# Find the value of k if points (k,3), (6, -2) and (-3,4) are collinear.

Last updated date: 21st Mar 2023

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Answer

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Hint: For solving this question, we should know the basics of the condition of collinearity. Further, for three points to be collinear, they should all lie on the same straight line. Thus, we would write a line equation through two points and satisfy the third point in the equation of line to get the value of k.

Complete step by step solution:

For solving the above given problem related to collinearity, we first start by writing the line equation through (6, -2) and (-3,4), since these two points are known. Line equation through two points $({{x}_{1}},{{y}_{1}})$ and $({{x}_{2}},{{y}_{2}})$ is given by –

$y-{{y}_{1}}=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}(x-{{x}_{1}})$

Thus, substituting the values, we get,

$y-(-2)=\dfrac{4-(-2)}{-3-6}(x-6)$

$y+2=\dfrac{6}{-9}(x-6)$

$y+2=-\dfrac{2}{3}(x-6)$ -- (1)

Now, for (k,3), (6, -2) and (-3,4) to be collinear, they should all lie on the same straight line. Now, we have written the line equation through points (6, -2) and (-3,4), we need to make sure that (k,3) should also satisfy this equation for the collinearity condition to hold true. Thus, we put (k,3) in place of (x,y) in equation (1).

$y+2=-\dfrac{2}{3}(x-6)$

$3+2=-\dfrac{2}{3}(k-6)$

$5=-\dfrac{2}{3}(k-6)$

-15 = 2k – 12

2k = -15+12

2k = -3

k =$-\dfrac{3}{2}$=-1.5

Hence, the value of k for points (k,3), (6, -2) and (-3,4) to be collinear is -1.5.

Note: We can also solve the same question by using the principle that slope between (k,3) and (6, -2), and (6, -2) and (-3,4) to be the same. The formula of slope between points $({{x}_{1}},{{y}_{1}})$ and $({{x}_{2}},{{y}_{2}})$, is given by $\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$. Thus, equating the slopes, we get –

$\dfrac{4-(-2)}{-3-6}$=$\dfrac{-2-3}{6-k}$

$\dfrac{6}{-9}$=$\dfrac{-5}{6-k}$

6(6-k) = 45

36-6k = 45

6k = -9

k = -1.5

Thus, we get the same answer from both the answers.

Complete step by step solution:

For solving the above given problem related to collinearity, we first start by writing the line equation through (6, -2) and (-3,4), since these two points are known. Line equation through two points $({{x}_{1}},{{y}_{1}})$ and $({{x}_{2}},{{y}_{2}})$ is given by –

$y-{{y}_{1}}=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}(x-{{x}_{1}})$

Thus, substituting the values, we get,

$y-(-2)=\dfrac{4-(-2)}{-3-6}(x-6)$

$y+2=\dfrac{6}{-9}(x-6)$

$y+2=-\dfrac{2}{3}(x-6)$ -- (1)

Now, for (k,3), (6, -2) and (-3,4) to be collinear, they should all lie on the same straight line. Now, we have written the line equation through points (6, -2) and (-3,4), we need to make sure that (k,3) should also satisfy this equation for the collinearity condition to hold true. Thus, we put (k,3) in place of (x,y) in equation (1).

$y+2=-\dfrac{2}{3}(x-6)$

$3+2=-\dfrac{2}{3}(k-6)$

$5=-\dfrac{2}{3}(k-6)$

-15 = 2k – 12

2k = -15+12

2k = -3

k =$-\dfrac{3}{2}$=-1.5

Hence, the value of k for points (k,3), (6, -2) and (-3,4) to be collinear is -1.5.

Note: We can also solve the same question by using the principle that slope between (k,3) and (6, -2), and (6, -2) and (-3,4) to be the same. The formula of slope between points $({{x}_{1}},{{y}_{1}})$ and $({{x}_{2}},{{y}_{2}})$, is given by $\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$. Thus, equating the slopes, we get –

$\dfrac{4-(-2)}{-3-6}$=$\dfrac{-2-3}{6-k}$

$\dfrac{6}{-9}$=$\dfrac{-5}{6-k}$

6(6-k) = 45

36-6k = 45

6k = -9

k = -1.5

Thus, we get the same answer from both the answers.

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