Question

The equation of the line joining the point $(3,5)$ to the point of intersection of the lines $4x+y-1=0$and $7x-3y-35=0$ is equidistant from the points $(0,0)$ and $(8,34)$.

• A. True
• B. False
• C. Nothing can be said
• D. None of these

Answer 1

The correct option is A

True

Explanation for the correct answer:

Finding the equation of line,

Let the required line be $AB$

Given equations are: $4x+y-1=0\dots \dots \left(1\right)$and $7x-3y-35=0\dots \dots \left(2\right)$

Finding the point of intersection of the two lines by multiplying both equations by $3$

$$\begin{gathered} 3\left(4x+y-1\right)=0 \\ 12x+3y-3=0\dots \dots \left(3\right) \end{gathered}$$

Adding equations $\left(2\right)$ and $\left(3\right)$ we get

$$\begin{gathered} 7x-3y-35+12x+3y-3=0 \\ 19x-38=0 \\ 19x=38 \\ x=2 \end{gathered}$$

Substituting the value of $x$ in equation$\left(1\right)$ we get

$$\begin{gathered} 4x+y-1=0 \\ 4\left(2\right)+y-1=0 \\ 8+y-1=0 \\ y+7=0 \\ y=-7 \end{gathered}$$

Therefore the point of intersection is $(2,-7)$

Equation of line $AB$is given by,

$$\begin{gathered} y-y_1=\frac{y_2-y_1}{x_2-x_1}\left(x-x_1\right) \\ y-3=\frac{5+7}{3-2}(x-5) \\ \Rightarrow (y-3)=12(x-5) \\ \Rightarrow 12x-y-31=0 \end{gathered}$$

The distance of $AB$ from $(0,0)$

$$\begin{gathered} d=\frac{\left|a{x}_1+b{y}_1+c\right|}{\sqrt{a^2+b^2}} \\ =\frac{\left|12\left(0\right)-y\left(0\right)-31\right|}{\sqrt{{12}^2+1^2}} \\ =\frac{31}{\sqrt{145}} \end{gathered}$$

The distance of $AB$ from $(8,34)$

$$\begin{gathered} d=\frac{\left|a{x}_2+b{y}_2+c\right|}{\sqrt{a^2+b^2}} \\ =\frac{\left|12\left(8\right)-y\left(34\right)-31\right|}{\sqrt{{12}^2+1^2}} \\ =\frac{31}{\sqrt{145}} \end{gathered}$$

We see that $AB$ is equidistant from both the points $(0,0)$ and $(8,34)$.

Therefore, the correct answer is Option (A).