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Home  >>  CBSE XI  >>  Math  >>  Straight Lines
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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)

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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.
answered Jul 9, 2014 by thanvigandhi_1
 

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