Question

A line which passes through $\left( {5,6} \right)$ and $\left( { - 3, - 4} \right)$ has an equation of
A.$5x + 4y + 1 = 0$
B.$5x - 4y - 1 = 0$
C.$5x - 4y + 1 = 0$
D.$5x + y - 1 = 0$
E.None of these

Answer 1

Hint: In the solution, first we have to find the slope of the straight line by using the points $\left( {5,6} \right)$ and$\left( { - 3, - 4} \right)$. Since the general form of the straight line is known, we need to substitute the value of the slope and the given points on that expression. This will give us the equation of the straight line.

Complete step by step solution:
The given points are $\left( {5,6} \right)$ and$\left( { - 3, - 4} \right)$.
Let $m$ be the slope of the line.
It is known that the slope of the straight line is$m = \dfrac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}$.
Substituting the value 5 for${x_1}$, 6 for${y_1}$, $ - 3$ for ${x_2}$, $ - 4$ for ${y_2}$.
$\begin{array}{l}m = \dfrac{{ - 4 - 6}}{{ - 3 - 5}}\\ \Rightarrow m = \dfrac{{ - 10}}{{ - 8}}\\ \Rightarrow m = \dfrac{5}{4}\end{array}$
The general form of the straight line is$y = mx + c$. Where $m$ is the slope of the line and $c$ is the$y$ intercept.
Now, we have to calculate the value of$c$.
Substituting the value $\dfrac{5}{4}$ for$m$, 5 for $x$ and 6 for$y$.
$\begin{array}{l}y = mx + c\\ \Rightarrow 6 = \dfrac{5}{4} \times 5 + c\\ \Rightarrow c = \dfrac{{25}}{4} - 6\\ \Rightarrow c = \dfrac{1}{4}\end{array}$

Thus the equation of the straight line is
$\begin{array}{l}y = mx + c\\ \Rightarrow y = \dfrac{5}{4}x + \dfrac{1}{4}\\ \Rightarrow 4y = 5x + 1\\ \Rightarrow 5x - 4y + 1 = 0\end{array}$
Hence, the correct option is C.

Note: The slope of a line is a measure of how fast the line goes up or goes down. Since the points from which the straight line passes are given. So we can find the slope by substituting the given points. Once we get the slop, we can find the $y$ intercept by using the value of the slop and the given points. Then finally we can get our required equation of the line by using the slope and the $y$ intercept. Here for calculating the$y$ intercept, instead of the point $\left( {5,6} \right)$ for $\left( {x,y} \right)$ we can put $\left( { - 3, - 4} \right)$ also.