Question

Find the equation of a line whose x-intercept is -3 and which is perpendicular to the line 3x + 5y – 8 = 0.

Answer 1

Hint : We will use the concept that the product of slopes of two perpendicular lines = -1. Also, we will consider the fact that, for a line $a{{x}_{1}}+b{{x}_{2}}+c=0$, its slope is given by $-\dfrac{a}{b}$. We will also use the general equation of line that is given by $\left( y-{{y}_{1}} \right)=m\left( x-{{x}_{1}} \right)$.

Complete step by step solution :
It is given in the question that we have to find the equation of a line whose x-intercept is -3 and which is perpendicular to the line 3x + 5y – 8 = 0.
We know that the slope of a line is given by, $-\dfrac{a}{b}$, where a is the x-intercept and b is the y-intercept.
So, in the given question, we have been given a line 3x + 5y – 8 = 0. So, the slope of this line will be $-\dfrac{3}{5}$.
Now, we know that the product of slopes of two perpendicular lines is -1. So, we can write,
${{m}_{1}}{{m}_{2}}=-1$
So, here let us consider ${{m}_{1}}$ as the slope of the line, 3x + 5y – 8 = 0 and let ${{m}_{2}}$ be the slope of the required line. So, we will get,
$\left( -\dfrac{3}{5} \right){{m}_{2}}=-1$
On dividing the whole equation by $\left( -\dfrac{3}{5} \right)$ we will get,
$\begin{aligned} & {{m}_{2}}=\dfrac{-1}{-\dfrac{3}{5}} \\\ & {{m}_{2}}=\dfrac{5}{3} \\\ \end{aligned}$
So, the slope of the required line is $\dfrac{5}{3}$. Also, we have been given the question that the x-intercept of the required line is -3. So, that line will pass through the point (-3,0).
We know that the general equation of the line is given by, $\left( y-{{y}_{1}} \right)=m\left( x-{{x}_{1}} \right)$.
So, on putting ${{y}_{1}}=0,{{x}_{1}}=-3,m=\dfrac{5}{3}$ in the general equation, we get,

& \left( y-0 \right)=\dfrac{5}{3}\left( x-\left( -3 \right) \right) \\\ & \left( y-0 \right)=\dfrac{5}{3}\left( x+3 \right) \\\ & y=\dfrac{5\left( x+3 \right)}{3} \\\ & y=\dfrac{5x+15}{3} \\\ \end{aligned}$$ On cross multiplying the above equation, we get, 3y = 5x + 15. On transposing 5x and 15 from RHS to the LHS, we get, 3y – 5x – 15 = 0 On multiplying both the sides with (-1), we get, 5x – 3y + 15 = 0 Thus, the equation of the line whose x-intercept is -3 and which is perpendicular to the line 3x + 5y – 8 = 0 is 5x – 3y + 15 = 0. **Note** : It is observed that many students put the value of ${{x}_{1}}=3$ in the general equation of line as a result, they may get the final answer as, 5x – 3y – 15 = 0, which is wrong. So, the students must make sure that they substitute the correct values of the variables. We can also transform the equation of the line, 3x + 5y – 8 = 0 in the form of y = mx + c and then find slope as m. So, we get 5y = -3x + 8, which can be written as, $y=-\dfrac{3}{5}+\dfrac{8}{5}$ and on comparing this with y = mx + c, we get the slope as ${{m}_{1}}=-\dfrac{3}{5}$. Then, we can solve the question further that is by finding the slope of the required line, ${{m}_{2}}$ by using the property, ${{m}_{1}}{{m}_{2}}=-1$ and then finding the equation of the required line.