Question

if the point( x, y,), ( 2, 3) and (- 3, 4) are collinear then

Answer 1

$\textbf{Given:}$

$\text{The points (x,y), (2,3) and (-3,4)}$

$\textbf{To find:}$

$\text{The relation connecting x and y}$

$\textbf{Solution:}$

$\text{We know that}$

$\textbf{Slope of the line joining \bf(x_1,y_1) and \bf(x_2,y_2) is \bf\dfrac{y_2-y_1}{x_2-x_1} }$

$\text{Let the given points be A(x,y), B(2,3) and C(-3,4)}$

$\text{Since the points are collinear, we have}$

$\text{Slope of AB=Slope of BC}$

$\dfrac{3-y}{2-x}=\dfrac{4-3}{-3-2}$

$\dfrac{3-y}{2-x}=\dfrac{1}{-5}$

$-5(3-y)=2-x$

$-15+5y=2-x$

$\text{Rearranging terms, we get}$

$\bf\,x+5y-17=0$

$\therefore\textbf{The required relation is x+5y-17=0}$

Find more:

If (a,0) (b,0) (1,1) are these three points in a line than find the value of 1/a + 1/b = ?​

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k का मान ज्ञात कीजिए यदि बिंदु a (2,3) b (4,k)c (6, - 3)संरेखी है हिंदी मीडियम

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Answer 2

Answer:

\textbf{Given:}Given:

\text{The points (x,y), (2,3) and (-3,4)}The points (x,y), (2,3) and (-3,4)

\textbf{To find:}To find:

\text{The relation connecting x and y}The relation connecting x and y

\textbf{Solution:}Solution:

\text{We know that}We know that

\textbf{Slope of the line joining $\bf(x_1,y_1)$ and $\bf(x_2,y_2)$ is $\bf\dfrac{y_2-y_1}{x_2-x_1}$}Slope of the line joining (x1,y1) and (x2,y2) is x2−x1y2−y1

\text{Let the given points be A(x,y), B(2,3) and C(-3,4)}Let the given points be A(x,y), B(2,3) and C(-3,4)

\text{Since the points are collinear, we have}Since the points are collinear, we have

\text{Slope of AB=Slope of BC}Slope of AB=Slope of BC

\dfrac{3-y}{2-x}=\dfrac{4-3}{-3-2}2−x3−y=−3−24−3

\dfrac{3-y}{2-x}=\dfrac{1}{-5}2−x3−y=−51

-5(3-y)=2-x−5(3−y)=2−x

-15+5y=2-x−15+5y=2−x

\text{Rearranging terms, we get}Rearranging terms, we get

\bf\,x+5y-17=0x+5y−17=0

\therefore\textbf{The required relation is x+5y-17=0}∴The required relation is x+5y-17=0