Question

For what value of y are the points P(1, 4), Q(3, y) and R (-3, 16) are collinear?

Answer 1

we have to find the value of y so that points P(1, 4) , Q(3, y) and R(-3, 16) are collinear.

solution : we know, three points are collinear, area of triangle formed by joining of these three points must be zero.

here P(1,4) , Q(3, y) and R(-3, 16) are collinear.

so, area of triangle PQR = 0

⇒1/2 [1(y - 16) + 3(16 - 4) + (-3)(4 - y)] = 0

⇒[y - 16 + 3 × 12 - 12 + 3y ] = 0

⇒[4y - 16 + 36 - 12 ] = 0

⇒4y + 8 = 0

⇒y = -2

Therefore the value of y = -2.

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Answer 2

$\textbf{Given:}$

$\textsf{The points P(1, 4), Q(3, y) and R (-3, 16) are collinear}$

$\textbf{To find:}$

$\textsf{The value of y}$

$\textbf{Solution:}$

$\textbf{Concept used:}$

$\boxed{\begin{minipage}{8cm} \mathsf{Slope\;of\;the\;line\;joining\;(x_1,y_1)\;and\;(x_2,y_2)}\\\mathsf{is\;m=\dfrac{y_2-y_1}{x_2-x_1}} \end{minipage}}$

$\textsf{Since the points P,Q and R are collinear, we have}$

$\mathsf{Slope\;\of\;PQ=slope\;of\;QR}$

$\mathsf{\dfrac{y-4}{3-1}=\dfrac{16-y}{-3-3}}$

$\mathsf{\dfrac{y-4}{2}=\dfrac{16-y}{-6}}$

$\mathsf{\dfrac{y-4}{1}=\dfrac{16-y}{-3}}$

$\mathsf{-3(y-4)=16-y}$

$\mathsf{-3y+12=16-y}$

$\mathsf{y-3y=16-12}$

$\mathsf{y-3y=16-12}$

$\mathsf{-2y=4}$

$\implies\boxed{\mathsf{y=-2}}$

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