Question

if (p,q) lies on line joining (3,-2) and (-5,3), Prove that 5p+8q+1=0

Answer 1

Question :

If (p,q) lies on line joining (3,-2) and (-5,3), Prove that 5p+8q+1=0.

Theory :

1) Slope of a line joining two points $\sf(x_1,y_1)\:and\:(x_2,y_2)$ is given by :

$\rm\:m=\dfrac{y_2-y_1}{x_2-x_1}$

2) Two point form :

The equation of line passing through two given points $\sf(x_1,y_1)\:and\:(x_2,y_2)$ is:

$\sf\:y-y_1=\dfrac{y_2-y_1}{x_2-x_1}\times(x-x_1)$

Solution :

Let the points be A ( 3,-2) and B (-5,3)

Then ,the slope of line AB is

$\tt\:m=\dfrac{3+2}{-5-3}=\dfrac{5}{-8}$

Now equation of line AB ,by two point form is

$\sf\:y-y_1=\dfrac{-5}{8}\times(x-x_1)$

Since (p,q) lines on the line AB, then equation of line is

$\tt\:(y-q)=\dfrac{-5}{8}\times(x-p)$

$\tt\:-8y+8q=5x-5p$

$\tt\:5p+8q=5x+8y$

Since (3,-2) is a point on the line , thus it statisfie the equation of line

$\implies\tt\:5p+8q=5(3)+8(-2)$

$\implies\tt\:5p+8q=15-16$

$\implies\tt\:5p+8q+1=0$

Hence Proved

Answer 2

Question

If (p,q) lies on line joining (3,-2) and (-5,3), Prove that 5p+8q+1=0.

Solution :

Let X( 3,-2) and Y ( -5,3)

Now slope of line XY

$\rm\:m=\dfrac{y_2-y_1}{x_2-x_1}$

$\rm\:m=\dfrac{-5}{8}$

Equation of line by two point form :

$\sf\:y-y_1=\dfrac{-5}{8}\times(x-x_1)$

It is given that (p,q) lie on the line XY

$\sf\implies\:y-p=\dfrac{-5}{8}\times(x-q)$

$\sf\implies\dfrac{y-p}{x-q}=\dfrac{-5}{8}$

Now croos multiply the terms

$\sf\:-8y+8q=5x-5p$

Since (-5,3) is a point from which the line XY passes then

$\sf\implies\:-8y+8q=5x-5p$

$\sf\implies\:-8(3)+8q=5(-5)-5p$

$\sf\implies\:-24+25+5p+8q=0$

$\implies\sf\:5p+8q+1=0$

Hence Proved