing through thr point (2, 3).Write the coordinates of other two points on this line​

Answer 1

$\large\underline\blue{\bold{Given \:  Question :-  }}$

• The slope of a line is -3. The line passing through the point (2, 3). Write the coordinates of other two points on this line.

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$\huge{AηsωeR} ✍$

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$\begin{gathered}\begin{gathered}\bf Given - \begin{cases} &\sf{slope \: of \: line \: is \: - 3} \\ &\sf{line \: passing \: through \: point \: (2, 3)} \end{cases}\end{gathered}\end{gathered}$

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$\begin{gathered}\begin{gathered}\bf To \: find :- \begin{cases} &\sf{two \: coordinates \: on \: the \: line} \end{cases}\end{gathered}\end{gathered}$

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$\begin{gathered}\begin{gathered}\bf Formula \: used:- \begin{cases} &\sf{\bf \:y - y_1 = m(x-x_1)} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf where= \begin{cases} &\sf{(x_1,y_1) \: is \: the \: point \: from \: where \: lne \: passed } \\ &\sf{m \: is \: the \: slope \: of \: line} \end{cases}\end{gathered}\end{gathered}$

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$\large\underline\blue{\bold{Solution :- }}$

It is given that line passes through the point (2, 3) and having slope, m = - 3.

So, equation of line is given by

$\bf \:y - y_1 = m(x-x_1)$

$\begin{gathered}\begin{gathered}\bf where= \begin{cases} &\sf{ (x_1,y_1) } = (2, 3) \\ &\sf{m = - 3} \end{cases}\end{gathered}\end{gathered}$

$\sf \:  ⟼y - 3 = - 3 \times (x - 2)$

$\sf \:  ⟼y - 3 = - 3x + 6$

$\sf \:  ⟼3x + y = 9\sf \:  ⟼(1)$

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☆ Now, to find the coordinates on line (1)

$\large\underline\red{\bold{❥︎Step :- 1 }}$

$\sf \:  ⟼❶ Substituting \: 'x = 0' \: in \: (1), we \: get$

$\begin{gathered}:\implies\:\:\sf{3\times{0}\:+\:y\:=\:9} \\ \end{gathered}$

$\begin{gathered}:\implies\:\:\sf{{0}\:+\:y\:=\:9} \\ \end{gathered}$

$\begin{gathered}:\implies\:\:\sf{\:y\:=\:9} \\ \end{gathered}$

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$\large\underline\red{\bold{❥︎Step :- 2 }}$

$\sf \:  ⟼ ❷ Substituting \: 'x = 1' \: in \: (1), \: we \: get$

$\begin{gathered}:\implies\:\:\sf{3\times{1}\:+\:y\:=\:9} \\ \end{gathered}$

$\begin{gathered}:\implies\:\:\sf{{3}\:+\:y\:=\:9} \\ \end{gathered}$

$\begin{gathered}:\implies\:\:\sf{\:y\:=\:6} \\ \end{gathered}$

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Hᴇɴᴄᴇ,

➢ Pair of points of the given equation are shown in the below table.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 9 \\ \\ \sf 1 & \sf 6 \end{array}} \\ \end{gathered}$

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Answer 2

Step-by-step explanation:

Given Question:−

The slope of a line is -3. The line passing through the point (2, 3). Write the coordinates of other two points on this line.

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\huge{AηsωeR} ✍AηsωeR ✍

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\begin{gathered}\begin{gathered}\begin{gathered}\bf Given - \begin{cases} &\sf{slope \: of \: line \: is \: - 3} \\ &\sf{line \: passing \: through \: point \: (2, 3)} \end{cases}\end{gathered}\end{gathered}\end{gathered}

Given−{

slopeoflineis−3

linepassingthroughpoint(2,3)

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\begin{gathered}\begin{gathered}\bf To \: find :- \begin{cases} &\sf{two \: coordinates \: on \: the \: line} \end{cases}\end{gathered}\end{gathered}

Tofind:−{

twocoordinatesontheline

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\begin{gathered}\begin{gathered}\bf Formula \: used:- \begin{cases} &\sf{\bf \:y - y_1 = m(x-x_1)} \end{cases}\end{gathered}\end{gathered}

Formulaused:−{

y−y

1

=m(x−x

1

)

\begin{gathered}\begin{gathered}\begin{gathered}\bf where= \begin{cases} &\sf{(x_1,y_1) \: is \: the \: point \: from \: where \: lne \: passed } \\ &\sf{m \: is \: the \: slope \: of \: line} \end{cases}\end{gathered}\end{gathered}\end{gathered}

where={

(x

1

,y

1

)isthepointfromwherelnepassed

mistheslopeofline

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\large\underline\blue{\bold{Solution :- }}

Solution:−

It is given that line passes through the point (2, 3) and having slope, m = - 3.

So, equation of line is given by

\bf \:y - y_1 = m(x-x_1)y−y

1

=m(x−x

1

)

\begin{gathered}\begin{gathered}\begin{gathered}\bf where= \begin{cases} &\sf{ (x_1,y_1) } = (2, 3) \\ &\sf{m = - 3} \end{cases}\end{gathered}\end{gathered}\end{gathered}

where={

(x

1

,y

1

)=(2,3)

m=−3

\sf \: ⟼y - 3 = - 3 \times (x - 2) ⟼y−3=−3×(x−2)

\sf \: ⟼y - 3 = - 3x + 6 ⟼y−3=−3x+6

\sf \: ⟼3x + y = 9\sf \: ⟼(1) ⟼3x+y=9 ⟼(1)

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☆ Now, to find the coordinates on line (1)

\large\underline\red{\bold{❥︎Step :- 1 }}

❥︎Step:−1

\sf \: ⟼❶ Substituting \: 'x = 0' \: in \: (1), we \: get ⟼❶Substituting

′

x=0

′

in(1),weget

\begin{gathered}\begin{gathered}:\implies\:\:\sf{3\times{0}\:+\:y\:=\:9} \\ \end{gathered}\end{gathered}

:⟹3×0+y=9

\begin{gathered}\begin{gathered}:\implies\:\:\sf{{0}\:+\:y\:=\:9} \\ \end{gathered}\end{gathered}

:⟹0+y=9

\begin{gathered}\begin{gathered}:\implies\:\:\sf{\:y\:=\:9} \\ \end{gathered}\end{gathered}

:⟹y=9

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\large\underline\red{\bold{❥︎Step :- 2 }}

❥︎Step:−2

\sf \: ⟼ ❷ Substituting \: 'x = 1' \: in \: (1), \: we \: get ⟼❷Substituting

′

x = 1

′

in(1),weget

\begin{gathered}\begin{gathered}:\implies\:\:\sf{3\times{1}\:+\:y\:=\:9} \\ \end{gathered}\end{gathered}

:⟹3×1+y=9

\begin{gathered}\begin{gathered}:\implies\:\:\sf{{3}\:+\:y\:=\:9} \\ \end{gathered}\end{gathered}

:⟹3+y=9

\begin{gathered}\begin{gathered}:\implies\:\:\sf{\:y\:=\:6} \\ \end{gathered}\end{gathered}

:⟹y=6

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Hᴇɴᴄᴇ,

➢ Pair of points of the given equation are shown in the below table.

\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 9 \\ \\ \sf 1 & \sf 6 \end{array}} \\ \end{gathered}\end{gathered}

x

0

1

y

9

6