Question

How do you write the equation in point slope form given that the two points are (3, 8) and (9, 0)?

Answer 1

To write the equation in point slope form given that two points are given by (3, 8) and (9, 0). We know that if we have given the two points say $\left( {{x}_{1}},{{y}_{1}} \right)\And \left( {{x}_{2}},{{y}_{2}} \right)$ then we can write the equation of straight line passing through these two points as follows: $y-{{y}_{1}}=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\left( x-{{x}_{1}} \right)$. This equation of a straight line is in a point slope form. Similarly, we can write the equation of a straight line passing through two points (3, 8) and (9, 0).

Complete step-by-step solution:
In the above problem, we have given two points (3, 8) and (9, 0) and we are asked to write the equation of a straight line passing through these two points and are in the point slope form.
The point – slope form for any two points say $\left( {{x}_{1}},{{y}_{1}} \right)\And \left( {{x}_{2}},{{y}_{2}} \right)$ is given as follows:
$y-{{y}_{1}}=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}\left( x-{{x}_{1}} \right)$ …………….. (1)
Now, comparing the two points given in the above problem to $\left( {{x}_{1}},{{y}_{1}} \right)\And \left( {{x}_{2}},{{y}_{2}} \right)$ we get,
$\begin{aligned} & {{x}_{1}}=3,{{y}_{1}}=8; \\\ & {{x}_{2}}=9,{{y}_{2}}=0 \\\ \end{aligned}$
Now, we are going to substitute the above values in eq. (1) we get,
$\begin{aligned} & \Rightarrow y-8=\dfrac{0-8}{9-3}\left( x-3 \right) \\\ & \Rightarrow y-8=\dfrac{-8}{6}\left( x-3 \right) \\\ \end{aligned}$
In the above equation, the numerator and denominator of the fraction written in the R.H.S will be divided by 2 and we get,
$\Rightarrow y-8=\dfrac{-8}{6}\left( x-3 \right)$
Hence, we have got the point slope form for the two points given in the above problem.

Note: You can even further simplify the equation of straight line which we have written in the above solution as follows:
$\Rightarrow y-8=\dfrac{-8}{6}\left( x-3 \right)$
Multiplying 6 on both the sides of the above equation we get,
$\begin{aligned} & \Rightarrow 6\left( y-8 \right)=-8\left( x-3 \right) \\\ & \Rightarrow 6y-48=-8x+24 \\\ \end{aligned}$
Adding 48 on both the sides of the above equation we get,

& \Rightarrow 6y=-8x+24+48 \\\ & \Rightarrow 6y=-8x+72 \\\ \end{aligned}$$ Adding 8x on both the sides of the above equation we get, $\Rightarrow 6y+8x=72$