Question

find the value of p for which the points are (8,p),(3,-4 and (2,-5) are collinear
​

Answer 1

Given:

Three points- (8,p),(3,-4) and (2,-5) which are collinear

To Find:

Value of p

Solution:

We know that,

• Area of triangle formed by three collinear points  is equal to zero.
• Area of triangle formed by three points (x₁,y₁), (x₂,y₂) and (x₃,y₃) is given by

$\sf{Area\:of\:\triangle=\Bigg|\dfrac{1}{2}\Big(x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})\Big)\Bigg|}$

Now, here

Let  

• A(8,p)⇒(x₁,y₁)

• B(3,-4)⇒(x₂,y₂)

• C(2,-5)⇒(x₃,y₃)

So,

$\sf{Area\:of\:\triangle ABC=\dfrac{1}{2}\Bigg|\Big(8(-4-(-5))+3(-5-p)+2(p-(-4))\Big)\Bigg|}$

​$\sf{Area\:of\:\triangle ABC=\dfrac{1}{2}\Bigg|\Big(8(-4+5)+3(-5-p)+2(p+4)\Big)\Bigg|}$

$\sf{Area\:of\:\triangle ABC=\dfrac{1}{2}\Bigg|\Big(8(1)+3(-5-p)+2(p+4)\Big)\Bigg|}$

$\sf{Area\:of\:\triangle ABC=\dfrac{1}{2}\Bigg|\Big(8-15-3p+2p+8\Big)\Bigg|}$

$\sf{Area\:of\:\triangle ABC=\dfrac{1}{2}\Bigg|\Big(1-p\Big)\Bigg|}$

$\sf{Area\:of\:\triangle ABC=\dfrac{1-p}{2}}----(1)$

Also,

$\sf{Area\:of\:\triangle ABC=0}------(2)$

From (1) and (2), we get

$\longrightarrow\mathrm{\dfrac{1-p}{2}=0}$

$\longrightarrow\mathrm{1-p=0}$

$\longrightarrow\mathrm{1=p}$

$\longrightarrow\mathrm{\pink{p=1}}$

Hence, the value of p is 1.