# State whether the following statement is true or false and justify : 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)

Toolbox:
• Equation of line joining two points $(x_1, y_1)$ and $(x_2, y_2)$ is $\large\frac{y-y_1}{y_2-y_1}$$= \large\frac{x-x_1}{x_2-x_1} • Distance of a line ax+by+c=0 from (x_1, y_1) is d = \bigg| \large\frac{ax_1+by_1+c}{\sqrt {a^2+b^2}} \bigg| Step 1 : Let us find the point of intersection of the lines 4x+y-1=0 and 7x-3y-35=0. ( \times 3 ) 4x+y=1 \qquad 7x - 3y=35 _________________ \qquad 12x+ \not{3}\not{y} = 3 \qquad 7x - \not{3}\not{y} = 35 _________________ \qquad 19x \qquad = 38 \Rightarrow x = 2 4(2)+y=1 \quad \Rightarrow y = -7 Hence the point of intersection is (2, -7) Hence the equation of the line joining the point (3,5) and (2, -7) is \large\frac{y-5}{-7-5}$$ = \large\frac{x-3}{2-3}$
$\large\frac{y-5}{-12}$$=\large\frac{x-3}{-1}$
$\Rightarrow -y+5 = -12x+36$
$\Rightarrow 12x-y=31$
Distance of this line from (0,0) is
$d_1 = \bigg| \large\frac{12(0)-1(0)-31}{\sqrt{12^2+1^2}} \bigg|$
$= \large\frac{31}{\sqrt{145}}$
$d_2 = \bigg| \large\frac{12(8)-1(34)-31}{\sqrt{12^2+1^2}} \bigg|$
$= \bigg| \large\frac{96-34-31}{\sqrt{145}} \bigg|$
$= \large\frac{31}{\sqrt {145}}$
$\therefore d_1 = d_2$
Hence it is a true statement.