Question

The vertices of a triangle OBC are (0, 0), (−3, −1) & (−1, 3) respectively Then the equation of the line parallel to BC which is at $\dfrac{1}{2}$ from the origin is

A. $\text{2x+2y-}\sqrt{\text{2}}=0$

B. $\text{2x+2y-}\sqrt{\text{2}}=0$

C. $\text{2x-2y+}\sqrt{\text{2}}=0$

D. None of these.

Answer 1

Hint: First, try to compute the slope of the line BC. As the required line is parallel to the side BC, therefore the slope will be the same. Now find the intercept to get the equation.

Complete step by step solution: 

Vertices of the triangle are:-

O (0, 0), B (−3, −1), C (−1, 3)

Slope of BC:-$\dfrac{\text{3+1}}{\text{-1+3}}\text{=2}$

Formula:-slope between $\left( {{x}_{1}},y, \right)$ &\[\left( {{x}_{2}},{{y}_{2}} \right)\] is $\text{m}=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$

Equation of line is given by $\text{y=mx+c}$

\[\therefore y=2x+c\]

Thus we get m = 2

Distance from the origin is given by

$\text{d=}\left| \dfrac{\text{c}}{\sqrt{\text{1+}{{\text{m}}^{\text{2}}}}} \right|$

\[\therefore \dfrac{\text{1}}{\text{2}}\text{=}\dfrac{\left| \text{c} \right|}{\sqrt{\text{5}}}\]

$\text{c=+}\dfrac{\sqrt{\text{5}}}{\text{2}}\text{,}\dfrac{\text{-}\sqrt{\text{5}}}{\text{2}}$

But c can’t take negative value hence required line equation is $\text{y=2x+}\dfrac{\sqrt{\text{5}}}{\text{2}}$

Therefore, Option (D) is the correct option.

Note: Though we get 2 values for C, we should only choose I value depending on which quadrants the point of the triangle lie Do not forget to take the mod value of C while calculating distance. because distance can’t be negative and this feature can be achieved by using the mod function or modulus.